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bamboo711
08-27-2008, 08:28 AM
Guys, I can't figure this question out: A spherical balloon with radius r inches has volume defined by the function below. Find a function that represents the amount of air required to inflate the balloon from a radius of r inches to a radius of r + 5 inches. (Give the answer in terms of π and r.)

oh and the originial formula is v= 4/3 pi r^2

ScovilleJenkins
08-27-2008, 08:37 AM
old v=4/3pi r^2
new v=4/3pi (r+5)^2 = 4/3 pi(r^2 +10r+25) = 4/3pir^2 + 40/3pi*r +100/3 pi

air required = new v -old v = 40/3 pi*r + 100/3 pi = (20/3)pi (2r+5)

its been a while since ive done something like this but i think its right

Douggo
08-27-2008, 08:38 AM
You're looking for the difference in volumes between radius (r+5) and radius r. Just plug them into the original formula and subtract. Not sure what "n" is supposed to be...

gj011
08-27-2008, 09:28 AM
Formula for sphere volume is:

V = 4/3pi r^3.

08-27-2008, 09:36 AM
The answer is the difference between the sphere with an 'r + 5' radius, and the sphere with the 'r' radius.
v = 4/3 pi (r + 5)^3 - 4/3 pi r^3
v = 4/3 pi [(r + 5)^3 - r^3] factor out 4/3 pi
v = 4/3 pi (r^3 + 15r^2 + 75r + 125 -r^3) expand (r + 5)^3
v = 4/3 pi (15 r^2 + 75 r + 125) r^3 and minus r^3 cancel

xtrakewl
08-27-2008, 09:41 AM

08-27-2008, 10:01 AM

22? Since the radius was stated in inches, and we're talking about volume, let's suppose that you mean 22 cubic inches. Sometimes the answer is a constant, like 22, and no variable r appears in the final answer. Rather than repeat my calculations above, let's see if '22' works if the original radius was 10 inches. After all, if the answer is 22, that means 22 works for ALL values of r.
The original volume would be 4/3 pi 10^3. Which equals 4,188.79 cubic inches, (rounding to the nearest hundredth). The increased sphere would have a volume equal to 4/3 pi 15^3. Which equals 14,137.17 cubic inches. The difference is nearly 9,000 cubic inches, so the suggestion of 22 cannot be correct.

xtremerunnerars
08-27-2008, 01:53 PM
I think he was just being funny.

YULitle
08-27-2008, 02:26 PM
I think he was just being funny.

Yeah, isn't that supposed to be the answer to The Question in Hitchhiker's Guide?

bamboo711
08-28-2008, 09:23 AM
Thanks for the help guys. I eventually got it because of you. Thanks.

I have another question:
Find each of the following functions.
f(x) = sqrt(2 - x)
g(x) = sqrt(x^2 - 1)

f+g
f-g
fg
f/g

I have no clue where to start. Help appreciated. Thanks!

08-28-2008, 10:25 AM
Who are you thanking? Me? Anyway, this is mostly a question to see if you know what the terminology means. For example, f + g = f(x) + g(x). See? So just replace f(x) with sqrt(2-x) and g(x) with sqrt(x^2 - 1). (You won't be able to combine like terms because of the "sqrt"s).

f - g, fg, and f/g, work the same way. It's just about knowing the terminology, because these don't simplify into something that looks different.

Sean Dugan
08-28-2008, 10:29 AM
Take a pin and puncture the balloon. Ergo, Volume = 0.

bamboo711
08-31-2008, 09:25 AM
Steady Eddy, you are a math whiz pal. You got both of those right so thanks a lot for the help everyone, but especially Eddy.

If you invest x dollars at 7% interest compounded annually, then the amount A(x) of the investment after one year is A(x) = 1.07x. Find each of the following.

A of A

A of A of A

A of A of A of A

Find a formula for the composition of n copies of A.

Thanks guys!

Lakoste
08-31-2008, 12:05 PM
Yeah, isn't that supposed to be the answer to The Question in Hitchhiker's Guide?

That is 42

Swissv2
08-31-2008, 12:13 PM
Steady Eddy, you are a math whiz pal. You got both of those right so thanks a lot for the help everyone, but especially Eddy.

If you invest x dollars at 7% interest compounded annually, then the amount A(x) of the investment after one year is A(x) = 1.07x. Find each of the following.

A of A

A of A of A

A of A of A of A

Find a formula for the composition of n copies of A.

Thanks guys!

While I am sure they don't mind helping you out, I would recommend that you do not continually ask for direct answers to your homework, or those who wish to respond to the question can give an example problem related to the question.

We will be happy to help out, but not do your homework for you.

bamboo711
08-31-2008, 05:58 PM
Honestly I got an answer but it's not correct. I can't come up with hte formula to calculate the percentages for each additional year. Any help?

Bundey
08-31-2008, 06:07 PM
Go play tennis

bamboo711
08-31-2008, 06:17 PM

Bundey
08-31-2008, 06:18 PM

What math is this?

sapient007
08-31-2008, 06:28 PM
Steady Eddy, you are a math whiz pal. You got both of those right so thanks a lot for the help everyone, but especially Eddy.

If you invest x dollars at 7% interest compounded annually, then the amount A(x) of the investment after one year is A(x) = 1.07x. Find each of the following.

A of A

A of A of A

A of A of A of A

Find a formula for the composition of n copies of A.

Thanks guys!

use this logic:

A = annual balance
x = interest

starting year = A
year 1 A(1 + x)
year 2 A(1 + x)2
etc
etc

bamboo711
09-01-2008, 12:12 PM
Wait so what? I multiple the year x annual balance (1 + interest) ????

An example maybe? Thanks!

bumfluff
09-01-2008, 12:33 PM
You familiar with exponential growth?

bumfluff
09-01-2008, 01:43 PM
Looking at the way you described the problem, I don't think that will help you with your understanding of what is being asked.

You say x is the initial investment and yet you have the function A(x), however x is a constant so A(x) isn't really a function.

A better way to look at it is to call y the amount you have in your account (I use y simply because if I were to graph the function then the amount would be on the y-axis so the use of x would be confusing and far too non-conformist for me) and yo the initial investment. We shall then say the amount in your account is equal to a function of the number of years (n) it has been in your account. so y = A(n). What you are essentially being asked is to find A(n) when the interest is 7%. I think you need to understand this first as it seems that you are not entirely sure what you are needing to do.

09-01-2008, 04:41 PM
Since A(x) = 1.07x
A of A = 1.07(1.07x) = 1.07^2 x
A of A of A = 1.07(1.07^2 x) = 1.07^3 x

so A1 of A2 of ... of An = 1.07^n x

so by using the exponents button on a scientific calculator, you can model compound interest.

bamboo711
09-05-2008, 04:27 PM
Thank you sir. I am having some difficulty with inverse functions:

If g(x) = 6 + x + e^x, find g-1(7).

AND

Find a formula for the inverse of the function.
f(x) = e^x^7

Any ideas? Thanks guys (Eddy)

meowmix
09-05-2008, 04:32 PM
First, what math are you in? You shouldn't be having a problem with this...

I won't give you the answer, I'll leave that for you to figure out. But for inverse functions, take the f(x)/g(x)/whatever(x) and put a y in place of it. Now for every x in the original equation, put a y and for every y, put an x. Solve for y. That's the inverse function. After that, it's just plug and chug.

bamboo711
09-05-2008, 04:56 PM
I'm in Calc.

I know how to do simple problems but I mean significantly harder with various e^x and such.

Like for y = e^x^7 I do ln to each side and get ln(y)= x^7.... then what?

Thanks!

topspin2
09-05-2008, 05:07 PM
Yeah, isn't that supposed to be the answer to The Question in Hitchhiker's Guide?

that be 42.

09-05-2008, 05:18 PM
Thank you sir. I am having some difficulty with inverse functions:

If g(x) = 6 + x + e^x, find g-1(7).

let g-1(7)=x, by definition, g(x)=7, replace g(x) with 7, and re-write.
7=6 + x + e^x
1=x + e^x
from here it seems obvious that x is zero. Maybe there is a deductive path, or maybe they expect you to be able to tell that it is zero just by looking at it.
Finally, let's check, g(0) = 6 + 0 + e^0 = 7. It checks!

AND

Find a formula for the inverse of the function.
f(x) = e^x^7

Any ideas? Thanks guys (Eddy)
let y = (e^x)^7, now we do the 'switcheroo'
x = (e^y)^7 and our job now is to solve for y
ln x = 7 ln(e^y)
ln x = 7 (y)
y = (1/7)(ln x)
so f-1(x) = (1/7)(ln x)

bamboo711
09-10-2008, 07:48 PM
Eddy it was the 7 root of ln (x) but I ended up figuring that part out. So thanks.

Any idea with this?

Consider the following limit.
lim_(x->-4) (4 x^2 + a x + a + 44)/(x^2 + 3 x - 4)
(a) Find the number a such that the limit exists.
(b) Find the value of the limit.

Thanks guys.

09-10-2008, 11:22 PM
Eddy it was the 7 root of ln (x) but I ended up figuring that part out. So thanks.

I wasn't wrong. They're the same thing. With a log you can make the exponent the coefficient and vice-versa. (1/7) as an exponent means to take the 7th root.

Any idea with this?

Consider the following limit.
lim_(x->-4) (4 x^2 + a x + a + 44)/(x^2 + 3 x - 4)
(a) Find the number a such that the limit exists.
(b) Find the value of the limit.
First check to see that the denominator is zero when you plug in a -4 for x. Yep, it is. Ok, can we factor the denominator? (x + 4)(x - 1). This means we want to factor the numerator in such a way that "x + 4" is a factor of it as well. Let's use synthetic division.
-4] 4 a (a + 44)
-16 -4a+64
4 a-16 -3a+108
So we end up with a remainder of "-3a + 108", for a to be a factor, this remainder should be zero. So -3a + 108 = 0 which implies a = 36.

So the answer to (a) is 36.
Now we can write this as: 4x^2 + 36x + 80
4(x^2 + 9x + 20), so 4(x +4)(x + 5)
Since (x + 4) is in both the numerator and denominator, they cancel.
[4(x + 5)] / [(x - 1)], since x approaches -4, we get
4/(-5) or -4/5. So thats the answer to (b)

09-15-2008, 02:51 PM
Is that the answer you wanted? I didn't figure that out for nothing, did I?

bamboo711
09-15-2008, 04:03 PM
Yeah dude that was exactly right. How are you so good at all this? haha math major or something.

Any clue with this?

For the limit below, find values of δ that correspond to the ε values.
lim_(x->0) (e^x - 1)/x = 1

0.5 _____
0.1 ______

AND

For the limit below, find values of δ that correspond to the M values.
lim_(x->pi/2) tan^2(x) = infinity
M
500
10,000

09-15-2008, 07:10 PM
Yeah dude that was exactly right. How are you so good at all this? haha math major or something.

Any clue with this?

For the limit below, find values of that correspond to the ε values.
lim_(x->0) (e^x - 1)/x = 1

0.5 _____δ
0.1 ______

AND

For the limit below, find values of δ that correspond to the M values.
lim_(x->pi/2) tan^2(x) = infinity
M
500
10,000

You are welcome sir. This is kind of fun for me. You guessed it, math major.

| (e^x - 1)/x - 1 | < .5
-.5 < (e^x - 1)/x -1 < .5
.5 < (e^x - 1)/x < 1.5 adding one to each part
.5x < (e^x -1) < 1.5x multiplying each part by x
.5x + 1 < e^x < 1.5x + 1 adding one to each part, again
solve .5x + 1 < e^x and also solve e^x < 1.5x + 1
x>-1.594 x<.762
so -1.594 < x - δ < .7627
so | x - δ | < .7627 is sufficient, since x => 0, we can say δ = .7627

tan^2 (x) = 500
tan(x)= radical(500) taking the square root of each side
arctan(radical 500) = x = 1.526

tan^2(x) = 10,000
tan(x)= 100
arctan(100)=x=1.5607966
BTW, the actual value of pi/2 is 1.570796, pretty close, and you see how it gets close to the actual value as M gets larger.

bamboo711
09-16-2008, 07:23 AM
Thanks man you really helped me out with the first problem. Yeah you're pretty good at math to be able to do all these. Now for the second problem, you have done something wrong somewhere. For the part that is 500 I know the answer is .0447 somehow. Any ideas? Oh and I can't figure out the IV and V below. I think the first one is +-1 but I am not sure. Any ideas? The rest of problem below is where you're supposed to obtain the info from.

A crystal growth furnace is used in research to determine how best to manufacture crystals used in electric components for the space shuttle. For proper growth of the crystal, the temperature must be controlled accurately by adjusting the input power. Suppose the relationship is given by the following equation, where T is the temperature in degrees Celsius and w is the power input in watts.
T(w) = 0.1w2 + 2.156w + 20

(a) How much power is needed to maintain the temperature at 201°C? (Give your answer correct to 2 decimal places.)

(b) If the temperature is allowed to vary from 201°C by up to ±1°C, what range of wattage is allowed for the input power? (Give your answer correct to 2 decimal places.)

(iv) What value of is given?
_______°C

(v) What is the corresponding value of ?
______watts

Thanks for the help.

09-16-2008, 10:38 AM
Thanks man you really helped me out with the first problem. Yeah you're pretty good at math to be able to do all these. Now for the second problem, you have done something wrong somewhere. For the part that is 500 I know the answer is .0447 somehow. Any ideas? Oh and I can't figure out the IV and V below. I think the first one is +-1 but I am not sure. Any ideas? The rest of problem below is where you're supposed to obtain the info from.

I only have time right now to answer this first part. I'll get to the 'furnace' problem later. Here's how they got the 0.0447, recall that the answer I got for M = 500 is 1.526. How close it that to pi/2? That's 1.5707 - 1.526 = 0.0447. For the second part you'd get 1.5707 - 1.5607 = 0.01. Sorry for not following through. Later, dude.

bamboo711
09-16-2008, 11:08 AM
Thanks man... that was right. Yeah I'm not exactly certain about the last one. I'm pretty sure the IV part is +/-1 but I'm not sure at all what to put in for v.

diredesire
09-16-2008, 02:07 PM

bamboo711
09-16-2008, 05:15 PM
I wish I had the solutions man. I did that problem like a week ago and just couldn't remember how to do it. haha. Steady Eddy is a beast at math though. Oh calculus......

diredesire
09-16-2008, 07:57 PM
Is this a high school or college course? What book are you using, out of curiosity? :shock:

09-18-2008, 09:12 AM
A crystal growth furnace is used in research to determine how best to manufacture crystals used in electric components for the space shuttle. For proper growth of the crystal, the temperature must be controlled accurately by adjusting the input power. Suppose the relationship is given by the following equation, where T is the temperature in degrees Celsius and w is the power input in watts.
T(w) = 0.1w2 + 2.156w + 20

(a) How much power is needed to maintain the temperature at 201°C? (Give your answer correct to 2 decimal places.)

(b) If the temperature is allowed to vary from 201°C by up to ±1°C, what range of wattage is allowed for the input power? (Give your answer correct to 2 decimal places.)

(iv) What value of is given?
_______°C

(v) What is the corresponding value of ?
______watts

Thanks for the help.
To solve (a) means that .1w^2 + 2.156w + 20 = 201
so, .1w^2 + 2.156w -181 = 0 (let's multiply by 10)
w^2 + 21.56w - 1810 = 0
the positive root is, w = 33.11 (rounding to two decimals)

For (b) proceed like above, but solve for 200, and 202 (+/- 1)
You will get 32.99 and 33.22 for your positive roots, respectively.

