Single Variable Calculus: Concepts and Contexts, Enhanced Edition

4th Edition

ISBN: 9781337687805

Author: James Stewart

Publisher: Cengage Learning

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Question

Chapter 4.7, Problem 30E

To determine

To calculate: The roots of the curve,

$y=−sinx$ .

Expert Solution & Answer

Answer to Problem 30E

The roots are

$x1≈4.5$

$x2≈4.493613$

$x3≈4.493409$ .

Explanation of Solution

Given information:

The infinitely many lines that are tangent to the curve $y=−sinx$ passes through the origin.

Formula used:

Newton’s Method:

We seek a solution of $f(x)=0$ , starting from an initial estimate $x=x1$ .

For $x=xn$ , compute the next approximation $xn+1$ by

$xn+1=xn−f(xn)f'(xn)$ and so on.

Equation of tangent line : $y−y1=m(x−x1)$

Slope of the tangent line : derivative of the function.

Calculation:

Consider the curve ,

$y=−sinx$

Now,

$f(x)=−sinxf'(x)=−cosx$

The line passes through the origin

$=y−y1=m(x−x1)=y−0=m(x−0)=y=mx$ (i)

While the tangent will touch the curve at some point. Let the x -coordinate be a.

Therefore , putting x-coordinate in equation $y=−sinx$ we get,

$y=−sina$ .

The point at which tangent touches the curve is $(a,−sina)$ .

Put it in equation (i)

$=−sina=ma$

Slope of the tangent :-

$m=−cosx$ .

At point $(a,−sina)$ the slope will be

$m=−cosa$

We get ,

$=−sina=(−cosa)a=−sina−cosa=a=sinacosa=a=tana=a=tana−a=0$

Sketching the graph of function $y=−sinx$ and $y=tanx−x$ . We observe the graph is symmetric.

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 4.7, Problem 30E

Figure 1.

Hence, we get the slope of the curve is $y=tanx−x$ . In other words $g(x)=tanx−x$ used to find out roots using Newton’s Method.

Therefore,

$g(x)=tanx−x$ and

$g'(x)=sec2(x)−1$

Now, let initial approximation be $x1=4.5$

For $n=1$

$x2=x1−g(x1)g'(x1)$

$x2=x1−tanx1−x1sec2(x1)−1$

$x2=(4.5)−tan(4.5)−4.5[sec(4.5)]2−1x2=(4.5)−4.637332055−4.522.5048486−1$

$x2=4.5−0.13733205521.5048486x2=4.5−0.006386097273x2=4.493613903$

The second approximation is $x2=4.493613$ .

For $n=2$

$x3=x2−g(x2)g'(x2)$

$x3=x2−tanx2−x2sec2(x2)−1$

$x3=(4.493613)−tan(4.493613)−4.493613[sec(4.493613)]2−1x3=(4.493613)−4.497726612−4.49361321.22954465−1$

$x3=4.493613−0.00411361220.22954465x3=4.493613−0.0002033467422x3=4.493409653$

The third approximation is $x3=4.493409$ .

Hence , the roots of the curve are :-

$x1≈4.5$

$x2≈4.493613$

$x3≈4.493409$

Chapter 4.7, Problem 29E

Chapter 4.7, Problem 31E

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 roots of the curve, y = − sin x .

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To calculate: The roots of the curve, y=−sinx<m:math><m:mrow><m:mi>y</m:mi><m:mo>=</m:mo><m:mo>−</m:mo><m:mi>sin</m:mi><m:mi>x</m:mi></m:mrow></m:math> .

Answer to Problem 30E

Explanation of Solution

Chapter 4 Solutions

To calculate: The roots of the curve,

$y=−sinx$ .