   Chapter 3.9, Problem 36E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# A faucet is filling a hemispherical basin of diameter 60 cm with water at a rate of 2 L/min. Find the rate at which the water is rising in the basin when it is half full. [Use the following facts: 1 L is 1000 cm3. The volume of the portion of a sphere with radius r from the bottom to a height h is V = π ( r h 2 − 1 3 h 3 ) as we will show in Chapter 6.)

To determine

To find: The rate of change of height of the water inside the basin when it is half full.

Explanation

Given:

A faucet is filling a hemisphere basin of diameter 60 cm.

The basin is filling with the water at a rate of 2L/min=2000mm3/min.

Formula used:

(1) Chain rule: dydx=dydududx

(2) Volume of the hemisphere of radius r is:V=23πr3

(3) Volume of the portion of sphere of radius r from the bottom and height h is:V=π(rh213h3)

Calculation:

Let us assume that V be the volume of the hemisphere of radius r. And h be the height of the water inside the hemisphere.

The volume of the hemisphere of radius r=30cm is,

V=23π(30)3=23π×27000=18000πcm3

Since the basin is half full, then

V=π(rh213h3)π(30×h213h3)=9000π(30×h213h3)=90003×30h2h33=9000

On further simplification the cubic equation in h is as follows.

3×30h2h33=900090h2h3=27000(h390h2)=27000h390h2+27000=0

From the numerical roots finder the three solutions of the above equations are as follows.

h1=19.58 ,h2=86.38 , and h3=15.96 .

But out these three values

h2=86.38 is greater than radius of the hemisphere, which is not possible.

h3=15

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