1. ## rate of change

here is the problem:
Use the formula V = (1/3)pi r2(h) for the volume of a right circular cylinder to find
a. the average rate at which the volume of a right circular cylinder changes with the radius r as r increases from 3 cm to 6 cm.
b. the instantaneous rate at which the volume of the right circular cylinder changes with r when r = 6 cm.

I tried to solve it at home and I arrived at the answers:

a. 3(pi)(h)

b. 4(pi)(h)

thanks

2. Originally Posted by mamen

here is the problem:
Use the formula V = (1/3)pi r2(h) for the volume of a right circular cylinder to find
a. the average rate at which the volume of a right circular cylinder changes with the radius r as r increases from 3 cm to 6 cm.
b. the instantaneous rate at which the volume of the right circular cylinder changes with r when r = 6 cm.

I tried to solve it at home and I arrived at the answers:

a. 3(pi)(h)

b. 4(pi)(h)

thanks
For 1. $\frac{V(6)-V(3)}{6-3}$

If this is what you did, your answer is right. Unless you made a mistake...You can see that the Average rate of change is nothing more than the slope of the secant line between the two points.

For 2. $V'(6)$

If you did this...

Keep in mind that Instantaneous implies derivative.

3. Originally Posted by VonNemo19
For 1. $\frac{V(6)-V(3)}{6-3}$

If this is what you did, your answer is right. Unless you made a mistake...You can see that the Average rate of change is nothing more than the slope of the secant line between the two points.

For 2. $V'(6)$

If you did this...

Keep in mind that Instantaneous implies derivative.
Here is my solution.
dy/dx=( 1/3 π(〖6)〗^(2 ) h- 1/3 π(〖3)〗^2 h)/(6-3)

dy/dx=( 1/3 π36h- 1/3 π9h)/2

dy/dx=( 1/3 π36h- 1/3 π9h)/2
dy/dx=(12πh-3πh)/3
dy/dx=9πh/3
=3πh
Solution:
dy/dx=2/3 πrh
At r = 6:
dy/dx=2/3 π6h
=4πh

so is my solution and answer correct?