Question

12 mark for review a trough is 9 feet long, and its cross section is in the shape of an isosceles right triangle with hypotenuse 2 feet, as shown above. water begins flowing into the empty trough at the rate of 2 cubic feet per minute. at what rate is the height h feet of the water in the trough changing 2 minutes after the water begins to flow? a decreasing at $\frac{2}{3}$ foot per minute b increasing at $\frac{2}{3}$ foot per minute c decreasing at $\frac{1}{6}$ foot per minute d increasing at $\frac{1}{6}$ foot per minute

Explanation:

Step1: Find the volume formula

For an isosceles - right - triangle cross - section of the trough, if the height of the water is $h$, the base of the water - filled triangle cross - section is also $h$ (since it's an isosceles right triangle). The length of the trough is $L = 9$ ft. The volume of water $V$ in the trough is $V=\text{Area of cross - section}\times\text{length}$. The area of the triangular cross - section of water $A=\frac{1}{2}h^{2}$, so $V=\frac{1}{2}h^{2}\times9=\frac{9}{2}h^{2}$.

Step2: Differentiate with respect to time

Differentiate both sides of the equation $V = \frac{9}{2}h^{2}$ with respect to time $t$ using the chain - rule. $\frac{dV}{dt}=9h\frac{dh}{dt}$.

Step3: Find the volume of water after 2 minutes

The rate of water flow is $\frac{dV}{dt}=2$ cubic feet per minute. After $t = 2$ minutes, $V=2\times2 = 4$ cubic feet.

Step4: Find the height of water when $V = 4$

Since $V=\frac{9}{2}h^{2}$ and $V = 4$, we have $\frac{9}{2}h^{2}=4$, then $h^{2}=\frac{8}{9}$, and $h=\frac{2\sqrt{2}}{3}$ ft.

Step5: Solve for $\frac{dh}{dt}$

We know that $\frac{dV}{dt}=2$ and $h=\frac{2\sqrt{2}}{3}$, and $\frac{dV}{dt}=9h\frac{dh}{dt}$. Substitute the values into the equation: $2=9\times\frac{2\sqrt{2}}{3}\times\frac{dh}{dt}$. First, simplify the right - hand side: $9\times\frac{2\sqrt{2}}{3}=6\sqrt{2}$. So, $2 = 6\sqrt{2}\frac{dh}{dt}$, and $\frac{dh}{dt}=\frac{1}{3\sqrt{2}}=\frac{\sqrt{2}}{6}\approx\frac{1.414}{6}\approx\frac{1}{6}$ ft/min. Since $\frac{dh}{dt}>0$, the height of the water is increasing.

Answer:

D. Increasing at $\frac{1}{6}$ foot per minute