Question

a water trough is 10 m long and has a cross - section in the shape of an isosceles trapezoid that is 20 cm wide at the bottom, 60 cm wide at the top, and has height 40 cm. if the trough is being filled with water at the rate of 0.3 m³/min, how fast (in m/min) is the water level rising when the water is 10 cm deep?

1. -/1 points

details my notes ask your teacher practice another
gravel is being dumped from a conveyor belt at a rate of 35 ft³/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. how fast (in ft/min) is the height of the pile increasing when the pile is 12 ft high? (round your answer to two decimal places.)

Explanation:

Step1: Find the cross - sectional area formula of the water in the trough

Let the water level be $h$ (in meters). The cross - section of the water is also a trapezoid. The length of the bottom of the water - filled trapezoid is $b_1 = 0.2$ m, and the length of the top of the water - filled trapezoid is $b_2=0.2 + \frac{0.6 - 0.2}{0.4}h=0.2 + h$ (using similar triangles). The length of the trough is $L = 10$ m. The volume of water $V$ in the trough is $V=A\times L$, where the cross - sectional area $A$ of the water is $A=\frac{(b_1 + b_2)h}{2}=\frac{(0.2+(0.2 + h))h}{2}=\frac{(0.4 + h)h}{2}=0.2h+0.5h^{2}$. So $V = L\times A=10\times(0.2h + 0.5h^{2})=2h+5h^{2}$.

Step2: Differentiate the volume equation with respect to time $t$

Differentiate $V = 2h+5h^{2}$ with respect to $t$ using the chain rule. $\frac{dV}{dt}=(2 + 10h)\frac{dh}{dt}$.

Step3: Substitute the given values to find $\frac{dh}{dt}$

We know that $\frac{dV}{dt}=0.3$ m³/min and $h = 0.1$ m. Substitute these values into the equation $\frac{dV}{dt}=(2 + 10h)\frac{dh}{dt}$. So $0.3=(2+10\times0.1)\frac{dh}{dt}$. First, simplify the right - hand side: $2 + 10\times0.1=2 + 1=3$. Then we have $0.3 = 3\frac{dh}{dt}$. Solve for $\frac{dh}{dt}$: $\frac{dh}{dt}=\frac{0.3}{3}=0.1$ m/min.

Answer:

$0.1$