   Chapter 2.8, Problem 27E

Chapter
Section
Textbook Problem

# A water trough is 10 m long and a cross-section has the shape of an isosceles trapezoid that is 30 cm wide at the bottom, 80 cm wide at the top, and has height 50 cm. If the trough is being filled with water at the rate of 0.2 m 3 / min , how fast is the water level rising when the water is 30 cm deep?

To determine

To find:

How fast is the water level rising when water is 30 cm deep?

Explanation

1. Concept:

We can plot the figure from the given situation

Here we have to find the rate at which the height of the water is rising in the trough.

Then we can consider the trapezoid and divide it into rectangles and triangles to write the formula of volume of trapezoid. After differentiating with respect to time, we can find the required rate.

2. Formula:

i. Volume of isosceles trapezoid = Area of trapezoid * length

ii. Power rule of differentiation:

ddxxn=n*xn-1

iii. Chain rule of differentiation:

ddtfx=ddxfx*dxdt

3. Given:

i.

dVdt=0.2 m3/min

ii. Length of trough, l = 10m

4. Calculations:

Let b1 and b2 be the length of lower base and upper base of isosceles trapezoid.

Volume of isosceles trapezoid is

V=Area of trapezoid ×length

V=12b1+ b2×h×l

V=12h0.3+b210

V=5h(0.3+ b2) ……….. (1)

If we draw the trapezoid and divide it into rectangles with base 0.3 =b1 as shown below, we will have two right triangles on the sides each one has a top length of 0.25m..

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