A water trough is 10 m long and a cross-section has the shape of an isosceles trapezoid that is 30 cm wide at the bottom, 80 cm wide at the top, and has height 50 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 30 cm deep?

Solution:

Given, the water trough is 10m long.

Cross-section of the trough is an isosceles trapezoid.

Bottom width,a = 30cm = 0.3 m

Top width, b = 80cm = 0.8 m

Height, h = 50cm = 0.5 m

We have to find how fast the water level is rising when the water is 30cm deep.

Volume of the trapezoid, V = (1/2)(a + b)hL

V = (1/2)(0.3 + 0.8)(0.5)(10)

= (0.5)(1.1)(5)

= 2.75 cubic m.

We know that

dV/dt = dA/dt × l

dA/dt = dA/dt × dh/dt = (h + 0.3) × dh/dt

So we get

dV/dt = (h + 0.3)l × dh/dt

Substituting the values

0.2 = (0.5 + 0.3) × dh/dt

dh/dt = 0.2/0.8

dh/dt = 0.25 m/min

Therefore, the water level is rising at 0.25 m/min.

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A water trough is 10 m long and a cross-section has the shape of an isosceles trapezoid that is 30 cm wide at the bottom, 80 cm wide at the top, and has height 50 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 30 cm deep?

Summary:

A water trough is 10 m long and a cross-section has the shape of an isosceles trapezoid that is 30 cm wide at the bottom, 80 cm wide at the top, and has height 50 cm. If the trough is being filled with water at the rate of 0.2 m³/min, the water level rising when the water is 30 cm deep is 0.25 m/min.