I'm not too sure what they mean in IV) and v), but I have some ideas. For IV) they might only mean what is +/- 1 for 201, answer 200 and 202.

For V), what are the w's associated with 200 and 202? answer 32.99 and 33.22, respectively.
Keep pacticing that math. Lot's of practice, that's the 'secret'. Just like in tennis!

bamboo711
09-19-2008, 06:16 AM
This is college calculus. My book is Stewart Multivariable maybe? I read that it's not a very good book.

Thanks Eddy, you helped me out greatly with that last question. I'm trying to get ahead of the game a little now, so how would you do a problem like this:

Consider the parabola y = 9x - x^2.
(a) Find the slope of the tangent line to the parabola at the point (1, 8).
_________
(b) Find an equation of the tangent line in part (a).
y=_________

09-19-2008, 06:35 AM
This is college calculus. My book is Stewart Multivariable maybe? I read that it's not a very good book.
I don't understand, the author's name is Stewart Multivariable?

Thanks Eddy, you helped me out greatly with that last question. I'm trying to get ahead of the game a little now, so how would you do a problem like this:

Consider the parabola y = 9x - x^2.
(a) Find the slope of the tangent line to the parabola at the point (1, 8 ).
_________
(b) Find an equation of the tangent line in part (a).
y=_________
These are easier than they sound. Be sure to know them for a test 'cause they're likely to ask it, and it's easy points if you know it. Let's write it as a function. f(x) = 9x - x^2
so f'(x) = 9 - 2x and f'(1) = 7.
So going back to algebra, we have x = 1 y = 8 m = 7. Use y = mx + b, or y - y1 = m(x - x1) to find the equation. You should get y = 7x + 1. Is that right? Well, the slope is obviously 7 when x is 1. And if x = 1, then y = 8. So it works on both counts. So..
(a) slope of the tangent line equals 7
(b) y = 7x + 1

bamboo711
09-19-2008, 08:00 AM
Thanks man. It's called James Stewart : Multivariable Calculus 6th edition. Good book?

Yeah that was really simple, I'll be able to remember how to do that.

If the tangent line to y = f (x) at (4,3) passes through the point (0,2), find the following.
(a) f (4)

The line would be y'=1/4x+ 2 I know, but where from there?

And this....

Consider the following curve.
y = 8 + 5x2 - 2x3.

(a) Find the slope of the tangent to the curve at the point where x = a.
m =

(b) Find the equation of the tangent line at the point (1, 11).
f(x) =

(c) Find the equation of the tangent line at the point (2, 12).
g(x) =

Thanks so much for the help.

09-19-2008, 09:21 AM
Thanks man. It's called James Stewart : Multivariable Calculus 6th edition. Good book?
That's the one I see in Arizona everywhere. I think it's a typical, 3 semester, calculus book. I think the best book I've seen is, "How to Ace Calculus; A Streetwise Guide". It's easier to follow, and it can predict what question you can expect of the tests!

If the tangent line to y = f (x) at (4,3) passes through the point (0,2), find the following.
(a) f (4)

The line would be y'=1/4x+ 2 I know, but where from there?

And this....

Consider the following curve.
y = 8 + 5x2 - 2x3.

(a) Find the slope of the tangent to the curve at the point where x = a.
m =

(b) Find the equation of the tangent line at the point (1, 11).
f(x) =

(c) Find the equation of the tangent line at the point (2, 12).
g(x) =

Thanks so much for the help.
For the first one, you don't have to do any math if you're alert. We're given the point (4,3), then they ask us to find f(4). It's not necessary to find the function, because f(4) has to be 3.

The next one: y = 8 + 5x^2 - 2x^3
y' = 10x - 6x^2
x = a
y' = 10a - 6a^2
(that's the answer to part a)

if a = 1, then y' = 4, so m = 4, x = 1, y = 11, gives us y = 4x + 7
(that's b)
if a = -1 then y' = -4, so m = -1, y = 12, x = 2, gives us y = -4x + 20
(that's c)

diredesire
09-19-2008, 10:21 AM
Bamboo, please send me an e-mail :)

dak95_00
09-19-2008, 05:53 PM

I am finding this thread humorous as I am a high school math teacher who is teaching Calculus for the first time in a few years.

Yahoo Answers is also very good to ask questions and get answers as there are many math nerds (no offense intended) there to answer many questions.

You might also try http://www.calc101.com too.

Morpheus
09-19-2008, 06:50 PM
That is 42

Correct, but what was the question again?

09-19-2008, 08:09 PM

I am finding this thread humorous as I am a high school math teacher who is teaching Calculus for the first time in a few years.
Humorous? Why? (I'm a retired H.S. math teacher)

Yahoo Answers is also very good to ask questions and get answers as there are many math nerds (no offense intended) there to answer many questions.

You might also try http://www.calc101.com too.
But he's asking me, not the other way around. Be like someone in a car saying, "Where's Glendale?" Me, "I don't know." Them, "Well, go to the gas station, they'll give you directions." I'm the one supplying answers, not asking questions.

bamboo711
09-19-2008, 09:00 PM

I really don't find this book that good at all to learn from, and I wish they had used a different book that would have taught in a different manner? I mean I consider myself pretty good at math (700 SAT 32 ACT on Math sections and I find the book not too useful.

You might find this thread humorous, but Steady Eddy is pretty sick at explaining his method and answers which I appreciate. Pretty much I don't intend on using math in any sense in the future, and I have to get through this class and make at least like a 3.5 in it, so Eddy is making it that much smoother. Thanks man.

More questions to come soon if that's okay.

09-19-2008, 09:21 PM
More questions to come soon if that's okay.
I'll keep my paper and pen at the ready. :)

Morpheus
09-20-2008, 04:22 AM
^^ I always use a pencil when I do math.

Bamboo, you will be surprised how often you use math. It is not just engineers or scientists that use it so I recommend you retain as much as possible. Make sure you really pay attention in probability and statistics.

bamboo711
09-20-2008, 08:36 AM
Steady Eddy, I see where you got everything in that problem except I don't see how you got the y = 4x + 7 and y = -4x + 20. I understand the -4 and 4 in that problem but I don't get where you got the 7 and 20 from. What am I overlooking? Thanks man.

09-20-2008, 08:42 AM
Steady Eddy, I see where you got everything in that problem except I don't see how you got the y = 4x + 7 and y = -4x + 20. I understand the -4 and 4 in that problem but I don't get where you got the 7 and 20 from. What am I overlooking? Thanks man.
You have to think way back to beginning algebra. Remember y = mx + b? m is the slope and b is the y intercept. So if you know the slope and just one point, you can find the equation. m = slope, x = the 1st # in the ordered pair, and y = the 2nd # in the ordered pair. Plug them in and solve for b. Another way is to use point slope form. That's y - y1 = m(x - x1) where x1 is the 1st # in the ordered pair and y1 is the 2nd # in the ordered pair. (Just leave "x" and "y" alone). Solve for y, and that will also give you the equation.

bamboo711
09-21-2008, 09:42 AM
Thanks Steady Eddy, there were similar problems to that and I figured them out with your help.

This following question (I'm having trouble taking derivatives of functions)

Consider the function below.
F(x) = (5 x)/(1+x^2)
(a) Find F '(2).

(b) Use the answer from part (a) to find an equation of the tangent line to the curve y = F(x) at the point (2, 2).
y =

Thanks man.

09-21-2008, 10:36 AM
^^ I always use a pencil when I do math.

Bamboo, you will be surprised how often you use math. It is not just engineers or scientists that use it so I recommend you retain as much as possible. Make sure you really pay attention in probability and statistics.
Most people use a pencil, like you do. I don't like pencils, they get dull and need to be sharpened, unless it's a mechanical pencil. But the mechanical pencil's point keeps breaking off. I'm too distracted using pencils. I use a pen, not because I don't make any mistakes. When I make a mistake, I just start over. Sometimes I go through lots of paper!

The second part of what you say is so true. I use math every day. Nearly all of it I do without paper or calculator. Just knowing about: slopes, probabilities, and expected values, changes how you see things.

Thanks Steady Eddy, there were similar problems to that and I figured them out with your help.

This following question (I'm having trouble taking derivatives of functions)

Consider the function below.
F(x) = (5 x)/(1+x^2)
(a) Find F '(2).

(b) Use the answer from part (a) to find an equation of the tangent line to the curve y = F(x) at the point (2, 2).
y =

Thanks man.
For the first one you have to use the dreaded 'quotient rule'. You'll get (5-5x^2)/(1+x^2)^2, when x = 2 we get -3/5.

(b) is not bad. We have m = -3/5, x = 2, and y = 2. Plug and chug and you get y = -3/5x +16/5.

If you have a graphing calculator, it can be worthwhile to plug these in and see how the line just grazes the function.

dak95_00
09-21-2008, 01:37 PM
^^^^^
If you're going to plug it into a graphing calculator (I'll assume TI-83/84), you might as well just type in the original function and then graph. Follow that by looking at the graph of the original and go to "Draw" and then "Tangent." Then just press "2" and "enter" and you'll get the equation of the tangent line given to you!

My 2 cents!

Quotient Rule

Hi/Ho

F'(Hi/Ho) = (HoDHi - HiDHo)/(HoHo) where D stands for derivative!

Zhou
09-21-2008, 01:39 PM
Quotient Rule

Hi/Ho

F'(Hi/Ho) = (HoDHi - HiDHo)/(HoHo) where D stands for derivative!

That was a cool trick our teacher taught us. It gave everyone a chuckle.

game set match 46 TIMES!!
09-21-2008, 01:52 PM
paying 0.11113957576 bucks if thy do my homework

09-21-2008, 03:43 PM
^^^^^
If you're going to plug it into a graphing calculator (I'll assume TI-83/84), you might as well just type in the original function and then graph. Follow that by looking at the graph of the original and go to "Draw" and then "Tangent." Then just press "2" and "enter" and you'll get the equation of the tangent line given to you!

My 2 cents!

Quotient Rule

Hi/Ho

F'(Hi/Ho) = (HoDHi - HiDHo)/(HoHo) where D stands for derivative!
I'm at the library and don't have my TI-83 with me. I'll try that as soon as I get home. Listening Bamboo? That's a handy feature for what you're doing. Thanks dak. (But you never answered, "Why is this humorous?")

dak95_00
09-21-2008, 06:08 PM
I just find it odd (no pun intended) that someone would look for math help at a tennis forum and that there is so much help to be found here. I would've never thought to look for it here.

Back to math and the calculator......
I always use the Nderiv feature and test my derivative against it. For example, if my function was y=3x^2 and I found my derivative to be y'=6x, I could test it by typing in (On the calculator) y1=3x^2, y2=NDeriv(y1, x,x), and y3=6x (my answer). Then I look at the graphs of y2 and y3 to be certain they are the same and then I know I am correct. To distinguish between the graphs, I turn off the graphing of y1 by unhighlighting its = sign and I bold highlight y3 by pushing enter once next to the /y3 so that when it graphs it will definitely be noticeable. It can be a lifesaver to test derivatives and your answers.

If you have TI-83/84 questions, I am probably your man. TI should put me on salary!

09-21-2008, 06:16 PM
I just find it odd (no pun intended) that someone would look for math help at a tennis forum and that there is so much help to be found here. I would've never thought to look for it here.

Back to math and the calculator......
I always use the Nderiv feature and test my derivative against it. For example, if my function was y=3x^2 and I found my derivative to be y'=6x, I could test it by typing in (On the calculator) y1=3x^2, y2=NDeriv(y1, x,x), and y3=6x (my answer). Then I look at the graphs of y2 and y3 to be certain they are the same and then I know I am correct. To distinguish between the graphs, I turn off the graphing of y1 by unhighlighting its = sign and I bold highlight y3 by pushing enter once next to the /y3 so that when it graphs it will definitely be noticeable. It can be a lifesaver to test derivatives and your answers.
I knew about the unhighlighting to turn off a graph. I learned this in "The TI 83 for Dummies". I bought my TI 83 at a pawn shop, so I don't have a manual. But the "For Dummies" book is much easier to understand anyway. I know there is so much the 83 can do than what I know about. So I don't have so much questions as I need, "Know an easier way to do that?" queries. If something can be done a cooler way, let me know.

If you have TI-83/84 questions, I am probably your man. TI should put me on salary!
Give it away, just like I do! (I also 'straighten-out' people on politics in 'rants & raves' for free. :) )

bamboo711
09-22-2008, 08:23 AM
Yo guys thanks for the help. Actually I have a ti-89 titanium but my school doesn't allow it because it does too much I guess. I just got a ti-84 silver today and that's the best calculator you can use on exams here. Can it perform function such as solving equations, factoring, expanding, limits, derivatives, and integrals? Thanks for the help with the derivative in the calculator I sure will try it out.

Oh and the quotient rule saying is awesome, so thanks for that.

More questions coming soon..... that's for sure.

Thanks guys.

09-22-2008, 09:18 AM

If you have TI-83/84 questions, I am probably your man. TI should put me on salary!
That's a good way to "think outside the box" on a test. I've got another, for you, Bamboo, or whoever might be taking a calclulus test. For solving derivatives, on a multiple choice test, let x = 2, and dx = .001., then find the fraction dy/dx, let x = 2 for the choices and simply take the one that almost matches! It's like cheating, because one choice is usually way closer than the others, and you get the answer without doing calculus. But it's not actually, because on a multiple choice test, you don't show your work, however you can get the correct answer, (without copying, of course), is ok.

And Bamboo, you're better off with the TI 84. If the 84 is all that's required, it is usually harder to learn how to use the technology for the 89, than just to solve it more conventionally with the 84. People think they can get A's just by spending on expensive calculators. They're better off just studying the material.

dak95_00
09-22-2008, 02:34 PM
Can it perform function such as solving equations, factoring, expanding, limits, derivatives, and integrals? Thanks for the help with the derivative in the calculator I sure will try it out.

Yes, but not as direct as a TI-89. I usually solve an equation by setting one side as y1 and the other as y2 and finding the intersection points or I set one side equal to zero and just enter the non-zero side as y1 and find the zeros.

Factoring: Well if I can find the zeros, I can manipulate them to become factors i.e x=3 is (x-3).

Nderiv will find derivatives at a value or show the graph but it won't give you the equation. The same can be said for integrals too.

Expanding......No way! Learn Pascal's Triangle, it saves time. I'll let Steady explain that one. Besides, why would I ever need to expand?

My advice, learn Logarithmic differentiation and stick to it. It makes most of the other rules obsolete.

09-22-2008, 04:47 PM
Can it perform function such as solving equations
Yes. Which is why, in the future, no one will learn math anymore, (just kidding). It's got a feature called "SOLVER". Go to "MATH", solver is at the bottom of that menu, so the fastest way to get there is to go up one item. Now you're at "SOLVER". Press enter and it will say "X = ", go up again, and it says "eqn:0 = " (You have to know how to set your equation equal to zero, but that's no big deal.) Type in the equation. Press enter. At the "X = " screen, pick a value pretty close to what you think X is, (it solves this using Newton's method, something you'll study in first semester calc.) Press enter and wait, you get the answer! Don't bother copying it down. The calculator automatically makes X equal to that answer until you change it. Love the solver.

bamboo711
09-23-2008, 08:41 PM
Just got this problem for my next homework assignment...... what in the world is this????? hahaahaha.

