Related Rates: The radius of a sphere is increasing at....

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katie9426

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I'm given the following: The radius of a sphere is increasing at a constant rate of 0.04cm/sec. [Note the formula for the volume of a sphere is v=(4/3)pi r^3]

I have to find:
a) at the time when the radius of the sphere is 10 cm, what is the rate of increase of its' volume?
At first I thought I was suppose to use the formula for instantaneous rates of change: lim as h-->0 of (Q(t-h) - Q(t))/h but I couldn't figure out how to make that work with the given information. So I took the derivative for the volume of a sphere and got v=4pi r^2, except I still am not sure how to end up with the rate of increasing volume?

Since I had trouble with a, I'm also having trouble with b and c!

b)At the time when the volume of the sphere is 36pi cm^3, what is the rate of increase of the area of a cross section through the center of the sphere?

c)At the time when the volume and the radius of the sphere are increasing at the same numerical rate, what is the radius?

Thank you for any help!
 
For part a:

\(\displaystyle \L\\V=\frac{4}{3}{\pi}r^{3}\)

\(\displaystyle \L\\\frac{dV}{dt}=4{\pi}r^{2}\frac{dr}{dt}\)

\(\displaystyle \L\\\frac{dV}{dt}=4{\pi}(10)^{2}(\frac{1}{25})\)
 
dV/dt isn't 2/50, dr/dt is. Actually, it should be 1/25. I remedied that.
 
Okay how is dr/dt (1/25)? So you took the derivative with respect to t? And after I solve that equation do i get the volume at time t?
 
I am sorry Katie, but if you can't see where 1/25 came from and don't know how to solve it when I have it all set up, you should see your professor. Good luck.
 
I'm given the following: The radius of a sphere is increasing at a constant rate of 0.04cm/sec. [Note the formula for the volume of a sphere is v=(4/3)pi r^3]
Click to expand...

Okay how is dr/dt (1/25)?
Click to expand...

it was given to you.

dr/dt = .04 = 4/100 = 1/25
 
All nevermind I see what you did. (.04 is dr/dt and you put it in a fraction). So I solved that out and got 16pi or approx. 50.265, right?

In part b would I set 36 cm^3 = (4/3)pi r^3 and solve for r? I did that and got 3/pi, but I'm not sure if thats correct.

Thanks for your help! Sorry to keep bothering you!
 
You're not bothering me. I wouldn't be here if I were bothered. You're on the right track. Except, the radius is not 3/Pi. Remember, the volume is 36Pi, not 36. Try again. Once you have the correct radius, use the formula for the area of a circle, differentiate w/r to t. That should give it to you.

You see, the cross-section of the sphere is a circle. BTW, you're correct for part a.
 
Okay thanks! I did that and got 9pi....i hope thats right!

For part C would you set volume and area equal to zero and solve?
 
No, 9Pi isn't correct. Sorry, Katie.

If we solve \(\displaystyle 36{\pi}=\frac{4}{3}{\pi}r^{3}\) for r, we get r=3

Now, use the area of a circle formula to find dA/dt.

\(\displaystyle \L\\A={\pi}r^{2}\)


For part c, no, you wouldn't use 0. Since dV/dt and dr/dt are the same, couldn't you divide through by dr/dt and get 1 on the left side and then solve for r?.

\(\displaystyle \L\\\frac{dV}{dt}=4{\pi}r^{2}\frac{dr}{dt}\)
 
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