The slope of the secant line PQ for x = 1.5 .

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 2.1, Problem 3E

(a)

(i)

To determine

To find: The slope of the secant line PQ for x=1.5.

Expert Solution

The slope of the secant line PQ when x=1.5 is 2.

Explanation of Solution

Given:

The equation of the curve is y=1(1x).

The point P(2,1) lies on the curve y.

Calculation:

The slope of the secant line between the points, P(2,1)  and Q(x,11x) is

m=11x(1)x2 (1)

Obtain the slope of the secant line PQ when x=1.5.

Substitute 1.5 for x in 1(1x).

1(1x)=1(11.5)=10.5=2

Substitute 1.5 for x and −2 for 1(1x) in equation (1).

m=2(1)1.52=2+10.5=10.5=2

Thus, the slope of the secant line PQ when x=1.5 is 2.

(i)

To determine

To find: The slope of the secant line PQ for x=1.5.

Expert Solution

The slope of the secant line PQ when x=1.5 is 2.

Explanation of Solution

Given:

The equation of the curve is y=1(1x).

The point P(2,1) lies on the curve y.

Calculation:

The slope of the secant line between the points, P(2,1)  and Q(x,11x) is

m=11x(1)x2 (1)

Obtain the slope of the secant line PQ when x=1.5.

Substitute 1.5 for x in 1(1x).

1(1x)=1(11.5)=10.5=2

Substitute 1.5 for x and −2 for 1(1x) in equation (1).

m=2(1)1.52=2+10.5=10.5=2

Thus, the slope of the secant line PQ when x=1.5 is 2.

(ii)

To determine

To find: The slope of the secant line PQ for x=1.9.

Expert Solution

The slope of the secant line PQ when x=1.9 is 1.111111.

Explanation of Solution

Substitute 1.9 for x in 1(1x).

1(1x)=1(11.9)=10.9=1.11111

Substitute 1.9 for x and 1.11111 for 1(1x) in equation (1).

m=1.11111(1)1.92=1.11111+10.1=0.111110.11.111111

Thus, the slope of the secant line PQ when x=1.9 is 1.111111.

(iii)

To determine

To find: The slope of the secant line PQ for x=1.99.

Expert Solution

The slope of the secant line PQ when x=1.99 is 1.010101.

Explanation of Solution

Obtain the slope of the secant line PQ when x=1.99.

Substitute 1.99 for x in 1(1x).

1(1x)=1(11.99)=10.99=1.01¯

Substitute 1.99 for x and 1.01¯ for 1(1x) in equation (1).

m=1.01¯(1)1.992=1.01¯+10.01=0.01¯0.011.010101

Thus, the slope of the secant line PQ when x=1.99 is 1.010101.

(iv)

To determine

To find: The slope of the secant line PQ for x=1.99.

Expert Solution

The slope of the secant line PQ when x=1.99 is 1.010101.

Explanation of Solution

Obtain the slope of the secant line PQ when x=1.999.

Substitute 1.999 for x in 1(1x).

1(1x)=1(11.999)=10.999=1.001¯

Substitute 1.999 for x and 1.001¯ for 1(1x) in equation (1).

m=1.001¯(1)1.9992=1.001¯+10.001=0.001¯0.0011.001001

Thus, the slope of the secant line PQ when x=1.999 is 1.001001.

(v)

To determine

To find: The slope of the secant line PQ for x=2.5.

Expert Solution

The slope of the secant line PQ when x=2.5 is 0.666667.

Explanation of Solution

Obtain the slope of the secant line PQ when x=2.5.

Substitute 2.5 for x in 1(1x).

1(1x)=1(12.5)=11.5=0.6¯

Substitute 2.5 for x and 0.6¯ for 1(1x) in equation (1).

m=0.6¯(1)2.52=0.6¯+10.5=0.3¯0.50.666667

Thus, the slope of the secant line PQ when x=2.5 is 0.666667.

(vi)

To determine

To find: The slope of the secant line PQ for x=2.1.

Expert Solution

The slope of the secant line PQ when x=2.1 is 0.909091.

Explanation of Solution

Obtain the slope of the secant line PQ when x=2.1.

Substitute 2.1 for x in 1(1x).

1(1x)=1(12.1)=11.1=0.90¯

Substitute 2.1 for x and 0.90¯ for 1(1x) in equation (1).

m=0.90¯(1)2.12=0.90¯+10.1=0.09¯0.10.909091

Thus, the slope of the secant line PQ when x=2.1 is 0.909091.

(vii)

To determine

To find: The slope of the secant line PQ for x=2.01.

Expert Solution

The slope of the secant line PQ when x=2.01 is 0.990099.

Explanation of Solution

Obtain the slope of the secant line PQ when x=2.01.

Substitute 2.01 for x in 1(1x).

1(1x)=1(12.01)=11.01=0.9900¯

Substitute 2.01 for x and 0.9900¯ for 1(1x) in equation (1).

m=0.9900¯(1)2.012=0.9900¯+10.01=0.0099¯0.010.990099

Thus, the slope of the secant line PQ when x=2.01 is 0.990099.

(viii)

To determine

To find: The slope of the secant line PQ for x=2.001.

Expert Solution

The slope of the secant line PQ when x=2.001 is 0.999001.

Explanation of Solution

Obtain the slope of the secant line PQ when x=2.001.

Substitute 2.001 for x in 1(1x).

1(1x)=1(12.001)=11.001=0.999000¯

Substitute 2.001 for x and 0.999000¯ for 1(1x) in equation (1).

m=0.999000¯(1)2.0012=0.999000¯+10.01=0.000999¯0.010.999001

Thus, the slope of the secant line PQ when x=2.001 is 0.999001.

(b)

To determine

To guess: The value of the slope of the tangent line to the curve at P(2,−1).

Expert Solution

The estimated value of the slope of the tangent line to the curve at P(2,1) is 1.

Explanation of Solution

The secant line is closer to the tangent line at P(2,1) when t=1.999 and t=2.001.

From part (a), the slope of the secant line PQ when x=1.999 is 1.001001 and the slope of the secant line PQ when x=2.001 is 0.999001.

The value of the slope of the tangent line to the curve at P(2,1) is closer to the average value of the slope of the secant lines near to P.

Take the average of the two secant lines when x=1.999 and x=2.001.

m=1.001001+0.9990012=2.0000022=1.0000011

Thus, the estimated value of the slope of the tangent line to the curve at P(2,1) is 1.

(c)

To determine

To find: The equation of the tangent line to the curve at P(2,−1).

Expert Solution

The equation of the tangent line is y=x3.

Explanation of Solution

Formula used:

The equation of the tangent line to the curve y=f(x) at the point (x1,y1) is,

yy1=m(xx1) (2)

Calculation:

From part (b), the slope of the tangent line to the curve is 1.

Substitute m=1 and (x1,y1)=(2,1) in equation (2).

y(1)=1(x2)y+1=(x2)

Isolate y.

y=x21y=x3

Thus, the equation of the tangent line to the curve at P(2,1) is y=x3.

Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!