Single Variable Calculus: Concepts and Contexts, Enhanced Edition

4th Edition

ISBN: 9781337687805

Author: James Stewart

Publisher: Cengage Learning

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Question

Chapter 4.6, Problem 57E

To determine

To calculate: The point $P$ on the line segment $AB$ so as to maximize the angle $θ$ .

Expert Solution & Answer

Answer to Problem 57E

The point $P$ be chosen on the line segment from the point $B$ to $x=4.50$ unit.

Explanation of Solution

Given information:

A point $P$ on the line segment $AB$

Formula used:

Trigonometric ratio:

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 4.6, Problem 57E , additional homework tip 1

$tanθ=pb$

Quadratic formula: let the quadratic equation is $ax2+bx+c=0,(a≠0)$ Thus, if $b2−4ac≥0$ then the roots of the quadratic equation is given by

$x=−b±b2−4ac2a$

Let $f$ be a differentiable function defined on an interval $I$ and let $a∈I$ .

Then

$x=a$ is a point of local maximum value of $f$ , if

$f′(a)=0$ and
$f′(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f′(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f′(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima

$x=a$ is a point of local maximum value of $f$ , if

$f′(a)=0$ and
$f′(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f′(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f′(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.

$f′(a)=0$ and If $f′(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

Calculation:

As per the given problem

Draw the diagram of a point $P$ on the line segment $AB$

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 4.6, Problem 57E , additional homework tip 2

Form the given figure the triangle $△BPC$ makes the angle $α$ and the triangle $△APD$ makes the angle $β$

Here, it is observed that

$θ=π−α−β$

Recall that,

Trigonometric ratio: 5

$tanθ=pb$

$tanα=2x and tanβ=53−x⇒α=tan−1(2x) and β=tan−1(53−x)$

$θ=π−tan−1(2x)−tan−1(53−x)$

Recall that,

Let $f$ be a differentiable function defined on an interval $I$ and let $a∈I$ .

Then

$x=a$ is a point of local maximum value of $f$ , if

$f′(a)=0$ and
$f′(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f′(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f′(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima

$x=a$ is a point of local maximum value of $f$ , if

$f′(a)=0$ and
$f′(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f′(x)<0$ at every point sufficiently close to and to the left of $a$ , and at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.

$f′(a)=0$ and If $f′(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

Differentiate with respect to $x$

$θ′(x)=π−tan−1(2x)−tan−1(53−x) =−(11+4x2×−2x2)−(11+25(3−x)2×−5(3−x)2)$

Solve for $θ′(x)=0$ , and simplified

$−(11+4x2×−2x2)−(11+25(3−x)2×−5(3−x)2)=0⇒2x2+4+5x2−6x+34=0⇒5x2−6x+34=−2x2+4⇒−2x2+12x−68=5x2+20⇒7x2−12x−88=0$

Recall that,

Quadratic formula: let the quadratic equation is $ax2+bx+c=0,(a≠0)$ Thus, if $b2−4ac≥0$ then the roots of the quadratic equation is given by

$x=−b±b2−4ac2a$

$x=+12±(−12)2−4×7×−882×7 =12±144+246414 =12±260814 =12±51.0614$

Solve for different sign’s

$x=12+51.0614 and x=12−51.0614 =4.50 and =−2.79$

Because length can’t be negative,

Therefore,

$x=4.50$

Conclusion:

Thus the point $P$ be chosen on the line segment from the point $B$ to $x=4.50$ unit.

Chapter 4.6, Problem 56E

Chapter 4.6, Problem 58E

Chapter 4 Solutions

Single Variable Calculus: Concepts and Contexts, Enhanced Edition

Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E Ch. 4.1 - Prob. 9E Ch. 4.1 - Prob. 10E

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The point P on the line segment A B so as to maximize the angle θ .

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To calculate: The point P<m:math><m:mi>P</m:mi></m:math> on the line segment AB<m:math><m:mrow><m:mi>A</m:mi><m:mi>B</m:mi></m:mrow></m:math> so as to maximize the angle θ<m:math><m:mi>θ</m:mi></m:math> .

Answer to Problem 57E

Explanation of Solution

Chapter 4 Solutions

To calculate: The point $P$ on the line segment $AB$ so as to maximize the angle $θ$ .