A water trough is 8 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 is the water level rising when the water is 30 cm deep?

1.785 cm / min

Step-by-step explanation:

We have to the following equation:

V = A * L

The volume is equal to the area times the length, now, if we derive with respect to time we have:

dV / dt = dA / dt * L

Now, the area of a trapezoid is:

A = (Upper base + Lower base) / 2 * height

Now we have:

Upper base = 40 + h

Lower base = 40

Replacing we have:

A = (40 + h + 40) / 2 * h

A = (h ^ 2) / 2 + 40 * h

Now, we derive and we are left:

dA / dt = d [(h ^ 2) / 2 + 40 * h] / dh * dh / dt

dA / dt = (h + 40) * dh / dt

we replace dA / dt:

dV / dt = [(h + 40) * dh / dt] * L

We have that dV / dt is equal to 0.1 m ^ 3 / min, we know that 1m ^ 3 is equal to 10 ^ 6 cm ^ 3, therefore:

dV / dt = 10 ^ 5 cm ^ 3 / min

in addition L is equal to 8 m or what is equal to 800 cm, and also h at the required moment is 30 cm

we replace everything and we have:

10 ^ 5 = 800 * (30 +40) * dh / dt

dh / dt = 10 ^ 5/56000

dh / dt = 1.785 cm / min

step-by-step explanation: