Related Rates (Calculus)?

Question

A swimming pool is 12 meters long, 6 meters wide, 1 meter deep at the shallow end, and 3 meters deep at the deep end. Water is being pumped into the pool at 1/4 cubic meter per minute, and there is 1 meter of water at the deep end.

(a) What percent of the pool is filled?
(b) At what rate is the water rate rising?

Answer 1

Let h be the depth of water at the deep end, in metres. When h is between 0 and 2, the water will extend to a distance of (6h) metres along the pool. (When h = 2, the water is 1m below the surface and has just reached the shallow end which is at this depth.)

So the cross-section of the water in the pool is a triangle with base (6h) m and height h m, and thus area (1/2) (6h) (h) = 3h^2 m^2. So the volume of water is 6 (3h^2) = 18h^2 cubic metres.

The total volume of the pool is the volume of this section when it is full (h=2: V = 18 (2^2) = 72 m^3) plus the top metre, which is a rectangular prism and has V = 12 (6) (1) = 72 m^3. So the total volume is 144 m^3.

a) At h = 1, V = 18 m^3, so the precentage filled is 18 / 144 * 100 = 12.5%.

b) V(h) = 18h^2, so dV/dh = 36h. Now dV/dt = dV/dh . dh/dt, and we're given dV/dt = 1/4 m^3/min, so we get
1/4 = 36h dh/dt
so at h = 1, dh/dt = 1/144 m/min = 6.94 mm/min.

Answer 2

(a)
1 meter in the depend means its one meter high 6 meters wide and it reached half way so 6 meters long
its just a triangular prism:
1x6x6/2 = 18cubic meters

total pool volume is (1+3)/2 x 6 x12 = 144cubic meters
18/144 = 12.5%

so the pool starts at 12.5% full

(b)dont understand really