Find an equation of the line that is both normal to the given parabola and also parallel to the given line.
y of p = 6x^2 - 5 x + 7
7x - 6y = 8

09-24-2008, 09:24 AM
Just got this problem for my next homework assignment...... what in the world is this????? hahaahaha.

Find an equation of the line that is both normal to the given parabola and also parallel to the given line.
y of p = 6x^2 - 5 x + 7
7x - 6y = 8
Don't be intimidated. It's not that hard. But it is annoying. What slope does it need to have? It's parallel to the line 7x - 6y = 8, let's put that into slope intercept form. -6y = -7x + 8, 6y = 7x - 8, y = 7/6 x - 4/3. So the slope has to be 7/6.

Another condition is that it has to be normal to the parabola p = 6x^2 - 5x +7. p' = 12x - 5. Recall that normal means perpendicular, so it crosses the parabola at a place where the parabola's slope is -6/7. Soooo -6/7 = 12x - 5
29/7 = 12x, x = 29/84. And y = 7043/1176, yikes!

Now use y - y1 = m(x - x1) or y = mx + b to find the equation of the line. I get y = 7/6 x + 19708/3528. At this point the # is so messy, I'm getting red flags. So I put it all into my TI 83 to see how it looks, and it looks correct. So I'd assert (with a 95% probability) that y = 7/6 x + 19708/3528, or y = 7/6 x + 5.5861678. (in decimal)

diredesire
09-24-2008, 09:33 AM
Just got this problem for my next homework assignment...... what in the world is this????? hahaahaha.

Find an equation of the line that is both normal to the given parabola and also parallel to the given line.
y of p = 6x^2 - 5 x + 7
7x - 6y = 8

Hey bamboo, did you get my e-mail?

dak95_00
09-24-2008, 05:55 PM
Don't be intimidated. It's not that hard. But it is annoying. What slope does it need to have? It's parallel to the line 7x - 6y = 8, let's put that into slope intercept form. -6y = -7x + 8, 6y = 7x - 8, y = 7/6 x - 4/3. So the slope has to be 7/6.

Another condition is that it has to be normal to the parabola p = 6x^2 - 5x +7. p' = 12x - 5. Recall that normal means perpendicular, so it crosses the parabola at a place where the parabola's slope is -6/7. Soooo -6/7 = 12x - 5
29/7 = 12x, x = 29/84. And y = 7043/1176, yikes!

Now use y - y1 = m(x - x1) or y = mx + b to find the equation of the line. I get y = 7/6 x + 19708/3528. At this point the # is so messy, I'm getting red flags. So I put it all into my TI 83 to see how it looks, and it looks correct. So I'd assert (with a 95% probability) that y = 7/6 x + 19708/3528, or y = 7/6 x + 5.5861678. (in decimal)

I agree w/ your work. I also agree that the numbers were not nice but I confirmed through my work too and by looking at the tangent line w/ slope
-6/7 and your normal line graphed on the TI-83 w/ a zoomsquare window. Looked good!

Oh, have you ever thought to just enter the equation in point-slope form as
y=m(x-x1)+y1 ? This is the way my text suggests it and I like it because there are fewer chances for mistakes and it still enters into the calculators easily.

bamboo711
09-24-2008, 06:22 PM
Yup man that was definitely right... I checked it myself and everything and entered the question and it was correct. A bit tough. Haha. Thanks.

A few more questions: lim x-->0 of tanx/x. I know the answer is 1 and I have 1/x*sinx/cosx and yet I still get undefined when I plug 0 in. Any ideas?

Oh, and some new problems that are giving me some trouble for some reason?

Differentiate the following function.
H (x) = (5x^2 + 3x^-4 )^ 3

^^^Why would it not just be 30x^5-36x^-13?

Differentiate the following function.
F(x) = (1/5 x)^3

^^^Why is that not (3/5x)^2

And finally....Find the points on the curve given below, where the tangent is horizontal. (Round the answers to three decimal places.)
y = 6x^3 + 3x^2 - 6x + 7
P1 = (-.768, ___) (smaller x-value)
P2 = (.434,___) (larger x-value)

^^^I found the X's but I can't seem to know what to put in the second blanks? I guessed 0 at first since that's the point, but its wrong.

Thanks for all the help.... and support guys.

09-24-2008, 11:12 PM
I agree w/ your work. I also agree that the numbers were not nice but I confirmed through my work too and by looking at the tangent line w/ slope
-6/7 and your normal line graphed on the TI-83 w/ a zoomsquare window. Looked good!

Oh, have you ever thought to just enter the equation in point-slope form as
y=m(x-x1)+y1 ? This is the way my text suggests it and I like it because there are fewer chances for mistakes and it still enters into the calculators easily.
That's a good idea. I'll try it. To get a y1 on there, I go VARS Y-VARS Y1, right?

Oh, and some new problems that are giving me some trouble for some reason?

Differentiate the following function.
H (x) = (5x^2 + 3x^-4 )^ 3

^^^Why would it not just be 30x^5-36x^-13?
H'(x) = 3(5x^2 + 3x^-4)^2 * (10x + -12x^-5) (chain rule)

Differentiate the following function.
F(x) = (1/5 x)^3

^^^Why is that not (3/5x)^2
F'(x) = 3(1/5 x)^2 * (1/5), so 3/125 x^2 (chain rule again)
or rewrite F(x) = 1/125 x^3, then F'(x) = 3/125 x^2 (same answer)
it's because (1/5 x)^3 not the same as 1/5 x^3.

And now I'm going to wuss out, 'cause I'm too tired to do math right.

bamboo711
09-25-2008, 05:55 PM
Thanks man.

lim x->0 tan(x)/x

Why is it 1? I got 1/x * sin(x)/ cos(x)

09-25-2008, 05:58 PM
Thanks man.

lim x->0 tan(x)/x

Why is it 1? I got 1/x * sin(x)/ cos(x)
Have you had L'Hospital's rule yet?

bamboo711
09-25-2008, 06:09 PM
You see, we're taught what a basic derivative is, and then are supposed to learn all this by ourselves I suppose. And no sadly I don't haha.

dak95_00
09-25-2008, 06:14 PM
Have you had L'Hospital's rule yet?

My guess is that they haven't yet.

I'd say it is much more like they were told to accept (Sin x)/x is 1 as the Lim x-->0 and do a product of limits because the Lim x-->0 of cos x is 1 and 1*1=1. I believ you can use the Sandwich (or Squeeze) Theorem to show (sin x)/x is 1 as lim x-->0. I just tell students to remember that one! I also tell them to pay attention to the coefficients as (all of these are the lim x-->0) because (sin 3x)/x =3 and (sin 4x)/2x =2 (4/2) and so on and so on!

We know L'Hospital's Rule is easier!

dak95_00
09-25-2008, 06:17 PM
Bamboo,

You did see how I manipulated 1/x * sin x/cos x to be (sin x )/x and (oops) 1/(cos x) which is the same as (tan x)/x.

bamboo711
09-25-2008, 06:18 PM
the limit of 0 might be 1 when cos is 0, but you still have the 1/x which when you plug in the 0 in the denominator you get undefined. What's the deal?

dak95_00
09-25-2008, 06:34 PM
Yes. It is called an indeterminate because the (Sin x) is also 0 and that means we have 0/0 with the (sin x)/x. Take a look at the graph of (sinx)/x and then press trace and insert 0.0000001 to demonstrate x-->0+ and also -0.0000001 to demonstrate x-->0- and you'll clearly "see" 1 is the answer there too.

L'Hospital's Rule is for indeterminate functions which is when you get a 0/0. It says that you just take the derivative of the numerator and denominator and you'll get the answer. That is, sin x is cos x and x is 1 leaving us w/ (cos x)/1 which as x-->0 is 1/1 or 1. How about that rule? Exciting huh!

bamboo711
09-26-2008, 09:47 AM
Sweet man.... that's really useful, I'll use that in the future for sure. Thanks.

lim x-->0 sin(x)/2x

^^^How is the answer to that 0? That's what my calculator shows but I don't see it. sin(0)=1 and 2(0)= 0 so undefined no?

Thanks for the help

YULitle
09-26-2008, 09:52 AM
Sweet man.... that's really useful, I'll use that in the future for sure. Thanks.

lim x-->0 sin(x)/2x

^^^How is the answer to that 0? That's what my calculator shows but I don't see it. sin(0)=1 and 2(0)= 0 so undefined no?

Thanks for the help

sin(0)=0, not 1

dak95_00
09-26-2008, 11:40 AM
[QUOTE=dak95_00;2740098]My guess is that they haven't yet.

I'd say it is much more like they were told to accept (Sin x)/x is 1 as the Lim x-->0 and do a product of limits because the Lim x-->0 of cos x is 1 and 1*1=1. I believ you can use the Sandwich (or Squeeze) Theorem to show (sin x)/x is 1 as lim x-->0. I just tell students to remember that one! I also tell them to pay attention to the coefficients as (all of these are the lim x-->0) because (sin 3x)/x =3 and (sin 4x)/2x =2 (4/2) and so on and so on!

Bamboo,

You have to look no further than the coefficients. 1x/2x = 1/2 so the lim x-->0 of (sin x)/2x = 1/2. <-- indeterminate (Thanks YU!)

09-26-2008, 11:53 AM
Your teacher won't accept this. But when you're stuck, try this, type sin(x)/(2x) into the TI-83, then .001 STO X. Bring back the expression, (use 2nd enter, 2nd enter), press enter and you get .4999999167. Sure looks like it's approaching 0.5 as x is approaching zero, doesn't it? Oh, and be sure you have you calculator in radians, and not degrees, (degrees will mess things up when you're taking calc).

The teacher wants proof, but once you know an answer, it's easier to find the proof, isn't it? And if you're taking a multiple choice test, proof isn't necessary, just so you get the right answer.

BTW, here's the proof: rewrite the expression as (1/2)[sin(x)/(x)]. The part in brackets equals one as x => zero, so it becomes (1/2)[1], which equals 1/2.

bamboo711
09-26-2008, 06:09 PM
Thanks....yeah I definitely need to learn all those semi-ridiculous rules. haha.

Find the points on the curve given below, where the tangent is horizontal. (Round the answers to three decimal places.)
y = 6 x 3 + 3 x 2 - 6 x + 7
P1 = (-.768 ,??? ) (smaller x-value)
P2 = ( .434, ???) (larger x-value)

I got the first part but can't seem to figure out or have a clue what goes in the second blanks.

dak95_00
09-26-2008, 06:22 PM
Find the points on the curve given below, where the tangent is horizontal. (Round the answers to three decimal places.)
y = 6 x 3 + 3 x 2 - 6 x + 7

Since it says to find the points on the curve, plug them into the curve i.e. original function given. You did good work finding the x values!

bamboo711
09-26-2008, 06:43 PM
You mean the -.768 and .434? Because at first I thought it was just referring to the point which would be 0 for the y. Thanks.

dak95_00
09-26-2008, 07:39 PM
Yes. Substitute those values in for x one at a time to get your y values into the original function. You will not get zero.

You would get zero if you substituted them into the derivative since the derivative represents the slope of the curve and the horizontal tangents have a slope of zero.

Into the original function, you are finding the local maximum and minimum points of the curve or the points where the slope is changing from positive to negative and back to positive. (Remember this as it will most likely be the next lesson!)

bamboo711
09-26-2008, 09:26 PM
I plugged -.768 and .434 into the original equation and the answer is still wrong for some reason? I got 7.12 and 5.45.

AND

Find equations of both lines through the point (2,-3) that are tangent to the parabola y = x^2 + x.
y = (smaller slope)
y = (larger slope)

^^^ I take the deriv (2x + 1) and where do I go from there?

Thanks

dak95_00
09-27-2008, 05:16 AM
I plugged -.768 and .434 into the original equation and the answer is still wrong for some reason? I got 7.12 and 5.45.

Thanks

Bamboo,
I got (-.768, 10.660) and (.434, 5.452) rounding to 3 decimal places.

More importantly: Do you know how to store values of x on the calculator so you can simply type in 6x^3 + 3x^2 - 6x + 7 to get those y values?

dak95_00
09-27-2008, 05:48 AM
Find equations of both lines through the point (2,-3) that are tangent to the parabola y = x^2 + x.
y = (smaller slope)
y = (larger slope)

^^^ I take the deriv (2x + 1) and where do I go from there?

Thanks

Bamboo,

This one gets the WTF? award! However, I think I got it. You'll have to let me know so I can give myself a pat on the back and pump my first like Tiger Woods. Otherwise, I'll just throw up my arms and say, "Phooey!"

Here is what I did:
I wrote out an equation of a line using 2x+1 as the slope and traveling through the point (2,3). I then set it equal to the original equation x^2+x. The tangent line equation would look like y=m(x-x1)+y1 or y=(2x+1)(x-2)+3. Now when I set it equal to the original equation, I get
(2x-1)(x-2)+3=x^2+x and I foiled out the left and added the three to get
2x^2-3x+1=x^2+x and then I made it all equal 0 so I was left w/
x^2-4x+1=0 and I used the Quadratic Formula to get my x values of

2+SquRt(3) and 2-SquRt(3) ??How else could I say Square Root??

Now, I plugged those into the derivative to get the actual slope and I used
(2,3) as the x and y values for my tangent line. I checked this on my calculator using the tangent line draw feature and more algebra and it worked because I got the same slope-intercept formula as I received when I used my 2+/-SquRt(3) for x and what I got when I plugged those into the original equation for y.

Bamboo,
Are these problems to some computer generated quiz? I ask because I used to help students at The Ohio State University with what was called Webwork for them. They were computer generated quizzes that students would enter their answers and they would immediately get feedback as to their correctness and could change them as necessary. These quizzes were also downloadable in a PDF form that I kept and use in my classes today. If these are similar, could I get copies to use in my classes? I love collecting materials to show my high school students what will be expected of them in college and to use as a problem of the day or week. Thanks if you could get me either electronic or hard copies through mail or email.

bamboo711
09-27-2008, 03:21 PM
Yo dude I tried it and got the two lines y= 1.536x+0.072 and 8.4641x-13.9282 and I got my answers wrong for some reason?

Also, a simple question that I know how to do but I am getting the answer wrong for some reason? Simple chain rule.

Differentiate the following function.
H (x) = (5x^2+3x^-4)^ 3

^^^Why would H'(x) not be 3(5x^2+3x^-4)^2*10x-12x^-5???

It's called webassign actually. I don't get it through pdf files but I do the questions online and submit them and see if I am getting the answers correctly. I could still possibly get them for you though. Thanks for the help.

bumfluff
09-27-2008, 03:45 PM
Well you are right, so unless you have read the question wrong then either you have entered it wrong, maybe you have used the wrong notation or the answer is wrong. All of these are possible, my philosophy: do the question, check your working throughly, if it is wrong query the answer next, if there is no answer then you have done all you can. If it is a test then make sure all your working is there and is clean and concise. In a written test I tend to use up a whole forest for my calculations.

About the whole sin x / x question that you posted earlier although it has been explained I don't think my way of looking at it has been mentioned.

Basically you can say that as x-->0, sin x --> x, the gradient at the origin of y = sin x is 45 which is the same as y = x. So when you have sin x / x it effectively becomes x / x.

dak95_00
09-27-2008, 04:40 PM
Yo dude I tried it and got the two lines y= 1.536x+0.072 and 8.4641x-13.9282 and I got my answers wrong for some reason?

Also, a simple question that I know how to do but I am getting the answer wrong for some reason? Simple chain rule.

Differentiate the following function.
H (x) = (5x^2+3x^-4)^ 3

^^^Why would H'(x) not be 3(5x^2+3x^-4)^2*10x-12x^-5???

