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This is problem 49 page 219: A man is in a boat 2 miles from the nearest point on the coast. He is to go to a point Q, located 3 miles down the coast and 1 mile inland. If he can row at 2 miles per hour and walk at 4 miles per hour, toward what point on the coast should he row in order to reach point Q in the least amount of time.

This is problem 49 page 219:

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Page 1: This is problem 49 page 219:

This is problem 49 page 219:

A man is in a boat 2 miles from the nearest point on the coast. He is to go to a point Q, located 3 miles down the coast and 1 mile inland. If he can row at 2 miles per hour and walk at 4 miles per hour, toward what point on the coast should he row in order to reach point Q in the least amount of time.

Page 2: This is problem 49 page 219:

Need a picture:

Q

1 mile

2 miles

3 miles

x

3 - x

Find x to minimize time

Page 3: This is problem 49 page 219:

Use the Pythagorean theorem twice to find the distance to row and walk

To row: 42 x

To walk: 1)3( 2 x

Page 4: This is problem 49 page 219:

The time to row is: milesxmiles

hour4

22

Notice I used unit analysis to find I had to divide by 2 and not multiply by 2

The time to walk is:

milesxmiles

hour1)3(

42

Page 5: This is problem 49 page 219:

The primary equation is the total time:

1)3(4

14

2

1 22 xxT

We don’t need a “secondary equation”. Why?

The primary equation only has one variable, there is no need to substitute for a second variable.

Page 6: This is problem 49 page 219:

1)3(4

14

2

1 22 xxT

To minimize, find the derivative = 0

This will be good chain rule practice

))1)(3(2(1)3(

2

14124

2

121' 2

122

12 xxxxT

Outside Inside Outside Inside

Do you see another chain rule here?

Page 7: This is problem 49 page 219:

))1)(3(2(1)3(

2

14124

2

121' 2

122

12 xxxxT

We need to solve this mess = 0

First, let’s clean it up a bit by multiplying constants

)3(1)3(4

1421' 2

122

12 xxxxT

Wow, let’s cheat and use MathPert to graph this and find the solution

Page 8: This is problem 49 page 219:

Click on this

Page 9: This is problem 49 page 219:

Now this

Page 10: This is problem 49 page 219:

Select “Type it in”

Page 11: This is problem 49 page 219:

note the use of ( ) everywhere and the use of negative exponents now

click OK

Page 12: This is problem 49 page 219:

Make sure this is the equation we want and it is. Click OK again

Page 13: This is problem 49 page 219:

Here is our graph, we want where it equals 0, ie, the x-intercept

click this about 4 times to get a better view

Page 14: This is problem 49 page 219:

This looks like the point

Page 15: This is problem 49 page 219:

Use the Point and Slope Tool

click it and center the crosshairs on the intercept

Page 16: This is problem 49 page 219:

y = 0 and x = 1

this is where the derivative is equal to 0!

We just use a ‘numerical” technique to solve the equation

Page 17: This is problem 49 page 219:

Just to make sure, evaluate at x = 1:

)3(1)3(4

1421' 2

122

12 xxxxT

You should get:

054

2

52

1

The man should aim for a point 1 mile down the coast

Page 18: This is problem 49 page 219:

)3(1)3(4

1421' 2

122

12 xxxxT

To review what I typed into MathPert:

1/2((x^2+4)^(-1/2)(x))-1/4((3-x)^2+1)(-1/2)(3-x))

“Count” the number of opening (( versus the number of closing )), they must match in any expression. I could also type, which is easier:

x/2sqrt(x^2+4)-(3-x)/4sqrt((3-x)^2+1)