Chapter 4.7, Problem 30E

To determine

To calculate: The roots of the curve,

  $y=-\sin x$ .

Answer to Problem 30E

The roots are

  $x_1\approx 4.5$

  $x_2\approx 4.493613$

  $x_3\approx 4.493409$ .

Explanation of Solution

Given information:

The infinitely many lines that are tangent to the curve $y=-\sin x$ passes through the origin.

Formula used:

Newton’s Method:

We seek a solution of $f(x)=0$ , starting from an initial estimate $x=x_1$ .

For $x=x_n$ , compute the next approximation $x_{n+1}$ by

  $x_{n+1}=x_n-\frac{f(x_n)}{f\text{'}(x_n)}$ and so on.

Equation of tangent line : $y-y_1=m(x-x_1)$

Slope of the tangent line : derivative of the function.

Calculation:

Consider the curve ,

  $y=-\sin x$

Now,

  $\begin{array}{l}f(x)=-\sin x\\ f\text{'}(x)=-\cos x\end{array}$

The line passes through the origin

  $\begin{array}{l}=y-y_1=m(x-x_1)\\ =y-0=m(x-0)\\ =y=mx\end{array}$ (i)

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

Therefore , putting x-coordinate in equation $y=-\sin x$ we get,

  $y=-\sin a$ .

The point at which tangent touches the curve is $(a,-\sin a)$ .

Put it in equation (i)

  $=-\sin a=ma$

Slope of the tangent :-

  $m=-\cos x$ .

At point $(a,-\sin a)$ the slope will be

  $m=-\cos a$

We get ,

  $\begin{array}{l}=-\sin a=(-\cos a)a\\ =\frac{-\sin a}{-\cos a}=a\\ =\frac{\sin a}{\cos a}=a\\ =\tan a=a\\ =\tan a-a=0\end{array}$

Sketching the graph of function $y=-\sin x$ and $y=\tan x-x$ . We observe the graph is symmetric.

  

  Figure 1.

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

Therefore,

  $g(x)=\tan x-x$ and

  $g\text{'}(x)={\sec }^2(x)-1$

Now, let initial approximation be $x_1=4.5$

For $n=1$

  $x_2=x_1-\frac{g(x_1)}{g\text{'}(x_1)}$

  $x_2=x_1-\frac{\tan x_1-x_1}{{\sec }^2(x_1)-1}$

  $\begin{array}{l}{x}_2=(4.5)-\frac{\tan (4.5)-4.5}{{\left[\sec (4.5)\right]}^2-1}\\ x_2=(4.5)-\frac{4.637332055-4.5}{22.5048486-1}\end{array}$

  $\begin{array}{l}{x}_2=4.5-\frac{0.137332055}{21.5048486}\\ x_2=4.5-0.006386097273\\ x_2=4.493613903\end{array}$

The second approximation is $x_2=4.493613$ .

For $n=2$

  $x_3=x_2-\frac{g(x_2)}{g\text{'}(x_2)}$

  $x_3=x_2-\frac{\tan x_2-x_2}{{\sec }^2(x_2)-1}$

  $\begin{array}{l}{x}_3=(4.493613)-\frac{\tan (4.493613)-4.493613}{{\left[\sec (4.493613)\right]}^2-1}\\ x_3=(4.493613)-\frac{4.497726612-4.493613}{21.22954465-1}\end{array}$

  $\begin{array}{l}{x}_3=4.493613-\frac{0.004113612}{20.22954465}\\ x_3=4.493613-0.0002033467422\\ x_3=4.493409653\end{array}$

The third approximation is $x_3=4.493409$ .

Hence , the roots of the curve are :-

  $x_1\approx 4.5$

  $x_2\approx 4.493613$

  $x_3\approx 4.493409$