Question

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, 90 cm wide at the top, and has height 50 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?

Answer 1

Let's start by finding the volume of water in the trough as a function of its depth.

At a depth of h cm, the cross-sectional area of the water is an isosceles trapezoid with bases of length b1 = 40 + (h/5) cm and b2 = 90 + (h/2) cm, and height 50 cm. The average width of the trapezoid is (b1 + b2)/2 = 65 + (3h/10) cm. Therefore, the volume of water in the trough at this depth is:

V(h) = 10 m x [(65 + (3h/10)) / 100 cm] x h cm

Simplifying this expression, we get:

V(h) = (13/2000) h^2 m^3

Now, we can use the chain rule to find the rate of change of V with respect to time t, given that dV/dt = 0.1 m^3/min:

dV/dt = (dV/dh) x (dh/dt) = (26/2000) h (dh/dt)

At the moment when the water is 10 cm deep, we have:

h = 10 cm

dV/dt = 0.1 m^3/min

Plugging in these values, we get:

0.1 m^3/min = (26/2000) x 10 cm x (dh/dt)

Solving for dh/dt, we get:

dh/dt = 0.1 m^3/min / (26/2000) / 10 cm

dh/dt ≈ 0.192 m/min

Therefore, the water level is rising at a rate of approximately 0.192 m/min when the water is 10 cm deep.