13-14 Water flowing into trapezoidal trough

Problem 13
A trapezoidal trough is 10 ft long, 4 ft wide at the top, 2 ft wide at the bottom and 2 ft deep. If water flows in at 10 ft³/min, find how fast the surface is rising, when the water is 6 in deep.

Solution 13

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Volume of water:
$V = \frac{1}{2}[ \, 2 + (2 + 2x) \, ]y (10)$

$V = 20y + 10xy$

From the figure:
$\dfrac{x}{y} = \dfrac{1}{2}$

$x = \frac{1}{2}y$

$V = 20y + 5y^2$

$\dfrac{dV}{dt} = 20\dfrac{dy}{dt} + 10y \dfrac{dy}{dt}$

when y = 6 in or 0.5 ft
$10 = 20\dfrac{dy}{dt} + 10(0.5) \dfrac{dy}{dt}$

$\dfrac{dy}{dt} = 0.4 \, \text{ ft/min}$ answer

Problem 14
For the trough in Problem 13, how fast the water surface is rising when the water is 1 foot deep.

Solution 14

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From the Solution 13
$\dfrac{dV}{dt} = 20\dfrac{dy}{dt} + 10y \dfrac{dy}{dt}$

When y = 1 ft
$\dfrac{dV}{dt} = 20\dfrac{dy}{dt} + 10y \dfrac{dy}{dt}$

$10 = 20\dfrac{dy}{dt} + 10(1) \dfrac{dy}{dt}$

$\dfrac{dy}{dt} = \frac{1}{3} \, \text{ ft/min}$ answer

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