It's called webassign actually. I don't get it through pdf files but I do the questions online and submit them and see if I am getting the answers correctly. I could still possibly get them for you though. Thanks for the help.

1st: I don't know why those two lines aren't working but as Bumfluff said it might be on how you are entering them. I noticed on 1 equation you went out 3 decimal places and the second you went out 4. Maybe that is why???

2nd: Your chain rule is correct. Again, I think it is in the notation. I would have put parenthesis around that last part because you want to multiply the first part by the whole thing and not just the 10x.
Here: 3(5x^2+3x^-4)^2*(10x-12x^-5) and in algebra the parenthesis make all the difference in the world.

bamboo711
09-28-2008, 08:13 AM
Thanks man.... I'll have to talk to my teacher about all this. Like I'm stuck at 86% but like I mean I know all the rest of my answers are right except this question which I need some help on. I hate having to enter in math through the computer (And I checked and I typed everything in right)

Find the equations of the tangent line, y1, and the normal line, y2 to the following curve at the given point P.
y = x^7+3e^x

P = (0, 3)

^^^ I derive it and get 7x^6+ 3e^x

Would the tangent line just be (7x^6+3e^x)(x)+3??? Is that right?
And what about the normal line? Would it just be (x^7+3e^x)(x) + 3?

Thanks.

09-28-2008, 10:53 AM
You've got the right derivative with y' = 7x^6 + 3e^x. But to get the line tangent to the line at the point (0, 3), plug in 0 for x. So y' = 3 = m

So with x = 0, y = 3, and m = 3 => y = 3x + 3
Normal to that, would have a slope of -1/3.
So with x = 0, y = 3, and m = -1/3 => y = -1/3x + 3

bamboo711
09-28-2008, 04:55 PM

lim x->3 (x-3)/ abs(x-3)

^^ Would it be undefined because it approaches -1 from the left and 1 from the right? How would I prove that without a graph?

THanks

09-28-2008, 05:39 PM

lim x->3 (x-3)/ abs(x-3)

^^ Would it be undefined because it approaches -1 from the left and 1 from the right? How would I prove that without a graph?

THanks
You got it. To prove it, first use lim x => 3 (subscript minus sign), and show that does not equal lim x => 3 (subscript plus sign). This is stuff you did in the first part of the calculus course.

bamboo711
09-29-2008, 08:06 AM
Thanks Steady Eddy, that was a wonderful explanation.

I'm not really sure what to do at all with this question.

Find the value of c so that f(x) is continuous on the entire real number line when

Piecewise:

(x-2) when x<=5
(cx-3) when x>5

09-29-2008, 08:17 AM
Thanks Steady Eddy, that was a wonderful explanation.

I'm not really sure what to do at all with this question.

Find the value of c so that f(x) is continuous on the entire real number line when

Piecewise:

(x-2) when x<=5
(cx-3) when x>5
So to be continuous both sides would have to equal the same thing when x = 5. So replace x w/ 5 and make an equality to solve.
(5 - 2) = (c5 - 3)
3 = 5c - 3
6 = 5c, so c = 6/5. So that's one where if you can figure out what the heck they're asking, then it's not that hard.

bamboo711
09-29-2008, 05:25 PM
Steady Eddy, as you see sometimes I'm having trouble seeing what the problem wants out of me. How are you knowing exactly what they are wanting so easily?

Also, this, I'm encountering some trouble especially due to the 2xsqrt(x) part which looks like a whole new product rule to me within the question?

Find the derivative of the function below in two ways.
F(x) = (x -2xsqrt(x)) /( sqrt(x))
(a) By using the Quotient Rule.
F '(x) =

(b) By simplifying first.
F '(x) =

dak95_00
09-29-2008, 05:52 PM
This problem stunk! The idea is to simplify using algebra 1st, otherwise you have a mountain of a problem instead of a relative mole hill.

1st: 2xsqrt(x) = 2x^(3/2) since sqrt(x) is x^(1/2) and making that change save you from an even larger mountain of monstrous work! I hope that is okay to do for this problem.

2nd: The quotient rule is not much fun so remember (HodHi-HidHo)/(HoHo) given Hi/Ho.

Hi = x-2x^(3/2)
Ho = x^(1/2)
dHi = 1-3x^(1/2)
dHo = (1/2)x^(-1/2)

and now we have the quotient rule as.....
(x^(1/2))(1-3x^(1/2))-(x-2^(3/2))((1/2)x^(-1/2))
-------------------------------------------------
(x^1/2)(x^1/2)

simpligying using algebra yields through the initial distributions

1x^(1/2)-3x-(1/2)x^(1/2)+x
---------------------------
x

now combining like terms in the numerator

(1/2)x^(1/2)-2x
----------------
x

Finally, if I divide each part of the numerator by the denominator....

(1/2)x^(-1/2) - 2 since you just subtract exponents.

I'll use a new post for the easy method!

dak95_00
09-29-2008, 05:57 PM
By simplifying first, I'll just divide the numerator by x^(1/2) or again, subtract exponents. And again, I am using x-2x^(3/2) as the numerator!

(x)/(x^(1/2) is x^(1/2) and (2x^(3/2))/(X^(1/2) is just 2x! This means that all of the garbage is really just .....

x^(1/2)-2x and the derivative of that is........

(1/2)x^(-1/2) - 2

Wow! That was much easier!!!!!! Now, 9/10 times you won't have a denominator that is a monomial you can easily divide but you should be able to convert it using negative exponents and use the product rule which IMO is easier!

dak95_00
09-29-2008, 06:00 PM
My turn.....

Given f(x)=x^2-kx, find where the line y=4x-9 is a tangent of f(x).

I'll say this. Without Bamboo's question from the other day, I might've struggled w/ this one much longer than I did. It took me under 5 minutes.

bumfluff
09-30-2008, 02:41 AM
Thanks for that one dak, it got me thinking of maths again as I haven't been doing enough of it recently.

For my solution I let x^2-kx equal 4x-9 and because when you make a quadratic it has to have equal roots, you solve k through b^2-4ac = 0.

Then you substitute your 2 values for k into f(x) differentiate each one let it equal the gradient of the line and find x then find y. There are two different points because there are two different values of k.

bamboo711
09-30-2008, 06:46 AM
Nice solution.

I don't get what I am doing wrong here:

The curve below is called a witch of Maria Agnesi. Find an equation of the tangent line to this curve at the point (-3, 1/10).
y=1/(1+x^2)

My wrong solution: so y' would be -2x/((x^(2)+1)^2)
y'(3)= 6/100
y-3 = .06(x+3)
y= .06x+ 3.18

Where am I going wrong? Thanks!

09-30-2008, 07:39 AM
y-3 = .06(x+3)

not "-3", should be "-1/10" because in the ordered pair (3, 1/10) the y coordinate is 1/10.

bamboo711
09-30-2008, 08:00 AM
So the answer should be y=0.06x+0.28???

shaheinthedream
09-30-2008, 08:30 AM
I don't mean to steal your post but I think this question is apropos...

A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20 m/s at an angle 5.0 degrees above the horizontal. The horizontal distance to the net is 7.0 m , and the net is 1.0 m high.

Does the ball clear the net? If so, by how much? If not, by how much does it miss?

09-30-2008, 09:31 AM
So the answer should be y=0.06x+0.28???
You got it!

I don't mean to steal your post but I think this question is apropos...

A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20 m/s at an angle 5.0 degrees above the horizontal. The horizontal distance to the net is 7.0 m , and the net is 1.0 m high.

Does the ball clear the net? If so, by how much? If not, by how much does it miss?
So we want to find the trajectory. You can do this on a TI 83 using parametric equations. Go to mode and set to Par, and Degree. Press "Y=" and let x = 20cos(5)T and y=2 + 20sin(5)T - 4.877T^2

When I set T = 0.352 I get x (distance) = 7.013 and y (height) = 2.005. So even slightly past your distance of 7 m, I show the height is still higher than 2 m. I'd say that it clears the net by a little more than a meter. Of course, I'm not figuring in air resistance or spin. But without much spin it's still pretty close to clearing the net by a meter.

bamboo711
09-30-2008, 02:18 PM
Differentiate.
g(t) = (5t - sqrt(t)) / t^(1/3)

^^^I expanded it out with quotient rule and is it just me or does this just not simplify and is ugly as anything?

Thanks guys.

09-30-2008, 03:18 PM
Differentiate.
g(t) = (5t - sqrt(t)) / t^(1/3)

^^^I expanded it out with quotient rule and is it just me or does this just not simplify and is ugly as anything?

Thanks guys.
I simplified first, but it still stays "ugly".

bamboo711
09-30-2008, 04:06 PM
Where do you simplify? Mine is ugly all over.

09-30-2008, 06:17 PM
Make it into two terms, with t^3 as a denominator for each term. (Yep, you can combine two terms with common denominators, so you can also go the other way, and write a fraction as two terms with common denominators. You'll do that alot in calc.)

Then since each term has a t in the numerator and the denominator, subtract the 1/3 from the exponent in each term. Now you can find the derivative without using the #\$@## chain rule!

dak95_00
10-01-2008, 12:39 PM
Bamboo,

This one gets the WTF? award! However, I think I got it. You'll have to let me know so I can give myself a pat on the back and pump my first like Tiger Woods. Otherwise, I'll just throw up my arms and say, "Phooey!"

Here is what I did:
I wrote out an equation of a line using 2x+1 as the slope and traveling through the point (2,3). I then set it equal to the original equation x^2+x. The tangent line equation would look like y=m(x-x1)+y1 or y=(2x+1)(x-2)+3. Now when I set it equal to the original equation, I get
(2x-1)(x-2)+3=x^2+x and I foiled out the left and added the three to get
2x^2-3x+1=x^2+x and then I made it all equal 0 so I was left w/
x^2-4x+1=0 and I used the Quadratic Formula to get my x values of

2+SquRt(3) and 2-SquRt(3) ??How else could I say Square Root??

Now, I plugged those into the derivative to get the actual slope and I used
(2,3) as the x and y values for my tangent line. I checked this on my calculator using the tangent line draw feature and more algebra and it worked because I got the same slope-intercept formula as I received when I used my 2+/-SquRt(3) for x and what I got when I plugged those into the original equation for y.

Bamboo,
Are these problems to some computer generated quiz? I ask because I used to help students at The Ohio State University with what was called Webwork for them. They were computer generated quizzes that students would enter their answers and they would immediately get feedback as to their correctness and could change them as necessary. These quizzes were also downloadable in a PDF form that I kept and use in my classes today. If these are similar, could I get copies to use in my classes? I love collecting materials to show my high school students what will be expected of them in college and to use as a problem of the day or week. Thanks if you could get me either electronic or hard copies through mail or email.

I screwed up!!!!!!
I used (2,3) instead of (2,-3)!!!!
This means that I should've gotten x=-1 and x=5 instead. I found this error as I was working this problem in class w/ my students today. My work and logic was correct, just not my numbers since I used 3 instead of -3.

bamboo711
10-01-2008, 03:19 PM
Dak..... haahahahah I asked my teacher in class about this question, and she showed me where this went wrong. Thanks for the help though you were almost there.

Derive:
y = e^(7sqrt(x))

Obviously a chain rule that is re-written to look like e^(7x^(1/2)). How would I do this?

Thanks guys.

10-01-2008, 03:37 PM
Dak..... haahahahah I asked my teacher in class about this question, and she showed me where this went wrong. Thanks for the help though you were almost there.

Derive:
y = e^(7sqrt(x))

Obviously a chain rule that is re-written to look like e^(7x^(1/2)). How would I do this?

Thanks guys.
You've pretty much answered it. Chain rule. Since e^x is its own derivative, y'=e^(7sqrt(x))*(7sqrt(x))' I'll leave it to you to find (7sqrt(x))' because I know you can do it.

BTW, did you know it doesn't matter what rule you use as long as you use it correctly? Let y = (x^3)(x^2), if I find y' by the product rule, that's y'=(x^3)'(x^2) + (x^3)(x^2)', so y'=(3x^2)(x^2) + (x^3)(2x), y'=(3x^4) + (2x^4), y'=5x^4.
If I would have simplified first, y = x^5, so by the product rule, y' = 5x^4. In this case the product rule was way easier, (it usually is), but don't get "paralysis by analysis" trying to figure out which rule to use. They all work if you use them correctly, but some are easier to use than others. In fact, if you have time, a good way to check your answers is to solve it using a different rule.

bamboo711
10-02-2008, 07:53 AM
Ehhhh I'm still struggling.

I have e^(7x^1/2)

f'= e^(7x^(1/2))*(7/2x^(-1/2)

^^^ Is that right? I don't think so but not sure.

10-02-2008, 09:52 AM
Ehhhh I'm still struggling.

I have e^(7x^1/2)

f'= e^(7x^(1/2))*(7/2x^(-1/2)

^^^ Is that right? I don't think so but not sure.

That's right.

dak95_00
10-02-2008, 03:19 PM
I can't wait until you learn implicit differentiation so we can then introduce you to logarithmic differentiation and end all of this confusion! Once you get the chain rule, you pretty much have the differentiation thing down!

You should really consider checking these on your calculator by following these steps. Enter as follows:

\Y1=original problem to differentiate
\Y2=nDeriv(Y1,x,x)

Now, take notice and make certain that the grapher is set up to only graph Y2 & Y3 and that Y3 is set up to graph in bold font by highlighting that back slash before the y3. To determine if you found the correct dervative, your bold (Y3) graph just needs to overlap your other skinny (Y2) graph.

For example:
\y1=x^2
\y2=nDeriv(y1,x,x)
\y3=2x

Now give it a try! It works for all of these derivative problems.

bamboo711
10-02-2008, 07:19 PM
Yeah that's the next section after chain rule

Yo dak I don't understand how to do it. Can I not just be in the main screen and find the derivative? what would I type in because I tried nDeriv(x^2,x,x) and it gave me 0 in my ti-84 silver. Thanks guys.

dak95_00
10-03-2008, 12:27 PM
Yo dak I don't understand how to do it. Can I not just be in the main screen and find the derivative? what would I type in because I tried nDeriv(x^2,x,x) and it gave me 0 in my ti-84 silver. Thanks guys.

If you do it on the main screen, you can only do it for a value to find the slope of a tangent line in which case you just type nDeriv(function, x, value). For example, nDeriv(x^2,x,5) would equal 10. Fairly worthless if you ask me.

Let's try it w/ the latest dervative problem (product rule).
The rule: uv=u'v+uv'

u=x
v=e^7x
u'=1
v'=7e^7x

y'=1(e^7x)+x(7e^7x) or factoring out the e^7x gives:
y'=e^7x(1+7x)

So, trying it on the calculator yields:
y1=e^(7x)
y2=nDeriv(y1,x,x)
y3=e^(7x)(1+7x)

Again, unhighlight the equal sign on y1 so it doesn't graph. Highlight the slash on y3 so it graphs in bold. Take a look at the graphs and you'll see they are identical. Try again. Use the VARS key to enter y1 into y2. You'll have to arrow over to Y-Vars and then to FUNCTION to find y1. You want your calculator to graph y2 and then y3 over top of it to check for correctness. If the same, you are correct. I just did it for your problem. The graph of the derivative hugs the x-axis and then the y-axis.

bamboo711
10-03-2008, 04:58 PM
Thanks man, your explanation was wonderful. The derivative thing only does it for a point? My friend told me that there are programs you can download that can find them for you, but of course, I have to show work on exams.

