an inverted cone is being drained of water at a rate of 30 cm^3 / min . it has a height that is 3 times its radius. how fast is the radius changing when the height of the water is 5 cm?

Question

I’m not sure if the height zeroes out because it’s not changing. Thank you

Answer 1

First things first we need to create a function for volume. The volume function for cones is normally:

V=(1/3)*H*R²*pi

However in this problem H=3*R so we must adjust the formula:

V= (1/3)*3*R*R²*pi = pi*R³

Now we need to derive this function, but with instead of deriving it based on R, like we normally would, we will derive in terms of T. While this sounds complicated all it does to our problem is gives us a variable that represents our change in radius.

So instead of:

dV= 3*pi*R²

We get:

dV= 3*pi*R²*dR

We can then rearrange the problem so we can get dR (the rate of change of the radius):

dR= dV/(3*pi*R²)

Finally we plug in our variable

dR= 30/(3*3.14*(5/3)²)=1.15