An inverted cone has a diameter of 42 in and a height of 15in. If the water flowing out of the container is 35pi in^3/sec , how fast is the depth of the water dropping when the height is 5in?

Answer 1

let the height of water be h in, let the radius of the water surface be r in
By similar triangles, r/h = 21/15 = 7/5
so r = 7h/5

V = (1/3)pi(r^2)h
= (1/3)pi(49h^2/25)(h
= (49/75)pi(h^3)
dV/dt = (49/25)pi(h^2)dh/dt
-35pi = (49/25)pi(25)dh/dt

solve for dh/dt

Answer 2

-5/7