   Chapter 3.8, Problem 36E

Chapter
Section
Textbook Problem

# Of the infinitely many fines that are tangent to the curve y = − sin x and pass through the origin, there is one that has the largest slope. Use Newton’s method to find the slope of that line correct to six decimal places.

To determine

To find:

The slope of line with largest slope and tangent to the y= -sinx pass through origin

Explanation

1) Concept:

Working rule of Newton’s method is,

Start with initial approximation x1  which is obtained from graph. Evaluate fx1and f '(x1) and apply the Newton’s formula. After the first iteration we get x2. Repeat the process with x1  replaced by x2. Calculate the approximations x1, x2, x3,  until the successive approximations xn, xn+1  agree the eight decimal places.

2) Formula:

Newton’s formula for nth approximation is xn+1=xn-fxnf'xn for n=1,2,3,

3) Given:

y=-sinx

4) Calculation:

y=-sinx

Differentiate given function with respect to x which gives the slope of tangent line to curve,

dydx=m=-cosx

Equation of line passing through origin is given by y=mx

Let (x, y) be any point on the curve, given y coordinate as y=-sin x. So substitute this point in equation of line,

-sinx=mx

Also, m=-cos x

-sinx=-cosx·x

Simplifying,

sinxcosx=x

tanx=x

Use Newton’s method to solve this equation, so

Let, fx=tanx-x

Differentiating, f(x) with respect to x,

f'x=sec2 x-1

Now plot the graph of f(x) to know the initial values,

The graph is symmetric about origin, from the graph we can say that the initial value occurs at x = 4.493, so take x1=4.493

So Newton’s formula for nth approximation becomes

xn+1=xn-fxnf'xn

For, n=1,

To find value of x2,

x2=x1-f(x1 )f'x2-1

x2=x1-fx1 f'x1=x1-f4

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