Find the first and second derivatives of the function.
y = e^7e^x
y'= My wrong guess was (7e^7e^x)*(7e^x)
y"= My wrong was (49e^7e^x)*(7e^x)(7e^x)+(7e^x)(7e^7e^x)(7e^x)

Mansewerz
10-03-2008, 05:05 PM
Ugh, i'm in ICA honors, but just following all the stuff on a message board post is headspinning because i'm used to seeing the actual to the x-power written in superscript(?). the caret symbols are hard to follow efficiently.

10-03-2008, 05:55 PM
Ugh, i'm in ICA honors, but just following all the stuff on a message board post is headspinning because i'm used to seeing the actual to the x-power written in superscript(?). the caret symbols are hard to follow efficiently.
Tell me about it. I've done tutoring on-line, and typing that stuff is the hardest thing to type! You have to use weird keys, and go back and forth between "shift" all the time. And the nature of the subject means that if you make one mistake, it's wrong. If I could do superscript it would be alot more clear(er?).

bamboo711
10-04-2008, 08:53 AM
Steady Eddy or Dak or someone that can help me with some calculus questions.... do you guys have email accounts that I can send you my questions and can straight cut and paste the questions I have so they are easy to read like on the actual webassign?

Thanks

dak95_00
10-04-2008, 12:13 PM
Bamboo,

I am dak95_00@yahoo.com to email me. Can you pull up your web quiz offline through a pdf to print out? If so, you could save them and send them to me as email attachments. Again, I'd love to have them.

Was it y=(e^7)(e^x) or was it y=e^(7e^x)? I'll pretend it was the first.

e^7 is a constant so the constant multiple rule applies or that just means we carry it along for the ride and just do the derivative of e^7 times e^7.

y'=(e^7)*(e^x) or the same thing! If the other, I'd have to do logarithmic differentiation.

bamboo711
10-04-2008, 12:57 PM
Sadly it was the other equation.... I just shot you an email.

bumfluff
10-04-2008, 01:40 PM
To solve y=e^(7e^x) you need to use logarithmic differentiation.

Take the natural log of both sides so you get:

ln y = (ln e)^(7e^x)

=> ln y = 7e^x * ln e

=> ln y = 7e^x (ln e is 1 obviously)

The derivative with respect to x is 1/y dy/dx

so 1/y dy/dx = 7e^x

so dy/dx = y * 7e^x

Substitute in y and simplify if possible and bob's your uncle.

10-04-2008, 09:29 PM
Substitute in y and simplify if possible and bob's your uncle.
I was able to follow until I came to this thing about having an uncle bob. On the net it explained,
"Used to describe the means of obtaining a successful result. For example, 'left over right; right over left, and bob's your uncle - a reef knot'."
It's a UK thing it seems. Well, I'll be a monkey's uncle.

bumfluff
10-05-2008, 02:44 AM
Ah yes, I am trying to win you over to the ways of saying maths by exposing you to all the great sayings taht us brits have.

Just wondering bamboo, have you been taught logarithmic differentiation because if you knew what it was you would probably have no trouble with this question and it seems a little harsh to give you an example like this without knowing what it is.

Also, the best way to check your derivatives is get a calculator that can do them. I have a silver casio calculator that can do definite integrals and derivatives, so if you work out your derivative and find the value at a point and then check it with the answer your calculator gives, you know when you are right and wrong. And its not an expensive calculator either.

HornedWildebeast
10-08-2008, 05:11 PM
Ok I know everyone hates helping me with my hw but I'll post it anyway. *Note* I am not looking for an answer. Just trying to figure out where I went wrong.

I'm trying to finish my Pre-Calc project and I cannot figure out what I am doing wrong. Here is the problem: I have a student loan for \$32,016. For loan 1, the interest rate is 5%. Loan 2 has an interest rate of 6.8%. I am putting \$400 each month toward the loan and am trying to figure out how long it will take to pay off each loan. Can anyone figure out what I am doing wrong? Obviously loan 1 will be payed off quicker but I keep getting loan 2 as the quicker payoff. Thanks for any help

Length of time it will take to pay off loan if use LOAN 1:

A=P(1+r/k)^kt

32,016=400(1+.05/12)^12t

80.04=1.0042^12t

Log 1.0042 80.04=12t

Log 80.04/ log 1.0042=12t

1045.65=12t

88 months=t

7 years, 4 months

Length of time it will take to pay off loan if use LOAN 2:

A=P(1+r/k)^kt

32,016=400(1+.068/12)^12t

80.04=1.0057^12t

Log 1.0057 80.04=12t

Log 80.04/ log 1.0057=12t

771.05=12t

64.25 months=t

5 years, 5 months

10-09-2008, 10:31 AM
Ok I know everyone hates helping me with my hw but I'll post it anyway. *Note* I am not looking for an answer. Just trying to figure out where I went wrong.

I'm trying to finish my Pre-Calc project and I cannot figure out what I am doing wrong. Here is the problem: I have a student loan for \$32,016. For loan 1, the interest rate is 5%. Loan 2 has an interest rate of 6.8%. I am putting \$400 each month toward the loan and am trying to figure out how long it will take to pay off each loan. Can anyone figure out what I am doing wrong? Obviously loan 1 will be payed off quicker but I keep getting loan 2 as the quicker payoff. Thanks for any help

Length of time it will take to pay off loan if use LOAN 1:

A=P(1+r/k)^kt

32,016=400(1+.05/12)^12t

80.04=1.0042^12t

Log 1.0042 80.04=12t

Log 80.04/ log 1.0042=12t

1045.65=12t

88 months=t

7 years, 4 months

Length of time it will take to pay off loan if use LOAN 2:

A=P(1+r/k)^kt

32,016=400(1+.068/12)^12t

80.04=1.0057^12t

Log 1.0057 80.04=12t

Log 80.04/ log 1.0057=12t

771.05=12t

64.25 months=t

5 years, 5 months
You've got the wrong formulas. For the PV of series of payments you need:

[1 - (1 + .05/12)^(-n)] / (.05/12) = 32016/400 and solve for n. I get n = 97 payments.
For the other one [1- (1 + .068/12)^-n)] / (.068/12) = 32016/400. I get n = 106. So it does take longer to pay off the second one. Better yet, if you have a TI 83, use the financial program. You'll find it under APPS, Finance, TVM Solver. Solve for "N".

HornedWildebeast
10-09-2008, 12:01 PM
You've got the wrong formulas. For the PV of series of payments you need:

[1 - (1 + .05/12)^(-n)] / (.05/12) = 32016/400 and solve for n. I get n = 97 payments.
For the other one [1- (1 + .068/12)^-n)] / (.068/12) = 32016/400. I get n = 106. So it does take longer to pay off the second one. Better yet, if you have a TI 83, use the financial program. You'll find it under APPS, Finance, TVM Solver. Solve for "N".

Yep that's exactly what I figured out this morning. I knew something wasn't right and of course it was the formula. Thanks for the help

bamboo711
10-10-2008, 06:39 AM

A stone is dropped into a lake, creating a circular ripple that travels outward at a speed of 50 cm/s. Find the rate at which the area within the circle is increasing.
(a) after 2s
(b) after 3s
(c) after 6s

10-10-2008, 08:02 AM

A stone is dropped into a lake, creating a circular ripple that travels outward at a speed of 50 cm/s. Find the rate at which the area within the circle is increasing.
(a) after 2s
(b) after 3s
(c) after 6s
We have to use implicit differentiation, because we are figuring how it is changing with respect to time, not just the radius. So recall that the area of a circle is given by the formula A = pi r^2. So A' = pi 2r * r'

(a) after 2s r = 100, and r' = 50 (always), so A' = pi 200 * 50 = 10,000 pi cm^2 per second
(b) after 3s r = 150 and r' = 50, A' = pi 300 * 50 = 15,000 pi cm^2 per second
(c) after 6s r = 300 and r' = 50, A' = pi 600 * 50 = 30,000 pi cm^2 per second

bamboo711
10-10-2008, 01:55 PM
Thanks Steady Eddy. How did you know to use implicit?

Having some problems with exponential decay as well.

A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes. The initial population of a culture is 58 cells.
(a) Find the relative growth rate.
k = cells per hour

(b) Find an expression for the number of cells after t hours.
P(t) =

(c) Find the number of cells after 7 hours.
billion cells

(d) Find the rate of growth after 7 hours.
billion cells per hour

(e) When will the population reach 20,000 cells?
h

Thanks man, any advice with exponential growths and decay?

HornedWildebeast
10-10-2008, 03:19 PM
Thanks Steady Eddy. How did you know to use implicit?

Having some problems with exponential decay as well.

A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes. The initial population of a culture is 58 cells.
(a) Find the relative growth rate.
k = cells per hour

(b) Find an expression for the number of cells after t hours.
P(t) =

(c) Find the number of cells after 7 hours.
billion cells

(d) Find the rate of growth after 7 hours.
billion cells per hour

(e) When will the population reach 20,000 cells?
h

Thanks man, any advice with exponential growths and decay?

P=A(1+/-r)^x/t

a= initial amount
r= rate of decay (in this case .5 because it is a half life)
t= time

Just plug and chug. Really helpful to have a TI-83 and graph it.

bamboo711
10-10-2008, 05:07 PM
What????????? Is this right????

10-10-2008, 06:00 PM
Thanks Steady Eddy. How did you know to use implicit?
To be honest, at first I didn't! But then I didn't know how to put in 2s, 3s, and all that. Then I thought, we're not doing dA/dr, we're do dA/dt, with respect to time, so dA/dt = dA/dr * dr/dt. You need the chain rule (in a sense) which is what implicit is based on. But I don't use dA/dt, I just call it A', and I call dr/dt r', and remember that r' is the rate of growth of the radius.
Having some problems with exponential decay as well.

A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes. The initial population of a culture is 58 cells.Means it doubles every 20 minutes, doesn't it?

(a) Find the relative growth rate.
k = cells per hour
I'm going to pass on this one. I hope the others make sense.
(b) Find an expression for the number of cells after t hours.
P(t) =
So that would be P(t) = 58*2^(3h)
(c) Find the number of cells after 7 hours.
billion cells
Why does it say "billion cells"? Oh well, p(7) = 58*2^(21) = 121,634,816
(d) Find the rate of growth after 7 hours.
billion cells per hour
P' = 58*2^(3t)*ln 2 * 3, so when t = 7, P' = 58*2^(21)*ln2 *3 = 252,932,489.3 per hour. Could that be?
(e) When will the population reach 20,000 cells?
h
20,000 = 58*2^(3t), by doing some boring algebra, I get t = 2.8099, and I checked that and it worked!
Thanks man, any advice with exponential growths and decay?They make decay sound harder than it is. If they say it's half life is 150 years, you want a constant k, such that k^150 = 1/2. That's it. You don't really have to use e, but you teacher will like it if you do. But you could just let k = 150th root of (1/2), something calculators can do without logs.
What????????? Is this right????
I'm kind of confused, because it said decay, and then described what I think is growth. But if it does double every 20 minutes, then I think what I have is ok.

bamboo711
10-11-2008, 07:47 PM
Yo Steady Eddy that's not right for some reason but I don't know why or what. These are pretty challenging. I think you're supposed to use y(0)e^kt maybe?

Anyways, here's another problem that might be sorta similar. I really hate these problems they are too tough :(

A bacteria culture grows with constant relative growth rate. After 2 hours there are 400 bacteria and after 8 hours the count is 50,000.
(a) Find the initial population.
P(0) = ____bacteria

(b) Find an expression for the population after t hours.
P(t) =_______

(c) Find the number of cells after 7 hours.
P(7) =______ bacteria

(d) Find the rate of growth after 7 hours.
P'(7) = ________bacteria/hour

(e) When will the population reach 200,000?
t = _______hours

10-12-2008, 12:07 PM
Yo Steady Eddy that's not right for some reason but I don't know why or what. These are pretty challenging. I think you're supposed to use y(0)e^kt maybe?

Anyways, here's another problem that might be sorta similar. I really hate these problems they are too tough :(

A bacteria culture grows with constant relative growth rate. After 2 hours there are 400 bacteria and after 8 hours the count is 50,000.
(a) Find the initial population.
P(0) = ____bacteria

(b) Find an expression for the population after t hours.
P(t) =_______

(c) Find the number of cells after 7 hours.
P(7) =______ bacteria

(d) Find the rate of growth after 7 hours.
P'(7) = ________bacteria/hour

(e) When will the population reach 200,000?
t = _______hours
If it's wrong, it's what you asked me. Address why you describe growth but the book says it's decay. :confused:
Briefly, here's how to do these. (I gotta go somewhere.) k*a^8/k*a^2 = a^6= 50,000/400 = 125. So a^2 = 5, and a = sqrt(5). Using that it's easy to find that k = 80 (the initial amount). So the formula P(t) = 80*sqrt(5)^t, will give you your answers. You can express that as 80*e^kt, of course, but it will give the same answers. Talk to ya tomorrow, gotta go!

meowmix
10-12-2008, 05:58 PM
Hey, can somebody on here do this problem please?

Find the second derivative of :

x^2-Y^2=16.

I'm getting that the answer is y^2-x all over y^3... and I can't figure out where I went wrong. I got every single problem on this hw section wrong, and I'm confused as heck. Can somebody just please do this problem? (btw, the correct answer is supposed to be -16/y^3)

Mdubb23
10-12-2008, 06:27 PM
I'm in 8th grade, but I take level 4 geometry. This obviously is extremely basic compared to most of the stuff on here, but I just wanted to confirm that the three ways to prove triangles congruent are if all three sides are congruent, if two sides and their shared angle are congruent, and if two angles and their shared side are congruent. (I have to write two-column proofs to prove triangles congruent. I bet you guys remember those days . ;))

10-13-2008, 07:37 AM
Hey, can somebody on here do this problem please?

Find the second derivative of :

x^2-Y^2=16.

I'm getting that the answer is y^2-x all over y^3... and I can't figure out where I went wrong. I got every single problem on this hw section wrong, and I'm confused as heck. Can somebody just please do this problem? (btw, the correct answer is supposed to be -16/y^3)
Rewrite it this way: y^2 = x^2 - 16 ==> y = sqrt(x^2 - 16) ==>y'=(1/2)(x^2 - 16)^(-1/2)*(2x) ==>y'=(x)(x^2 - 16)^(-1/2) ==>y"=(x^2 - 16)^(-1/2)+(x)(-1/2)(x^2 - 16)^(-3/2)*(2x) ==>y"=(x^2 - 16)^(-1/2) - x^2(x^2 - 16)^(-3/2). Here's where the trouble probably came. Give each term the same denominator by multiplying the first one by (x^2 - 16). On the top the x^2s cancel, and you just have -16 on top. On the bottom you have (x^2 - 16)^(3/2), recall that from the beginning y = (x^2 - 16)^(1/2), so you can write that as y^3. So y" = (-16/y^3). It was algebra that made that one hard, not calculus.
I'm in 8th grade, but I take level 4 geometry. This obviously is extremely basic compared to most of the stuff on here, but I just wanted to confirm that the three ways to prove triangles congruent are if all three sides are congruent, if two sides and their shared angle are congruent, and if two angles and their shared side are congruent. (I have to write two-column proofs to prove triangles congruent. I bet you guys remember those days . ;))
Yep, the three are known as: SSS, SAS, and ASA. AAA would prove two triangles similiar, but says nothing about how big they are. SSA can sometimes be an impossible triangle, orther times it might be ambiguous, meaning that it could be either of two possible triangles.

