   Chapter 2.1, Problem 9E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# The point P(1, 0) lies on the curve y = sin(l0π/x).(a) If Q is the point (x, sin(10π/x)), find the slope of the secant line PQ (correct to four decimal places) for x = 2, 1.5, 1.4, 1.3, 1.2, 1.1, 0.5. 0.6, 0.7, 0 .8, and 0.9.Do the slopes appear to be approaching a limit?(b) Use a graph of the curve to explain why the slopes of the secant lines in part (a) arc not close to the slope of the tangent line at P.(c) By choosing appropriate secant lines, estimate the slope of the tangent line at P.

(a)

To determine

To find: The slope of the secant line PQ for the given values of x.

Explanation

Given:

The equation of the curve is y=sin(10πx).

The point P(1, 0) lies on the curve y.

The Q is the point (x,sin(10πx)).

Calculation:

The slope of the secant lines between the points, P(1, 0) and Q(x,sin(10πx)) is

mPQ=sin(10πx)0x1 (1)

Obtain the slope of the secant line PQ for the value of x=2.

Substitute 2 for x in sin(10πx),

sin(10πx)=sin(10π2)=sin(5π)=0

Substitute Q(x,sin(10πx))=(2,0) in equation (1),

mPQ=sin(10πx)0x1=0021=01=0

Thus, the slope of the secant line PQ for the value of x=2 is 0.

Obtain the slope of the secant line PQ for the value of x=1.5.

Substitute 1.5 for x in sin(10πx),

sin(10πx)=sin(10π1.5)=sin(203π)=0.866025

Substitute Q(x,sin(10πx))=(1.5,0.866025) in equation (1),

mPQ=sin(10πx)0x1=0.86602501.51=0.8660250.51.7321

Thus, the slope of the secant line PQ for the value of x=1.5 is 1.7321.

Obtain the slope of the secant line PQ for the value of x=1.4.

Substitute 1.4 for x in sin(10πx),

sin(10πx)=sin(10π1.4)=0.43388

Substitute Q(x,sin(10πx))=(1.4,0.43388) in equation (1),

mPQ=sin(10πx)0x1=0.4338801.41=0.433880.41.0847

Thus, the slope of the secant line PQ for the value of x=1.4 is 1.0847.

Obtain the slope of the secant line PQ for the value of x=1.3.

Substitute 1.3 for x in sin(10πx).

sin(10πx)=sin(10π1.3)=0.82298

Substitute Q(x,sin(10πx))=(1.3,0.82298) in equation (1).

mPQ=sin(10πx)0x1=0.8229801.31=0.822980.32.7433

Thus, the slope of the secant line PQ for the value of x=1.3 is 2.7433.

Obtain the slope of the secant line PQ for the value of x=1.2.

Substitute 1.2 for x in sin(10πx),

sin(10πx)=sin(10π1.2)=0.86602

Substitute Q(x,sin(10πx))=(1.2,0.86602) in equation (1),

mPQ=sin(10πx)0x1=0.8660201.21=0.866020.24.3301

Thus, the slope of the secant line PQ for the value of x=1.2 is 4.3301.

Obtain the slope of the secant line PQ for the value of x=1.1.

Substitute 1.1 for x in sin(10πx),

sin(10πx)=sin(10π1

(b)

To determine

To explain: The slopes of the secant lines in part (a) are not close to the slope of the tangent line at P by using a graph.

(c)

To determine

To estimate: The slope of the tangent line to the curve at P(1, 0).

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Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th 