Water is poured in a cone at 8cm/s. h = 12m,r = 6m.how fast is the water level rising when water is 4m deep?

Question

water is pouring into a conical cistern at the rate of 8cubic m per second. if the height of the cistern is 12m and the radius of its circular opening is 6m, how fast is the water level rising when the water is 4m deep?

Answer 1

Volume of a cone is (1/2) * pi * r^2 * h.

The radius is half the height, so at 4m of height the radius is 2m, making the volume at that point (16/3)*pi m^3. Add 8 m^3 of water and you have 24.75516082 m^3.
The height of the water has risen to 4+k, so the radius has expanded to 2+(1/2)k. This gives you the generic formula:
(1/3) * pi * (2+.5k)^2 * (4+k) = 24.75516082 or

k^3 + 6k^2 + 12k - 8.75516082 = 0. Now we need only solve for k.

k = 2/3 --> 2.207802143
k = 1/2 --> -1.13016082

We will extrapolate a solution as if the function were linear. The slope is (2.207802143+1.13016082)/(2/3-1/2) = 20.027777778 and point-slope gives us
(y-2.207802143)=20.027777778(x-2/3) --> y = 20.02777778x - 11.14404971

if y = 0, then x = .556429666. If we plug that value into the formula with the k's above, we are off by only .048, which is pretty close.

Thus, the rate of the change in the water level is about .5565 m/s.

A more exact solution could be found by using Newton's Method on k^3 + 6k^2 + 12k - 8.75516082 = 0.

Answer 2

.6366 m/s

Vcone = π/3*r(t)^2*h(t) where r(t) is the instantaneous radius and h(t) = the instantaneous height

r/6 = h/12 or r= h/2, so
Vcone = (π/3)*(h/2)^2*h = (π/12)h^3 now
dV/dt = (π/4)h^2(dh/dt) and dV/dt = 8cm/sec

So, dh/dt = 32/πh^2
When h = 4, dh/dt = 2/π = .6366 m/s