BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 6, Problem 5P
To determine

To show: The depth of the water decreases at a constant rate without regarding the shape of the bowl.

Expert Solution

Answer to Problem 5P

The depth of the water decreases at a constant rate.

Explanation of Solution

Given information:

Water in an open bowl evaporates at a rate proportional to the area of the surface of the water.

Calculation:

The rate of decrease of the volume is proportional to the area of the surface.

dVdt=kA(x)

Here, k is the positive constant and A(x) is the area of the surface at the water depth of x.

Consider that the rate of change of depth with respect to time as dxdt.

Apply Chain rule as shown below.

dVdt=dVdxdxdt

Substitute kA(x) for dVdt in the above Equation.

dVdxdxdt=kA(x) (1)

Expression for the volume of water up to a depth x as shown below.

V(x)=0xA(s)ds

Here, A(s) is the area of the cross section at the water depth of s.

Differentiate both sides of the Equation with respect to x.

dVdx=A(x)

Substitute A(x) for dVdx in Equation (1).

A(x)dxdt=kA(x)dxdt=kdxdt=constant

Therefore, the depth of the water decreases at a constant rate without regarding the shape of the bowl. Hence, it is proved.

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