   Chapter 4.7, Problem 49E ### 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

# EXAMPLE 4 A man launches his boat from point A on a bank of a straight river, 3 km wide, and wants to reach point B, 8 km downstream on the opposite bank, as quickly as possible (see Figure 7). He could row his boat directly across the river to point C and then run to B, or he could row directly to B, or he could row to some point D between C and B and then run to B. If he can row 6 km/h and run 8 km/h, where should he land to reach B as soon as possible? (We assume that the speed of the water is negligible compared with the speed at which the man rows.)Solve the problem in Example 4 if the river is 5 km wide and point B is only 5 km downstream from A.

To determine

To find: The direction how the man should row in the river.

Explanation

Given:

Width of the river=5 km

Speed of the rowing of the man = 6 km/h

Speed of running of the man= 8 km/h

Formula used:

Pythagoras Theorem: If c denotes the length of the hypotenuse and a and b denote the length of the other two sides of a right angle triangle, then the Pythagorean theorem can be expressed as the Pythagorean equation: a2+b2=c2

Calculation:

In Figure 1,

Let x be the distance from C D then the running distance is |DB|=5x

Consider right angle triangle ACD ,

AC=5CD=x

By Pythagoras theorem the rowing distance as |AD|=(x2+25)

Now, use

time=distancerate

Then, the rowing time is x2+256 and the running time is 5x8

So, the total time is,

T(x)=x2+256+5x8 For 0x5

Differentiate T with respect to x ,

dTdx=2x6×2(x2+25)18=x6(x2+25)18

For critical points,

dTdx=0x6(x2+25)18=08x=6(x2+25)64x2=36(x2+25)

Simplify that,

28x2=36×25x=36×2528=6×52×7=157

However, this makes x>5

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