To determine

To find: The rate of change of height of the water level inside the trough when the water is 6 inches deep.

Answer

The rate of change of the height of the water level is dhdt=45  ft/min.

Explanation

Given:

Let the trough is being filled with water at a rate of 12  ft3/min.

That is, dVdt=12  ft3/min.

The length of the trough is 10 ft and its ends have the shape of isosceles triangles that are 3 ft across at the top and have a height of 1 ft.

Formula used:

(1) Chain rule: dydx=dydu⋅dudx

(2) Volume of the trough of height h and width b and length 10​​ ft is V=12h×b×(10)​​  ft3.

Calculation:

Let V be the volume of the water inside the trough and h be the height of the water inside the trough and b be the width of the water inside the trough as shown in the figure-1 given below.

Since the volume and height of the water inside the trough are changes with time t,  the volume and height are the function of the time t.

Obtain dhdt when water is 6 inches deep.

Obtain h in terms of b.

Apply the properties of similar triangles in the above Figure 1,

bh=31

Thus, b=3h.

Substitute b=3h in V.

Then, the volume of the trough is,

V=12h×(3h)(10)=15h2​​ft2

Differentiate V with respect to the time t,

ddt[V]=ddt[15h2]dVdt=15(2h)dhdt                                                                [Qdydx=dydu⋅dudx]dhdt=130h⋅dVdt

Substitutes h=6​​  inches =12  ft and dVdt=12 in dhdt,

dhdt=130×12(12)=2×1230=2430=45  ft/min

Therefore, the rate of change of the height of the water level is dhdt=45  ft/min.