#### To determine

**To find:** The rate of change of height of the water level inside the trough when the water is 30 cm deep.

#### Answer

The rate of change of the height of the water level is dhdt=130 m/min.

#### Explanation

**Given:**

Let the trough is being filled with water at a rate be dVdt=0.2 m3/min.

The length of the trough is 10 m and the cross has the shape of an isosceles trapezoidal, which is 30 wide at bottom and 80 cm wide at top.

**Formula used:**

(1) Chain rule: dydx=dydu⋅dudx

(2) The area of the trapezoidal of base b1 and top b2 and height l is. A=12[b1+b2]×height.

**Calculation:**

Let V be the volume of the water inside the trough and h be the height of the water inside the trough and a be the distance between the rectangle and the right slant line at the surface of the water as shown in the figure-1 given below.

Find the area of the trapezoidal trough.

A=12[0.3+(0.3+2a)]h [Q b1=0.3 and b2=0.3+2a ]

Therefore, volume the trough with length 10 m is, V=10A.

Substitute A=12[0.3+(0.3+2a)]h in V.

V=10×12[0.3+(0.3+2a)]h=5[0.3+(0.3+2a)]h

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 30 cm deep.

First find h in terms of a.

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

ah=0.250.5a=12h

Substitutes a=12h in V,

V=5[0.3+(0.3+2(12h))]h=5[0.6+h]h=3h+5h2

Differentiate V with respect to the time t,

ddt[V]=ddt[3h+5h2]dVdt=3dhdt+10hdhdtdVdt=(3+10h)dhdtdhdt=1(3+10h)⋅dVdt

Substitutes h=30 cm =0.3 m and dVdt=0.2 in dhdt,

dhdt=1(3+10×0.3)(0.2)=1(3+3)(0.2)=0.26=130 m/min

Therefore, the rate of change of the height of the water level is dhdt=130 m/min.