Thread: An inverted cone has a diameter of 42 in and a height of 15in. If the water flowing o

An inverted cone has a diameter of 42 in and a height of 15in. If the water flowing out of the vertex of the container at a rate of 35 /sec , how fast is the depth of the water dropping when the height is 5in?

Originally Posted by yoman360

An inverted cone has a diameter of 42 in and a height of 15in. If the water flowing out of the vertex of the container at a rate of 35 /sec , how fast is the depth of the water dropping when the height is 5in?

Volume of the cone V = 1/3*π*r^2*h.....(1)
You want to find dh/dt. So you have to find r in terms of h.
r and h are proportional tp R and H, where R = 21" and H = 15".
So R/H = r/h.
So r = (R/H)*h.
Substitute this value in the eq.(1) and find dV/dt.
dV/dt is given. Find dh/dt.

Originally Posted by sa-ri-ga-ma

Volume of the cone V = 1/3*π*r^2*h.....(1)
You want to find dh/dt. So you have to find r in terms of h.
r and h are proportional tp R and H, where R = 21" and H = 15".
So R/H = r/h.
So r = (R/H)*h.
Substitute this value in the eq.(1) and find dV/dt.
dV/dt is given. Find dh/dt.

I followed your steps and this is what I get:




simplify so



plug in h and dv/dt



solve for dh/dt i get



the problem is that the answer key says the answer is in/sec

what did i do incorrectly?

Originally Posted by yoman360

I followed your steps and this is what I get:

plug in h

Never mind. h=5 and H= 15
I figured it out I was suppose to plug in 5 here instead if 15 then solving for dh/dt i got 5/7 in/sec

Thanks for the help