Calculating Volume of Right Circular Cylinder: Radii 3 to 6 cm

[charliemagne]

Homework Statement

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)

Are my answers correct?

thanks

 

[Mark44]

Homework Statement

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)

Are my answers correct?

thanks

Without working the problem for myself, I don't know. Show us how you got these answers. In this problem is only the radius changing, or are both radius and height changing?

 

[LCKurtz]

Homework Statement

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)

Are my answers correct?

thanks

That isn't the formula for the volume of a cylinder. It is the formula for the volume of a cone.

 

[charliemagne]

Without working the problem for myself, I don't know. Show us how you got these answers. In this problem is only the radius changing, or are both radius and height changing?

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

 

[LCKurtz]

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

Your formula for volume is V = (1/3)pi r²(h). There is no y or x in this equation so dy/dx makes no sense. You are talking about V as a funtion of r and apparently h is a constant. To get the average rate of change as r changes from 3 to 6 you want to calculate:

V_{ave} = \frac {V(6)-V(3)}{(6-3)}

That is what you actually started to calculate, but you mis-labeled, it isn't a derivative, and you need to check your arithmetic.

For part (b) you want dV/dr. Calculate it, but don't put the numbers in for r until your have the derivative simplified. I think you should get 24 pi h. No y or x anywhere in your answer.

 

[Mark44]

Solution:

dy/dx=( 1/3 π(〖6)〗^(2 ) h- 1/3 π(〖3)〗^2 h)/(6-3)

What are those things around the 6 and the 3?