question a trough has ends shaped like isosceles triangles, and has a length of 9m, width 3m and height 5m…

question a trough has ends shaped like isosceles triangles, and has a length of 9m, width 3m and height 5m, as shown in the image below. water is being pumped into the trough at a rate of 5m³/min. at what rate does the height of the water change when the water is 3m deep? submit an exact answer. note that the volume of a trough with length l, width w, and height h is given by v = 1/2·l·w·h. provide your answer below dh/dt = □ m/min feedback more instruction submit

question a trough has ends shaped like isosceles triangles, and has a length of 9m, width 3m and height 5m, as shown in the image below. water is being pumped into the trough at a rate of 5m³/min. at what rate does the height of the water change when the water is 3m deep? submit an exact answer. note that the volume of a trough with length l, width w, and height h is given by v = 1/2·l·w·h. provide your answer below dh/dt = □ m/min feedback more instruction submit

Answer

Explanation:

Step1: Relate width and height of water

Since the cross - section is an isosceles triangle, the ratio of width to height of the whole trough is $\frac{3}{5}$. For the water in the trough, if the height of water is $h$ and width is $w$, then $w=\frac{3}{5}h$. The length of the trough $l = 9$. The volume of water $V=\frac{1}{2}lwh=\frac{1}{2}\times9\times\frac{3}{5}h\times h=\frac{27}{10}h^{2}$.

Step2: Differentiate volume with respect to time

Differentiate $V=\frac{27}{10}h^{2}$ with respect to time $t$ using the chain - rule. $\frac{dV}{dt}=\frac{27}{10}\times2h\frac{dh}{dt}=\frac{27}{5}h\frac{dh}{dt}$.

Step3: Substitute known values

We know that $\frac{dV}{dt}=5$ and $h = 3$. Substitute these values into the equation $\frac{dV}{dt}=\frac{27}{5}h\frac{dh}{dt}$. So, $5=\frac{27}{5}\times3\times\frac{dh}{dt}$.

Step4: Solve for $\frac{dh}{dt}$

First, simplify the right - hand side: $\frac{27}{5}\times3\times\frac{dh}{dt}=\frac{81}{5}\frac{dh}{dt}$. Then, solve for $\frac{dh}{dt}$: $\frac{dh}{dt}=\frac{25}{81}$ m/min.

Answer:

$\frac{25}{81}$