Friday

August 26, 2016
Posted by **Mishaka** on Tuesday, January 17, 2012 at 7:37pm.

A. At the time when the radius of the sphere is 12 meters, what is the rate of increase in its volume?

B. At the time when the volume of the sphere is 36pi cubic meters, what is the rate of increase in its surface area?

C. Express the rate at which the volume of the sphere changes with respect to the surface area of the sphere (as a function of r).

Parts A and B I have figured out (A. = 28.8pi cubic meters per second; B. = 1.2pi squared meters per second) But I can't quite figure out C. How do I relate surface area and volume together into one equation???

- Calculus (Parts A and B done, just help with C) -
**Damon**, Tuesday, January 17, 2012 at 7:53pmV = (4/3)pi r^3

dV/dr = 4 pi r^2 which happens to be the area

so

dV/dt = 4 pi r^2 dr/dt

the differential change of volume is the surface area * dr

PS

I wonder how you did parts a and b without knowing that - Calculus (Parts A and B done, just help with C) -
**Mishaka**, Wednesday, January 18, 2012 at 3:32pmI did notice this when I took the derivative for part A of the volume. But it seemed too simple to just plug in this value, I wanted to make sure that I was certain about the placement. Thank you!!!

- Calculus (Parts A and B done, just help with C) -
**:)**, Wednesday, August 12, 2015 at 2:16pmdr/dt = .05

v = (4/3)pi r^3

dv/dt = (4/3)pi(3) r ^2 (dr/dt) <---- this step is where the mistake was. you needed to have differentiated the volume with respect to time. to do this use the power rule on the right side. the derivative of (4/3)pi r^3 is (4/3)pi(3) r ^2 (dr/dt). use the power rule to get this and multiply by radius differentiated with respect to time. (quick power rule for r^3: bring the 3 to the front and diminish the 3 by 1, giving you 3r^2)

then plug in dr/dt, or r(t) as what you wrote and get (4/3) pi (3)(12)^2 (.05) as your answer

= 28.8 pi

---------------------------------------...

B.

A = 4 pi r^2

V = 4/3 pi r^3

dr/dt = .05

you need to find the change in surface area (i used "A" as the variable) at the moment when the volume is 36 pi. The only other given number you have is dr/dt= .05

so you have the right volume but you need to find what the radius was when the volume equaled 36 pi.

you plug in 36 pi into the equation for volume where "V" is (to solve for "r"):

36 pi = (4/3) pi r^3

divide pi from each side.. 36= (4/3)r^3

multiply each side by (3/4) to get rid of the (4/3).... 27 = r^3

find the cube root of 27 and get.... r = 3

now you know what the radius is equal to at the moment when the volume is 36 pi.

differentiate the equation for surface area now with respect to time: A = 4 pi r^2

dA/dt = 4 pi 2 r (dr/dt)

plug in your value for r = 3 and (dr/dt)= .05

dA/dt = 4 pi 2 r (dr/dt)

dA/dt = 4 pi 2 (3)(.05)

= 1.2 pi - Calculus (Parts A and B done, just help with C) -
**:)**, Wednesday, August 12, 2015 at 2:35pmGiven:

dr/dt = 0.05 m/s

V = 4/3 πr³

dV/dt = 4πr² (dr/dt)

A = 4πr²

dA/dt = 8πr (dr/dt)

- - - - - - - - - - - - - - -

First question:

What is dV/dt when r = 12?

dV/dt = 4π(12 m)² (0.05 m/s)

dV/dt = 28.8π m³/s

dV/dt ≈ 90.478 m³/s

- - - - - - - - - - - - - - -

Second question:

When V = 36π, what is dA/dt?

Using volume formula,

36π = 4/3 πr³

27 = r³

r = 3

dA/dt = 8π(3m) (0.05 m/s)

dA/dt = 1.2π m²/s

dA/dt ≈ 3.77

- - - - - - - - - - - - - - -

Third question:

Find dV/dt in terms of A(r) (and I'm assuming only in terms of A(r)).

A(r) = 4πr²

dV/dt = 4πr² (dr/dt)

Since you're given dr/dt = 0.05, plug it in:

dV/dt = 4πr² (0.05)

Also, since A(r) = 4πr², just replace this term in dV/dt with A(r):

dV/dt = 0.05 A(r) - Calculus (Parts A and B done, just help with C) -
**ps**, Saturday, May 21, 2016 at 5:03amThe volume of sphere is increasing at the rate of 1200c.cm/sec. the rate increase in its surface area when the radius is 10cm is