Related Rates - Volume and Height

[NOVAMath]

Homework Statement

"A man pours water into a conical glass (with radius 9cm and height 6cm) at a rate of 8cm^3.

At what rate is the height of the water changing per second when the height of the glass is 2/3rd's full of water?

NOTE: The answer is supposedly 2/9(pi) cm/sec. I am supposed to show that this is right.

Homework Equations

V= 1/3*(pi)*r^2*h

h = (3V)/(2*(pi)^2)

The Attempt at a Solution

I tried to differentiate the h equation that I found above. Then, after differentiating, I subbed in the radius (6) and the Volume (48). Note that I found the radius to be 6, and volume to be 48 when the height is 2/3rds (and thus, 4). However, after I diff and then plug, I always come up with a negative answer.

Thanks for any help.

 

[NateTG]

Homework Equations

V= 1/3*(pi)*r^2*h

h = (3V)/(2*(pi)^2)

I really don't see how you came up with the second equation, unless it's the result of making mistakes while trying to solve the first one for h. (It should be clear to you that the volume should be proportional to the cube of the height -- and not linear with it.)

In order to solve this problem, you will need to use a relationship between the radius of the cone and the height of the cone.

 

[NOVAMath]

Sorry, that second equation should be:

3V/((pi)(r^2))

What two equations dealing with the radius and height should I be comparing?

 

[Sourabh N]

V is proportional to h*r^2 and r and h are proportional to each other (since it is a cone), so essentially V is proportional to h^3.

 

[rocomath]

Use similar triangles to get rid of radius.

\frac r h = \frac 9 6

Solve for radius and plug it back into your original equation.