Mdubb23
10-13-2008, 12:17 PM
Rewrite it this way: y^2 = x^2 - 16 ==> y = sqrt(x^2 - 16) ==>y'=(1/2)(x^2 - 16)^(-1/2)*(2x) ==>y'=(x)(x^2 - 16)^(-1/2) ==>y"=(x^2 - 16)^(-1/2)+(x)(-1/2)(x^2 - 16)^(-3/2)*(2x) ==>y"=(x^2 - 16)^(-1/2) - x^2(x^2 - 16)^(-3/2). Here's where the trouble probably came. Give each term the same denominator by multiplying the first one by (x^2 - 16). On the top the x^2s cancel, and you just have -16 on top. On the bottom you have (x^2 - 16)^(3/2), recall that from the beginning y = (x^2 - 16)^(1/2), so you can write that as y^3. So y" = (-16/y^3). It was algebra that made that one hard, not calculus.

Yep, the three are known as: SSS, SAS, and ASA. AAA would prove two triangles similiar, but says nothing about how big they are. SSA can sometimes be an impossible triangle, orther times it might be ambiguous, meaning that it could be either of two possible triangles.

Hey, thanks, Eddy, that helps a lot. When you taught, did you accept students just putting down SSS as a reason? My teacher sure doesn't. :roll:

10-13-2008, 05:23 PM
Hey, thanks, Eddy, that helps a lot. When you taught, did you accept students just putting down SSS as a reason? My teacher sure doesn't. :roll:
That was how they did it when I learned it back in Minnesota. And that was still how they were doing it in Arizona when I did my student teaching.

Mdubb23
10-13-2008, 05:42 PM
That was how they did it when I learned it back in Minnesota. And that was still how they were doing it in Arizona when I did my student teaching.

Alright, thanks again for your help. When I have my next question, I'll know where to look. ;)

meowmix
10-13-2008, 06:57 PM
Hey steadyeddy, thanks! I realized in class today where I went wrong... it wasn't the algebra that screwed me over, it was the fact that I didn't finish the problem. I didn't plug the original equation into my second derivative... which is why I got every single problem wrong... Kind of stupid now that I think about it...

10-13-2008, 08:53 PM
Hey steadyeddy, thanks! I realized in class today where I went wrong... it wasn't the algebra that screwed me over, it was the fact that I didn't finish the problem. I didn't plug the original equation into my second derivative... which is why I got every single problem wrong... Kind of stupid now that I think about it...
Glad to hear that you found the trouble. 'Cause when you miss every problem, (and I've done that), it's a big 'uh oh'.

bamboo711
10-14-2008, 05:44 PM

Find dy/dx by implicit differentiation.
xy = (4 + x^(2)y)^(1/2)

bamboo711
10-15-2008, 04:16 PM
Hey scratch the last problem, how do you do this related rates one? Thanks.

An upright cylindrical tank with radius 4 m is being filled with water at a rate of 3 m^3/min. How fast is the height of the water increasing? (Round the answer to four decimal places.)

meowmix
10-15-2008, 07:10 PM
Hey, by any chance, do you live near me? I JUST had a hw question that was almost identicle to that one... That or you use the same textbook?

I'm in CR School District, using Calculus of a Single Variable, 8th edition. You?

10-15-2008, 11:19 PM
Hey scratch the last problem, how do you do this related rates one? Thanks.

An upright cylindrical tank with radius 4 m is being filled with water at a rate of 3 m^3/min. How fast is the height of the water increasing? (Round the answer to four decimal places.)
To do related rates, I'd recommend not thinking about that dx/dt stuff. Just take a formula, and use implicit differentiation. So in your example, we need the formula for the volume of a cylinder.

V = pi*r^2 * h

We know r, so replace the r with 4.

V = 16 pi * h
V' = 16 pi * h'
Know what V' is? It's the rate the volume is increasing, i.e. 3.
3 = 16 pi * h', and they want to know the rate at which h' is increasing.
h' = 3/(16 pi) = 0.0597 meters per minute.

That one was so easy, it's confusing. I'm serious. Know these three for related rates.
The pythagorean theorem. a^2 + b^2 = c^2
So 2aa' + 2bb' = 2cc' which simplifies to aa' + bb' = cc'
a, b, and c are lengths, a', b', and c' are those sides rates of change.

They'll blow up a balloon. The volume of a sphere is V = 4/3 pi r^3.
So V' = 4 pi r^2 r'. They'll give you r, and then either V' or r', so you plug and chug.

They'll fill a cone with water. The volume of a cone is V = pi r^2 * h / 3.
They'll give you a hint about the ratio of the radius to the height. For example, suppose the height is 7 times that of the radius. Then it becomes
V = 7 pi r^3. V' = 21 pi r^2 * r'. Your job then is to use basic algebra to find V' or r'.

Knowing those three examples almost exhausts all the related rates riddles they can throw at you. After a bit of practice, you'll see that these can seem way easier than they did at first glance. Even I can do 'em, (sometimes).

bamboo711
10-16-2008, 09:09 AM
Thanks for that advice Steady Eddy. That question was rather simple so I don't have a clue why I couldn't get it, it was just confusing. Any ideas with this one? It's a little tougher I do believe.

Water is leaking out of an inverted conical tank at a rate of 10000 cm^3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank.
_____________cm^3/min

10-16-2008, 11:10 AM
Thanks for that advice Steady Eddy. That question was rather simple so I don't have a clue why I couldn't get it, it was just confusing. Any ideas with this one? It's a little tougher I do believe.

Water is leaking out of an inverted conical tank at a rate of 10000 cm^3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank.
_____________cm^3/min
See? Like I said, they'll ask about a cone. On this one, let's worry about the leak last. What's the ratio of the radius to the height of the cone. Since the diameter is 4 m, the radius must be 2 m. Since the height is 6 m, the height is 3 times the radius. So h = r/3. Recall that the volume of a cone is V = pi r^2 h/3. So V = pi (h/3)^2 h/3, V = pi h^3 / 27. V' = pi h^2 h'/9. Since h = 2 m = 200 cm...V' = pi (40,000)(20)/9. So if there was no leak V' would be 27,925 cm^3/min. (Just in case: h' = 20, and 1 m = 100 cm).

But 27,925 cm^3/min can't be the answer, 'cause it's leaking 10,000 cm^3/min. So if must be pumping 10,000 units more, i.e. 37,925 cm^3/min.

bamboo711
10-16-2008, 04:56 PM
Steady Eddy, that answer is not correct. I tried it and it's wrong. Maybe something that has to do with the derivative?

I know v= pi(h/2)^2(h/3)
Then I expand and get V= pih^3/12

Take the derivative and I get ((pih^2)/4) * dh/dt = dv/dt. Is that right?

I have no clue where to go from there. What to do man?

benz1
10-16-2008, 05:16 PM
I dont get any of this stuff at all.

dak95_00
10-16-2008, 07:50 PM
See? Like I said, they'll ask about a cone. On this one, let's worry about the leak last. What's the ratio of the radius to the height of the cone. Since the diameter is 4 m, the radius must be 2 m. Since the height is 6 m, the height is 3 times the radius. So h = r/3. Recall that the volume of a cone is V = pi r^2 h/3. So V = pi (h/3)^2 h/3, V = pi h^3 / 27. V' = pi h^2 h'/9. Since h = 2 m = 200 cm...V' = pi (40,000)(20)/9. So if there was no leak V' would be 27,925 cm^3/min. (Just in case: h' = 20, and 1 m = 100 cm).

But 27,925 cm^3/min can't be the answer, 'cause it's leaking 10,000 cm^3/min. So if must be pumping 10,000 units more, i.e. 37,925 cm^3/min.

Sorry guys. I've been busy.

I think you are off by a decimal place. I got 279,252.68 so that would be 289252.68 when you add the 10000. How about that one? Nice tutorial Steady. Definitely cones, cylinders, and triangles.

10-16-2008, 09:43 PM
Sorry guys. I've been busy.

I think you are off by a decimal place. I got 279,252.68 so that would be 289252.68 when you add the 10000. How about that one? Nice tutorial Steady. Definitely cones, cylinders, and triangles.
Sure looks like that's the problem, doesn't it? I used the computer calculator that's under 'accessories'. I should have used by trusty TI 83, I'm much better with that. Bamboo, does that fix it?

bamboo711
10-17-2008, 04:16 PM
That's right! Thanks! Mind explaining how you got to this answer step by step? I'm having problems comprehending how to take the derivative of the equation and why you put (h/3) as the radius as the radius and not h/2. Anyways just really trying to understand this problem. 87 on my first calc exam! Not bad right? It'll be a 90 with the homework and inclass quizzes probably. Thanks guys!

10-17-2008, 06:56 PM
That's right! Thanks! Mind explaining how you got to this answer step by step? I'm having problems comprehending how to take the derivative of the equation and why you put (h/3) as the radius as the radius and not h/2. Anyways just really trying to understand this problem. 87 on my first calc exam! Not bad right? It'll be a 90 with the homework and inclass quizzes probably. Thanks guys!
The whole deal with a "cone", related rates problem is you have two letters: r and h, and you've got to whittle them down to just one letter. Once you know the ratio of the height to the radius you can do this, because all lower levels of fluid in the cone creates a triangle of similar proportions to the cone. In this example we are told that h = 6 and the diameter = 4. So what is the ratio of the height to the radius? Since the diameter is 4, the radius is 2, and 6/2 is 3. We had V = pi r^2 h/3. Replace r with h/3 and V = pi (h/3)^2 h/3. V = pi (h^2/9)h/3, V = pi h^3/27. Do implicit differenciation and then replace r with 2.

Congrats on your exam score. :)

bamboo711
10-17-2008, 07:18 PM
Thanks I understand where you got those numbers from now.
This one is tricky I believe:
A particle is moving along the curve below.
y = sqrt(x)
As the particle passes through the point (4,2), its x-coordinate increases at a rate of 6 cm/s. How fast is the distance from the particle to the origin changing at this instant?
__________________ cm/s

10-18-2008, 08:44 AM
Thanks I understand where you got those numbers from now.
This one is tricky I believe:
A particle is moving along the curve below.
y = sqrt(x)
As the particle passes through the point (4,2), its x-coordinate increases at a rate of 6 cm/s. How fast is the distance from the particle to the origin changing at this instant?
__________________ cm/s
Let's use the pythagorean theorem. d^2 = x^2 + y^2, so 2dd' = 2xx' + 2yy'. This simplifies to dd' = xx' + yy'. Now d is sqrt(20) because 4^2 + 2^2 = 20. x = 4 x'=6 (given), and y = 2 y' =3/2. I better explain why y' = 3/2. y = x^(1/2), dy/dt = dy/dx * dx/dt = (1/2) x^(-1/2) dx/dt. Since dx/dt = 6 and x = 4, we have dy/dt = (1/2)(1/2)(6)=3/2.

sqrt(20)*d' = 4*6 + 2*3/2, 2sqrt(5)*d' = 27, so d'=13.5/sqrt(5)=6.03738

Let's check that by approximating the answer, let dt = .0001, so we'd want to figure the distance from (0,0) to (4.0006, sqrt(4.0006)). sqrt( 4.0006^2 + 4.0006) - sqrt(20) = .000603738. That's the change in the distance, and dividing that by the change in the time, dt = .0001, yields, 6.03738, accurate to 4 decimals.

bamboo711
10-19-2008, 01:19 PM
Thanks for that Steady Eddy, I really don't like related rate at all. I'm trying to get ahead of the class a little and am trying to understand the concepts of the next lesson.

Any idea with these for examples for me?

Find the linearization of the function given below at the specified point.
f(x)=4 ln(x) at a = 1
___________

AND

Find the differential of each function.
y = s/(1+5s)
dy=________

10-19-2008, 03:50 PM
Thanks for that Steady Eddy, I really don't like related rate at all. I'm trying to get ahead of the class a little and am trying to understand the concepts of the next lesson.

Any idea with these for examples for me?

Find the linearization of the function given below at the specified point.
f(x)=4 ln(x) at a = 1
f'(x) = 4/x. f'(1) = 4 f(1) = 4*0 = 0.
So with m = 4, x = 1, and y = 0 the equation of the line is y = 4x - 4.
___________

AND

Find the differential of each function.
y = s/(1+5s)
dy=________
So they want dy/ds? If so, use the quotient rule to get: [(1 + 5s) - (5s)] / (1 + 5s)^2 = 1/(1 + 5s)^2

bamboo711
10-20-2008, 05:18 PM
You forgot the ds at the end Steady Eddy but that was almost completely right.

The edge of a cube was found to be 30 cm with a possible error in measurement of 0.2 cm. Use differentials to estimate the maximum possible error, relative error, and percentage error in computing the volume of the cube and the surface area of the cube. (Round relative errors to four decimal places and percentage errors to two decimal places.)

Max error Relative error Percentage error
Volume ____ cm^3 ______ ___%
Surface Area ___cm^2 ______ ___%

10-20-2008, 10:00 PM
You forgot the ds at the end Steady Eddy but that was almost completely right.
Oh, I see. I asked if they wanted dy/ds? To turn this into dy, simply to the algebra trick of multiplying each side by ds. That would make one side "dy" and then ds would move to the other side.

The edge of a cube was found to be 30 cm with a possible error in measurement of 0.2 cm. Use differentials to estimate the maximum possible error, relative error, and percentage error in computing the volume of the cube and the surface area of the cube. (Round relative errors to four decimal places and percentage errors to two decimal places.)

Max error Relative error Percentage error
Volume ____ cm^3 ______ ___%
Surface Area ___cm^2 ______ ___%
Since I'm sleepy, I'm hoping it's enough to say that the volume of a cube is x^3, while the surface area equals 6x^2, where x is the side of the cube.

bamboo711
10-27-2008, 05:02 PM
Find the local maximum value of f using both the First and Second Derivative Tests.
f(x) = x + sqrt(7-x)

10-27-2008, 08:14 PM
Find the local maximum value of f using both the First and Second Derivative Tests.
f(x) = x + sqrt(7-x)Have you been on a math holiday? Welcome back. The first derivative of f(x) = x + (7 - x)^(1/2), is f'(x) = 1 + (1/2)(7 - x)^(-1/2)(-1). When does that equal zero? When 1 = (1/2)(1/radical(7 - x)) ==> radical(7 - x) = (1/2) ==> 7 - x = 1/4 so x = 6 3/4. Now is that a relative maximum or minimum?

Take the 2nd derivative, f''(x) = (1/2)(1/2)(7 - x)^(-3/2)*(-1). What a mess! But it's clear still that it's zero when x is 7. And when x is smaller than 7, (x - 7) is negative, the odd root of a negative is negative. So when x is 6.75, f'' is negative, or concave downward, so it's a relative maximum.

Dak, can you verify this on the TI 83?

game set match 46 TIMES!!
10-28-2008, 08:15 PM
this isnt a sticky yet im surprised haha...

bamboo711
10-30-2008, 11:52 AM
I didn't stop doing math.... just could get all of it. Thanks for that Eddy I got the same answer.

Now to optimization :((((( so scary!

A boat leaves a dock at 4:00 P.M. and travels due south at a speed of 20 km/h. Another boat has been heading due east at 15 km/h and reaches the same dock at 5:00 P.M. How many minutes after 4:00 P.M. were the two boats closest together? Round your answer to the nearest minute.

