# Math Help

Discussion in 'Odds & Ends' started by bamboo711, Aug 27, 2008.

1. ### bamboo711Rookie

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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

2. ### ScovilleJenkinsRookie

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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

3. ### DouggoSemi-Pro

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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...

4. ### gj011Banned

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Formula for sphere volume is:

V = 4/3pi r^3.

5. ### Steady EddyHall of Fame

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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

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7. ### Steady EddyHall of Fame

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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.

8. ### xtremerunnerarsHall of Fame

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I think he was just being funny.

9. ### YULitleHall of Fame

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

10. ### bamboo711Rookie

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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!

11. ### Steady EddyHall of Fame

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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.

12. ### Sean DuganRookie

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Take a pin and puncture the balloon. Ergo, Volume = 0.

13. ### bamboo711Rookie

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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!

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That is 42

15. ### Swissv2Hall of Fame

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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.

16. ### bamboo711Rookie

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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?

17. ### BundeyProfessional

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Go play tennis

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19. ### BundeyProfessional

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What math is this?

20. ### sapient007Semi-Pro

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use this logic:

A = annual balance
x = interest

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

21. ### bamboo711Rookie

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Wait so what? I multiple the year x annual balance (1 + interest) ????

An example maybe? Thanks!

22. ### bumfluffSemi-Pro

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You familiar with exponential growth?

23. ### bumfluffSemi-Pro

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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.

24. ### Steady EddyHall of Fame

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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.

25. ### bamboo711Rookie

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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)

26. ### meowmixHall of Fame

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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.

27. ### bamboo711Rookie

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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!

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that be 42.

29. ### Steady EddyHall of Fame

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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!

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)

30. ### bamboo711Rookie

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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.

31. ### Steady EddyHall of Fame

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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.

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)

32. ### Steady EddyHall of Fame

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Is that the answer you wanted? I didn't figure that out for nothing, did I?

33. ### bamboo711Rookie

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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

Last edited: Sep 15, 2008
34. ### Steady EddyHall of Fame

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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.

35. ### bamboo711Rookie

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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.

36. ### Steady EddyHall of Fame

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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.

37. ### bamboo711Rookie

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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.

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39. ### bamboo711Rookie

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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......

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Is this a high school or college course? What book are you using, out of curiosity? :shock:

41. ### Steady EddyHall of Fame

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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!

42. ### bamboo711Rookie

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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=_________

43. ### Steady EddyHall of Fame

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I don't understand, the author's name is Stewart Multivariable?

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

44. ### bamboo711Rookie

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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.

45. ### Steady EddyHall of Fame

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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!

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)

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Bamboo, please send me an e-mail

47. ### dak95_00Professional

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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.

48. ### MorpheusProfessional

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Correct, but what was the question again?

49. ### Steady EddyHall of Fame

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Humorous? Why? (I'm a retired H.S. math teacher)

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.

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