A water trough is 10 m long and has a cross-section in the shape of an isosceles trapezoid that is 40 cm wide at the bottom, 100 cm wide at the top, and has height 60 cm. If the trough is being filled with water at the rate of 0.1 m3/min, how fast (in m/min) is the water level rising when the water is 10 cm deep

0.238 m/min

Explanation:

The volume of water in the trough V =Ah' where A = area of cross-section = area of isosceles trapezoid = 1/2(a + b)h where a = length of bottom of isosceles trapezoid = 40 cm = 0.4 m, b =  length of top of isosceles trapezoid = 100 cm = 1 m and h = height of isosceles trapezoid = 60 cm = 0.6 m. So,

A = 1/2(a + b)h = 1/2(0.4 m + 1 m)0.6 m = (1.4 m)0.3 m = 0.42 m² and h' = height of water level in trough = H - h" where H = length of trough = 10 m and h" = depth of water level in trough = 10 cm = 0.1 m

So, V = Ah'

V = A(H - h") = A(10 - h")

Now, the rate of change of volume of the trough with respect to time dV/dt = d[A(10 - h")]/dt

dV/dt = -Adh"/dt

dh"/dt = -dV/dt/A

Since dV/dt = 0.1 m³/min, substituting the other variables into the equation, we have

dh"/dt = -dV/dt/A

dh"/dt = -0.1 m³/min/0.42 m²

dh"/dt = -0.238 m/min

This is the rate at which the depth is decreasing

Since the height h' = 10 - h"

dh'/dt = d(10 - h")/dt

= -dh"/dt

= -(-0.238 m/min)

= 0.238 m/min

So the water level is increasing at a rate of 0.238 m/min

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