Question

a conical container has a radius of 10 fts and height of 15 ft. soda is poured into the container at the rate of 20TT (pi) cubic ft by per min. at what rate is the height if the soda increasing when the height of the soda is 4 ft

Answer 1

what have you done so far?

Answer 2

I drew the picture of the question and wrote the formula. I do not know what to do from there

Answer 3

Have you defined your variables?

Answer 4

I do not know what to do

Answer 5

Let h be the height of the liquid and r be the radius at that height.
Volume of the liquid V = 1/3 * pi * r^2 * h (1)
We are interested in the rate of increase of h.
So keep h and eliminate r in the volume equation.
We need to find r in terms of h.

The triangles OCD and OAB are similar.
Therefore, r/h = 10/15 = 2/3
r = 2/3 * h (2)
Substitute r = 2/3 * h in equation (1)
V = 1/3 * pi * 4/9 * h^2 * h
V = 4/27 * pi * h^3

Take the derivative with respect to time t:
dV/dt = 4/27 * pi * 3 * h^2 * dh/dt

dV/dt = 20 * pi cubic feet / minute

20 * pi = 4/27 * pi * 3 * h^2 * dh/dt

Simplifies to: dh/dt = 45 / h^2
When h = 4 feet, dh/dt = 45 / 4^2 = 2.8125

dh/dt = 2.8125 feet / minute

Answer 6

thank you