Nellie
10-30-2008, 12:31 PM
x=-15+15t
y=-20t

(t=time in hours)

(1) Distance= sqrt (x^2+y^2), or

sqrt (625t^2-450t+225)

or

If you want to minimize, the square root, it is the same as minimizing the positive distance squared, so you are trying to minimize

625t^2+450t+225

to get 25(5t-3)^2=0

or t=3/5, or 36 minutes

Check, according to the equation 1, this distance would be sqrt(180), or about 13.4 km.

10-30-2008, 09:22 PM
or t=3/5, or 36 minutes

Check, according to the equation 1, this distance would be sqrt(180), or about 13.4 km.
Thanks Nellie!

bamboo711
10-31-2008, 06:12 AM
Thanks Nellie, that was a nice solution and I understand it better. Now for this problem, I know that the best optimization is a square but can someone show me the real steps to obtain the answer to this problem? Thanks guys.

Find the dimensions of a rectangle with area 1000 m^2 whose perimeter is as small as possible. (Give your answers in increasing order, to the nearest meter.)

FlamEnemY
11-01-2008, 02:51 AM
Find the dimensions of a rectangle with area 1000 m^2 whose perimeter is as small as possible. (Give your answers in increasing order, to the nearest meter.)

Let's call the sides of the rectangle side "a" and side "b". The area is S, the perimeter is P.

In this case we have

S=ab >> 1000=ab >> a=1000/b

We use this equation of "a" in the equation of the perimeter

P=2a+2b=2000/b +2b=(2000+2b^2)/b

We find the first derivative of P in order to find the minimum value.

P'=[(2000+2b^2)'b -b'(2000+2b^2)]/b^2

b^2 is always >0, it doesnt affect the sign (+ or -) of the equation so I won't use it in the equation anymore.

P'=4b^2-2000-2b^2=2b^2-2000
P'=0 in order to find the minimum/maximum value (in this case minimum).

Then b^2=1000 >> b=10x10^1/2
a=1000/b=100/10^1/2=10x10^1/2

There you go. Square :)

11-01-2008, 04:45 AM
Let's call the sides of the rectangle side "a" and side "b". The area is S, the perimeter is P.

In this case we have

S=ab >> 1000=ab >> a=1000/b

We use this equation of "a" in the equation of the perimeter

P=2a+2b=2000/b +2b=(2000+2b^2)/b
If we skip this last step, it might make things a bit easier. Leave it as P = 2000/b + 2b.
P' = -2000/b^2 +2, so solve 1 = 1000/b^2 (for optimization, set equal to zero), hence b^2 = 1000, b = sqrt(1000), so a = sqrt(1000) also.

bamboo711
11-02-2008, 07:54 AM
Thanks guys, pretty simple, I guess just I don't know what direction to go with the problems really well. What about this one?

Find the dimensions of the rectangle of largest area that has its base on the x-axis and its other two vertices above the x-axis and lying on the parabola. (Round your answers to the nearest hundredth.)
y = 6 - x^2
_______units (width)
__________units (height)

11-02-2008, 10:13 AM
Thanks guys, pretty simple, I guess just I don't know what direction to go with the problems really well. What about this one?

Find the dimensions of the rectangle of largest area that has its base on the x-axis and its other two vertices above the x-axis and lying on the parabola. (Round your answers to the nearest hundredth.)
y = 6 - x^2
_______units (width)
__________units (height)
area = length * width, or x * y, or x * (6 - x^2)
area = 6x - x^3
solve 0 = 6 - 3x^2 ==> x = sqrt(2)
The width is 2 rad(2), or 2.83
the height is 4
(The area is 8 rad(2), or 11.31)

FlamEnemY
11-03-2008, 11:58 AM
Food for thought:

We have a right square pyramid. We know the lateral surface of the pyramid is S (S is a given constant) and the sides of the pyramid are at an angle of ά degrees towards the base of the pyramid. Find tgά=? so that the pyramid has maximum volume.

We solved this at school, so I don't really need it, but it's quite tricky and may prove to be a test for one's math skills :)

bamboo711
11-03-2008, 12:53 PM
Thanks Steady Eddy, I understand. Final problem for a few days.

A baseball team plays in a stadium that holds 54,000 spectators. With ticket prices at \$10, the average attendance had been 49,000. When ticket prices were lowered to \$8, the average attendance rose to 51,000.
(a) Find the demand function, assuming it to be linear.
p(x) =_______

(b) How should ticket prices be set to maximize revenue?
\$ ________

Thanks for all the help guys. This optimization and related rates are tough!

11-04-2008, 03:28 PM
Food for thought:

We have a right square pyramid. We know the lateral surface of the pyramid is S (S is a given constant) and the sides of the pyramid are at an angle of ά degrees towards the base of the pyramid. Find tgά=? so that the pyramid has maximum volume.

We solved this at school, so I don't really need it, but it's quite tricky and may prove to be a test for one's math skills :)
This sounded so similar to finding the largest volume of a cylinder for a given surface area, that I thought it would be just as easy. (For that, make the height equal to the diameter). When I try it with the pyramid, I get sin a and cos a, but I can't isolate a. There must be some trick to it that I'm missing. BTW, my guess is that a = pi/4 or 45 degrees, correct?

Thanks Steady Eddy, I understand. Final problem for a few days.

A baseball team plays in a stadium that holds 54,000 spectators. With ticket prices at \$10, the average attendance had been 49,000. When ticket prices were lowered to \$8, the average attendance rose to 51,000.
(a) Find the demand function, assuming it to be linear.
p(x) =_______

(b) How should ticket prices be set to maximize revenue?
\$ ________

Thanks for all the help guys. This optimization and related rates are tough!
We have two points, (10, 49) and (8, 51), the slope = -1, so y = -x +59. More literally, y = -1000x + 59,000. But revenue is the product of x and y. r = -1000x^2 +59,000x, which would be optimal at 29.5. Recall that the stadium only holds 54,000, so we can't go over that. This means, any extreme lowballing strategy needs a 2nd look. But we're knocking the prices way up, so it stands at \$29.50 per ticket.

bamboo711
11-05-2008, 04:00 PM
Thanks man.

For this integral:

Find the most general antiderivative of the function. Use C for any needed constant. (x > 0)
f(x) = 2e^x + 5sec^2(x)

Why if F(x) not e^2x + 5tan^2(x)???????

11-05-2008, 09:45 PM
Thanks man.

For this integral:

Find the most general antiderivative of the function. Use C for any needed constant. (x > 0)
f(x) = 2e^x + 5sec^2(x)

Why if F(x) not e^2x + 5tan^2(x)???????
Check an anti-derivative by taking its derivative. (e^2x)' = (e^2x)*(2), not 2e^x, (you solve it by using the chain rule, and e^x is it's own derivative.)

Also, (5tan^2(x))' = 10 tan(x)*(tan^2(x))' by the chain rule. When I first tried learning these I had to do 'u substitutions' in order to get it to work. It's a pain at first, but it really gets you to "see" what works and what doesn't. These functions have other functions nested inside them. One really has to pay attention to the order of operations.

FlamEnemY
11-06-2008, 05:47 AM
This sounded so similar to finding the largest volume of a cylinder for a given surface area, that I thought it would be just as easy. (For that, make the height equal to the diameter). When I try it with the pyramid, I get sin a and cos a, but I can't isolate a. There must be some trick to it that I'm missing. BTW, my guess is that a = pi/4 or 45 degrees, correct?

Actually the answer is tga=square of 2 :) I can try to write the solution if you want.

11-06-2008, 10:39 AM
Actually the answer is tga=square of 2 :) I can try to write the solution if you want.
I'd appreciate that.

FlamEnemY
11-07-2008, 06:55 AM
Ok, let's say the pyramid is ABCDH, ABCD being the base square.
The side of the square is "b"

We build HK - median in triangle BCH, K part of BC. I don't know how this line is called in english when used in a pyramid, but it is used to find the lateral surface. We call it "k"
The lateral surface S=pk (p is half the perimeter of the base)

so S=2bk

if O is the center of ABCD, then HO is the height of the pyramid.
In triangle OKH:
HK=k
OK=b/2 (O is center of the square -> it is midpoint for every line paralell to the side of the square)
This way we can use cosά in triangle OKH to make an equation for k and b:
b/2=kcosά ; b=2kcosά

So S=2x(2kcosά)x(k)
S=4k^2 cosά
So k=square of (S/4cosά)
b=2kcosά
We make an equation of b using only S and cosά
In triangle OKH: h^2=k^2-b^2
we also have an equation of h using only S and cosά

We find h, then use it in the formula V=(b^2 h)/3

Now, if everything is OK, we'll have V=S x square(S) x square(cosά sin^2ά) / 6
we make funktion f(ά)=cosά sin^2ά . We need only cosά and sinά because S is constant and only ά affects the end result.

f(ά)=cosά (1-cos^2ά)

f '(ά)=sinά(2-3sin^2ά) (actually the first derivative is tricky to find and it requires several steps, but I write only the end result)

F '(ά)=0 => sinά=0 and sinά=square(2/3), which is the argument needed for max f(ά).

So sinά=square(2/3), tgά=sinά/cosά=square(2)

11-07-2008, 06:04 PM
Ok, let's say the pyramid is ABCDH, ABCD being the base square.
The side of the square is "b"

We build HK - median in triangle BCH, K part of BC. I don't know how this line is called in english when used in a pyramid, but it is used to find the lateral surface. We call it "k"
The lateral surface S=pk (p is half the perimeter of the base)

so S=2bk

if O is the center of ABCD, then HO is the height of the pyramid.
In triangle OKH:
HK=k
OK=b/2 (O is center of the square -> it is midpoint for every line paralell to the side of the square)
This way we can use cosά in triangle OKH to make an equation for k and b:
b/2=kcosά ; b=2kcosά

So S=2x(2kcosά)x(k)
S=4k^2 cosά
So k=square of (S/4cosά)
b=2kcosά
We make an equation of b using only S and cosά
In triangle OKH: h^2=k^2-b^2
we also have an equation of h using only S and cosά

We find h, then use it in the formula V=(b^2 h)/3

Now, if everything is OK, we'll have V=S x square(S) x square(cosά sin^2ά) / 6
we make funktion f(ά)=cosά sin^2ά . We need only cosά and sinά because S is constant and only ά affects the end result.

f(ά)=cosά (1-cos^2ά)

f '(ά)=sinά(2-3sin^2ά) (actually the first derivative is tricky to find and it requires several steps, but I write only the end result)

F '(ά)=0 => sinά=0 and sinά=square(2/3), which is the argument needed for max f(ά).

So sinά=square(2/3), tgά=sinά/cosά=square(2)
Now I don't feel bad. Here's one. A canoeist is paddling upstream. He travels a mile from where he started, when his hat blows off. Since it's an old hat, he lets it go and keeps paddling upsteam. After 10 minutes it occurs to him that his hat contains his plane ticket, so he goes after it. By the time he catches up with his hat, he sees that he's back to where he started. Show why this means the stream was 3 mph.

dave333
11-07-2008, 06:23 PM
Okay, I'm kinda hijacking this thread with this problem I'm doing...epsilon delta BS crap that is useless compared to good ol' limit rules. My course is doing a rigorous review of limits, mostly concerning proofs with epsilon delta.

I'm supposed to, given epsilon, find the maximum c (delta) so that the limit is basically true.

example: given that e (we'll just use e as epsilon) is .01, find c so that lim x-> 4 (x=4). Basically, find the maximum delta so this is true. Obviously with the definition of epsilon delta, in this case, epsilon=delta and so c=.01

I now have to do this rather annoying one: lim x-> 0 (sinx/x=1) given that e=0.01, find c (delta).
I've gotten to .99<sinx/x<1.01 but I'm a bit stuck here. I set sinx/x=.99 and sinx/x=1.01 and find x in each case, but then I'm not sure what to do from there.

And could someone expalin wtf the point of epsilon delta is for? As far as I am aware, the epsilon delta proof was invented by some French guy many many many years after Newtwon invented calculus.

11-08-2008, 06:02 AM
Okay, I'm kinda hijacking this thread with this problem I'm doing...epsilon delta BS crap that is useless compared to good ol' limit rules. My course is doing a rigorous review of limits, mostly concerning proofs with epsilon delta.

I'm supposed to, given epsilon, find the maximum c (delta) so that the limit is basically true.

example: given that e (we'll just use e as epsilon) is .01, find c so that lim x-> 4 (x=4). Basically, find the maximum delta so this is true. Obviously with the definition of epsilon delta, in this case, epsilon=delta and so c=.01

I now have to do this rather annoying one: lim x-> 0 (sinx/x=1) given that e=0.01, find c (delta).
I've gotten to .99<sinx/x<1.01 but I'm a bit stuck here. I set sinx/x=.99 and sinx/x=1.01 and find x in each case, but then I'm not sure what to do from there.

And could someone expalin wtf the point of epsilon delta is for? As far as I am aware, the epsilon delta proof was invented by some French guy many many many years after Newtwon invented calculus.
It was invented by Wierstrauss, (sp?), in the 1800s. He was German. It was said that the foundations of calculus weren't rigorous, and so Wierstrauss came up with the delta epsilon stuff. For example 1/x goes to zero, as x approaches infinity. It never really becomes zero, but it can be as small as we please. We can say that all 1/x is less than some number, when x is larger than some other number. So it's really more philosophical. Delta/epsilon proofs don't help you solve problems, it's about proving things. I'll be back later, I gotta get to work.

bamboo711
11-08-2008, 11:41 AM
I can't figure this out.
Find f.
f'(x) = 1/(sqrt(1 - x^2))
f(0.5) = 2

I got arcsin(x)+ (4-(pi/3)) but that's not right.

11-08-2008, 04:06 PM
I can't figure this out.
Find f.
f'(x) = 1/(sqrt(1 - x^2))
f(0.5) = 2

I got arcsin(x)+ (4-(pi/3)) but that's not right.
I think you're close. If f'(x) = 1/sqrt(1 - x^2), then f(x) = arcsin(x) + C
So arcsin(.5) + C = 2
pi/6 + C = 2, so C = 2 - pi/6.
Finally f(x) = arcsin(x) + 2 - pi/6.

Let's check, if f(x) = arcsin(x) + 2 - pi/6, then f'(x) = 1/sqrt(1-x^2), good.
And f(.5) = arcsin(.5) + 2 - pi/6 = pi/6 + 2 - pi/6 = 2 good.

f = arcsin(x) + 2 - pi/6

bamboo711
11-08-2008, 06:26 PM
Thanks man,
Silly mistake.

A stone is dropped from the top of a 250 m tower. (Acceleration due to gravity is -9.8 m/s^2. Ignore air resistance. Give your answers correct to two decimal places.)

(a) Find the distance of the stone above ground level at time t.
h(t)=___________

(b) How long does it take the stone to reach the ground?
______s

(c) With what speed does it strike the ground?
______m/s

(d) If the stone is thrown downward with a speed of 6 m/s, how long does it take to reach the ground?
_____s

11-08-2008, 08:42 PM
Thanks man,
Silly mistake.
They're easy to do in math.

A stone is dropped from the top of a 250 m tower. (Acceleration due to gravity is -9.8 m/s^2. Ignore air resistance. Give your answers correct to two decimal places.)

