Water is being poured into a conical tank that has a height of 4.5m and an upper radius of 2.7m. If the height of the water is increasing at a rate of 11.5cm/s at the moment when the water's height is 1.24m, at what rate, in m^3/min, is the water being added.

Question

I understand that V = π*r^2*h/3, just don't know how to set up the derivative's for a question like this and solve.

Answer 1

Two things:
- we would want to get rid of one of the variables on the right side
- since we are given dh/dt, and we are looking for dV/dt, we have to differentiate with respect to t
Given: dh/dt = 11.5 m/s when h = 1.24
Find: dV/dt at that moment.

At a given time of t seconds, let the height of the water level be h,
let the radius of the water level be r
h/r = 4.5/2.7
h = (5/3)r or r = (3/5)h
then V = (1/3)π r^2 h = (1/3)π (9/25)h^2 (h) = (9/75)π h^3

now in dV/dt = (9/25)π h^2 dh/dt
put in our given:
dV/dt = (9/25)π(1.24)^2 (11.5) m^3/s = appr 20 m^3/s

(I could have replaced the h and have my V = ... in terms of r, but that would
require extra steps to find dr/dt for that moment)

Answer 2

Good work, but dh/dt is cm/s, not m/s.
or, using the product rule,
v = π/3 r^2 h
dv/dt = π/3 (2rh dr/dt + r^2 dh/dt)
r = 3/5 h, so dr/dt = 3/5 dh/dt
3/5 * 1.24 = 0.744
dv/dt = π/3 (2*0.744*1.24*(3/5 * 11.5/100) + 0.744^2 * 11.5/100) = 0.199983
or, 0.20 m^3/s

Answer 3

thanks oobleck, i ended up doing dV/dt = 3pi/25 *3h^2 * dh/dt -> 3pi/25 * 3(1.24)^2 * (0.115m/s) = 0.20m^3/s. I like ur way of writing out ``dv/dt = π/3 (2rh dr/dt + r^2 dh/dt)``. Thanks for your input.

Answer 4

Another way of looking at it is to first order
dh/dt = dV/dt / surface area