Related Rates- CALC 123

Question

A water trough is 5 m long and has a cross-section in the shape of an isosceles trapezoid that is 30 cm wide at the bottom, 90 cm wide at the top, and has height 60 cm. If the trough is being filled with water at the rate of 0.2 m³/min how fast is the water level rising when the water is 50 cm deep?

Answer 1

dV/dt = 0.2 m³/min = 200000 cm³/min

Length = 5m = 500cm

Let h = height of water in cm

Formulate the volume of water as it relates to water height. Looking at the cross section of the trough, there is a rectangular component (first term) and two triangular components (2nd term). For the 3^rd dimension, the depth is a constant 500cm :

V = 30*h*500 + (2*(1/2)h) *((1/2)h)*500

V = 15000h + 250h²

Find rate of change of water volume wrt h:

dV/dh = 500h + 15000

Now we need to work the rate of change of h (wrt time) into this. Using the chain rule we can relate dV/dt to dh/dt:

dV/dt = (dV/dh) • (dh/dt)

Solving for dh/dt:

dh/dt = (dV/dt) / (dV/dh)

Plug in known values:

dh/dt = 200000 / (500h + 15000)

Plug in h = 50cm:

dh/dt = 200000 / (500(50) + 15000) = 5 cm/min