(a) Find the distance of the stone above ground level at time t.
h(t)=___________

(b) How long does it take the stone to reach the ground?
______s

(c) With what speed does it strike the ground?
______m/s

(d) If the stone is thrown downward with a speed of 6 m/s, how long does it take to reach the ground?
_____s
(a) h(t) = 250 - 9.8*t^2
(b) 0 = 250 - 9.8*t^2
t = sqrt(250/9.8 ) = 5.05 seconds
(c) 5.05*19.6 = 98.98 m/s (the deritivate of t^2 is 2t, hence the 19.6 #)
(d) 0 = 250 - 6t - 9.8t^2, so t = 4.75 seconds

I think those are the answers. Better double-check everything I say!

bamboo711
11-09-2008, 04:51 PM
It wasn't right Steady Eddy but thanks for trying man. I eventually figured most of it out. Want to see or are you not in desire?

Find f.
f ''(x) = 8 + cos(x)
f(0) = -1
f(pi/2) = 0
f(x) =__________

11-09-2008, 06:03 PM
It wasn't right Steady Eddy but thanks for trying man. I eventually figured most of it out. Want to see or are you not in desire?

Find f.
f ''(x) = 8 + cos(x)
f(0) = -1
f(pi/2) = 0
f(x) =__________
Did I get a, b, and c, but not d? I'll just move on to this one.
f''(x) = 8 + cos(x)
f'(x) = 8x + sin(x) + C
f(x) = 4x^2 - cos (x) + Cx + D
f(0) = 0 - 1 + 0 + D = -1, so D is zero, let's just ignore it.
f(pi/2) = pi^2 - 0 + C(pi/2) = 0 implies C is -2pi
f(x) = 4x^2 - cos(x) - 2pi(x). Could that be right? Let's check.
f(0) = 0 - 1 - 0 = -1, good.
f(pi/2) = pi^2 - 0 - pi^2 = 0, good.
and f'(x) = 8x + sin(x) - 2pi
so f''(x) = 8 + cos(x), good. So let's go with f = 4x^2 - cos(x) - 2pi(x)

bamboo711
11-10-2008, 03:59 PM
Thanks Steady Eddy, I need help with this one. I hate these.

Do the following.
a = 10
http://img522.imageshack.us/img522/3690/51001altoq2.gif (http://imageshack.us) http://img522.imageshack.us/img522/51001altoq2.gif/1/w222.png (http://g.imageshack.us/img522/51001altoq2.gif/1/)

(a) By reading values from the given graph of f, use five rectangles to find a lower estimate and an upper estimate for the area under the given graph of f from x = 0 to x = 20.

A =____(lower estimate)
A=_______ (upper estimate)

(b) Find new estimates using ten rectangles in each case.

A=_______ (lower estimate)
A=___________ (upper estimate)

11-11-2008, 07:10 AM
Thanks Steady Eddy, I need help with this one. I hate these.

Do the following.
a = 10
http://img522.imageshack.us/img522/3690/51001altoq2.gif (http://imageshack.us) http://img522.imageshack.us/img522/51001altoq2.gif/1/w222.png (http://g.imageshack.us/img522/51001altoq2.gif/1/)

(a) By reading values from the given graph of f, use five rectangles to find a lower estimate and an upper estimate for the area under the given graph of f from x = 0 to x = 20.

A =____(lower estimate)
A=_______ (upper estimate)

(b) Find new estimates using ten rectangles in each case.

A=_______ (lower estimate)
A=___________ (upper estimate)
That's just tedious work. All I can say is that since a = 10, each 'hashmark' must be two. Because if you count by twos you get to 10 when you reach the point marked "a".

Irvin
11-11-2008, 09:00 AM
Guys, I can't figure this question out: A spherical balloon with radius r inches has volume defined by the function below. Find a function that represents the amount of air required to inflate the balloon from a radius of r inches to a radius of r + 5 inches. (Give the answer in terms of π and r.)

oh and the originial formula is v= 4/3 pi r^2

Not enough information to determine. We would also need to know the altitude, temperature, and inward pressure of the ballon.

Irvin

Irvin
11-11-2008, 09:05 AM
Wouldn't it be easier to put a graduated beaker under water and pump air into with a hand pump. Measure the displacement of the air then pump up the ballon. Multiple the number of pumps by the average volume of air and you know the answer.

Work smarter not harder.

Irvin

11-11-2008, 08:02 PM
wow my h.s calc teacher was the worst ever =.=
basically we had the book and he told us wat page to read and do.
wished he actually taught us
steady eddy i bet ur class would have been fun to be in

11-12-2008, 05:43 AM
wow my h.s calc teacher was the worst ever =.=
basically we had the book and he told us wat page to read and do.
wished he actually taught us
steady eddy i bet ur class would have been fun to be in
Thanks. Sounds like s/he didn't understand the material, so just told you guys to read the book.

bamboo711
11-13-2008, 04:28 PM
Consider the given function.
f(x) = -1-1/4 x
Evaluate the Riemann sum for 2 ≤ x ≤ 4 , with six subintervals, taking the sample points to be left endpoints.
L6 =__________

Why is it not just ((4-2/6))= 1/3
1/3(f(2)+f(7/3)+f(8/3)+f(3)+f(10/3)+f(... -3.417????

I'm so confused

11-14-2008, 11:01 AM
Consider the given function.
f(x) = -1-1/4 x
Evaluate the Riemann sum for 2 ≤ x ≤ 4 , with six subintervals, taking the sample points to be left endpoints.
L6 =__________

Why is it not just ((4-2/6))= 1/3
1/3(f(2)+f(7/3)+f(8/3)+f(3)+f(10/3)+f(... -3.417????

I'm so confused
The six subintervals would be:
1) 2 - 2 1/3
2) 2 1/3 - 2 2/3
'till you get to 3 2/3 - 4
Then it's alot of boring plug and chug. I'm not very good at math when I get bored. If you got the answer wrong, just use these 6 #'s and add 'em up. If your answer guide is right, you should get the same sum. Persistence pays off on these.

bamboo711
11-14-2008, 06:46 PM
Thanks, the stupid app still didn't let me get the right answer but whatev. Having some trouble with this one that looks like a quoetient rule and its not a derivative so i'm confused.

Evaluate the integral.
integral from(1) to (16) of (x-1)/sqrt(x) dx

Thanks guys

11-15-2008, 08:00 AM
Thanks, the stupid app still didn't let me get the right answer but whatev. Having some trouble with this one that looks like a quoetient rule and its not a derivative so i'm confused.

Evaluate the integral.
integral from(1) to (16) of (x-1)/sqrt(x) dx

Thanks guys
Let's break this up into two terms: x/sqrt(x) - 1/sqrt(x). Using exponents, we can write these two terms as: x^1/2 - x^(-1/2). Now let's integrate each of these seperately. That's not hard, we just add one to the exponent, and make the reciprocal of that exponent the coefficient.
2/3 x^3/2 - 2 x^1/2. Mentally check that taking the derivative of each of those terms makes it what it should be.
The integral goes from 1 to 16, so 2/3 (64 - 1) - 2(4 - 1) = 42 - 6, but what's that? Just kidding, 42 - 6 = 36, and when I do this on my casio fx-115ES I get the same answer, so I'll go with it. 36.

bamboo711
11-19-2008, 06:15 AM
Thanks Steady Eddy that helped a lot,

The velocity function v(t) (in meters per second) is given for a particle moving along a line.
v(t) = 3t - 8, 0 <= t <= 4
(a) Find the displacement d1 traveled by the particle during the time interval given above.
d1 = ___________m

(b) Find the total distance d2 traveled by the particle during the time interval given above.
d2 = ____________m

FlamEnemY
11-22-2008, 12:22 AM
Now I don't feel bad. Here's one. A canoeist is paddling upstream. He travels a mile from where he started, when his hat blows off. Since it's an old hat, he lets it go and keeps paddling upsteam. After 10 minutes it occurs to him that his hat contains his plane ticket, so he goes after it. By the time he catches up with his hat, he sees that he's back to where he started. Show why this means the stream was 3 mph.

Sorry for the delay. However, here is the solution:

The starting point is A, the point he loses his hat is B and the point he decides to turn back is C.

A-------------B-----C
<~~~~

The time that the guy needs to get from C to A (The point where he turns around to the point of catching the hat) is "t"
The speed of the canoeist without stream is Vo, paddling upstream is Vo-Vstr,
downstream its Vo+Vstr.
AB=1 mile=(t+1/6h)Vstr ; t+1/6h because this is the time the hat travelled - ten minutes(1/6 hours) plus the time the canoenist needed for CA.
BC=1/6h(Vo-Vstr)
AC=t(Vstr+Vo)

AB+BC=AC
tVstr+1/6Vstr+1/6Vo-1/6Vstr=tVo+tVstr
1/6Vo=tVo
t=1/6h=10min
AB=1 mile=(1/6+1/6)Vstr=1/3Vstr
V stream=3 mph

Capt. Willie
11-22-2008, 12:31 AM
Math was never my strong point but I do know 7x13=28

FlamEnemY
11-23-2008, 05:14 AM
Here's something tricky:

Find the maximum value of f(x)=tgx + cotgx ; pi/2<x<pi

There are basically two ways of figuring this out :)

11-23-2008, 07:52 AM
Sorry for the delay. However, here is the solution:

The starting point is A, the point he loses his hat is B and the point he decides to turn back is C.

A-------------B-----C
<~~~~

The time that the guy needs to get from C to A (The point where he turns around to the point of catching the hat) is "t"
The speed of the canoeist without stream is Vo, paddling upstream is Vo-Vstr,
downstream its Vo+Vstr.
AB=1 mile=(t+1/6h)Vstr ; t+1/6h because this is the time the hat travelled - ten minutes(1/6 hours) plus the time the canoenist needed for CA.
BC=1/6h(Vo-Vstr)
AC=t(Vstr+Vo)

AB+BC=AC
tVstr+1/6Vstr+1/6Vo-1/6Vstr=tVo+tVstr
1/6Vo=tVo
t=1/6h=10min
AB=1 mile=(1/6+1/6)Vstr=1/3Vstr
V stream=3 mph
You da man! There's a tricky way to do this too. Since the current moves both the hat and the canoe, he paddles away from it at the same rate he paddles towards the hat. So if he paddled away for 10 minutes, it takes 10 minutes to get back to it. So it took 20 minutes for the hat to float 1 mile. This means the currents is going 3 mph. :) I like how if you don't notice that, regular algebra still gives you the same answer. (I also like how it doesn't matter what the canoeist's rate is, he can go 4 mph or 40 mph, it still comes out if the current is 3 mph!)
Math was never my strong point but I do know 7x13=28
You look strong to me.
Here's something tricky:

Find the maximum value of f(x)=tgx + cotgx ; pi/2<x<pi

There are basically two ways of figuring this out :)

Is "g" a constant?

FlamEnemY
11-23-2008, 09:37 AM
You da man! There's a tricky way to do this too. Since the current moves both the hat and the canoe, he paddles away from it at the same rate he paddles towards the hat. So if he paddled away for 10 minutes, it takes 10 minutes to get back to it. So it took 20 minutes for the hat to float 1 mile. This means the currents is going 3 mph. :) I like how if you don't notice that, regular algebra still gives you the same answer. (I also like how it doesn't matter what the canoeist's rate is, he can go 4 mph or 40 mph, it still comes out if the current is 3 mph!)

Ah yes. Relative speed and distance. That's a lot smarter :)

Is "g" a constant?

By cotgx I meant the reciprocal of tgx (trigonometry). We are probably working with different symbols, hence the misunderstanding.

11-23-2008, 11:03 AM
Here's something tricky:

Find the maximum value of f(x)=tgx + cotgx ; pi/2<x<pi

There are basically two ways of figuring this out :)
I'm not feeling tricky today, so I just used my graphing calulator. I get that the maximum occurs when x is approximately 2.3561957. At this value, f(x) is -2.

(I'm thinking that the question is asking what is the maximum value of f(x) = tan (x) + cot (x).)

Or this way. By taking f'(x) and setting it equal to zero. I get that x = pi/4, 3pi/4,... The only value in the domain is 3pi/4 which is 2.35619449... How's that?

FlamEnemY
11-23-2008, 11:28 AM
I'm not feeling tricky today, so I just used my graphing calulator. I get that the maximum occurs when x is approximately 2.3561957. At this value, f(x) is -2.

(I'm thinking that the question is asking what is the maximum value of f(x) = tan (x) + cot (x).)

Or this way. By taking f'(x) and setting it equal to zero. I get that x = pi/4, 3pi/4,... The only value in the domain is 3pi/4 which is 2.35619449... How's that?

Yes, this is f(x). I use tg instead of tan and cotg instead of cot. w/e.

Yes, -2 is the correct answer. But there is easier way to find it :)

----
for pi/2< x <pi tanx is < 0 for every x in this interval

a + 1/a >= 2 for every a>0

so a + 1/a <=-2 for every a<0

But f(x)=tanx + 1/tanx.

f(x) <= -2 :) max f(x)=-2

11-24-2008, 04:54 PM
Yes, this is f(x). I use tg instead of tan and cotg instead of cot. w/e.

Yes, -2 is the correct answer. But there is easier way to find it :)

----
for pi/2< x <pi tanx is < 0 for every x in this interval

a + 1/a >= 2 for every a>0
so a + 1/a <=-2 for every a<0

But f(x)=tanx + 1/tanx.

f(x) <= -2 :) max f(x)=-2
The part highlighted is true, but not immediately obvious. It asserts that 2 is the minium value for f(a) = a + a^(-1). That's true since f'(a) = 1 - a^(-2), so we'd need to solve 1 = a^(-2), so a = 1, hence f(1) = 2, so 2's the minimum. But if we go through all this, is it still an easier way? :confused:

11-24-2008, 05:11 PM
So I'm watching TV and they're solving 11 1/4 / 1 5/6. They do it the following way. 45/4 / 11/6.
45/4 * 6/11
45/2 * 3/11 = 135/22 = 6 3/22. And throughout this, the teacher kept stating, "There are no shortcuts, you have to do every step." Well...you don't.

One is better off being uninstructed I believe. What is being asked? Division is repeated subtraction, so it's asking "How many times can I subtract 1 5/6 from 11 1/4?" So, how many times can we subtract an amount a bit smaller than 2 from an amount just bigger than 11? Quickly, I'd say '5, maybe even 6 times'. (And in some real life circumstances, this is sufficient). Let's just say how many times can I subtract "2 minus 1/6" from "11 and 1/4"? I can see right away that if I increase the divisor 6 fold it becomes 12 minus 1, or 11. Take 11 away from 11 1/4, leaves 1/4. So the answer is 6 with remainder 1/4 over (2 - 1/6). (Weird fraction, let's fix it by timesing it by 6). Answer 6 (6/4) over 11, or 6 (3/2) over 11, finally 6 3/22. Same answer, but accomplished without any rules, just common sense. It might seem like more trouble, but that's because each step was explained. Also, this permitted you to see that 6 is very close to being correct right away. When you do it "by the steps" you might be way off and never know the difference.

I never use the procedures they drilled me on in school. There's always easier ways, and often I don't even need scratch paper to perform them. Learning the steps made it hard, not easier, to understand what I was trying to do.

fluffy Beaver
11-24-2008, 08:54 PM
Can anyone give me a set with cardinality

"aleph-null5" aside from just saying aleph-null#.

FlamEnemY
11-25-2008, 12:09 AM
The part highlighted is true, but not immediately obvious. It asserts that 2 is the minium value for f(a) = a + a^(-1). That's true since f'(a) = 1 - a^(-2), so we'd need to solve 1 = a^(-2), so a = 1, hence f(1) = 2, so 2's the minimum. But if we go through all this, is it still an easier way? :confused:

Every fraction + it's reciprocal is greater than 2, so the minimum value is for a=1 -> f(a)=2.

Like you said, some common sense and it's easy to solve ;)