Calculus: Graphical, Numerical, Algebraic

3rd Edition

ISBN: 9780133688399

Author: Finney, Ross L.

Publisher: Pearson/Prentice Hall

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Concept explainers

Rate of Change

The relation between two quantities which displays how much greater one quantity is than another is called ratio.

Slope

The change in the vertical distances is known as the rise and the change in the horizontal distances is known as the run. So, the rise divided by run is nothing but a slope value. It is calculated with simple algebraic equations as:

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Question

Chapter 3.1, Problem 20E

(a)

To determine

The lines that are tangent to the curve at x=4 .

(a)

Expert Solution

Answer to Problem 20E

The required equation of the tangent line is y=14x+1 .

Explanation of Solution

Given information:

The curve: y=x

Formula used:

Point-slope form of the equation of a line:

(y−y1)=m(x−x1)

Slope of a curve is:

f′(x)=limx→af(x)−f(a)x−a

The given curve is y=x .

Write the given definition using the above function as:

f′(a)=limx→ax−ax−a

Substitute 4 for a in the above limit and simplify.

f′(4)=limx→4x−4x−4=limx→4(x−2)(x+2)(x−2)=limx→41(x+2)

Substitute 4 for x in the above limit and simplify.

limx→41(x+2)=14+2=12+2=14

Also, substitute 4 for x in the curve y=x .

y=4=2

So, the slope of the tangent line is 14 and the point through which the line passes is (4,2) .

To find the equation of the tangent, use point-slope formula of an equation of a line (y−y1)=m(x−x1) .

Substitute 14 for m , 2 for y1 and 4 for x1 in the above equation and simplify.

(y−2)=14(x−4)y−2=14x−14⋅4y−2+2=14x−1+2y=14x+1

Hence, the required equation of the tangent line is y=14x+1 .

(b)

To determine

The lines that are normal to the curve at x=4 .

(b)

Expert Solution

Answer to Problem 20E

The required equation of the tangent line is y=−4x+18 .

Explanation of Solution

Given information:

The curve: y=x

Formula used:

Point-slope form of the equation of a line:

(y−y1)=m(x−x1)

Slope of a curve is:

f′(x)=limx→af(x)−f(a)x−a

The given curve is y=x .

Write the given definition using the above function as:

f′(a)=limx→ax−ax−a

Substitute 4 for a in the above limit and simplify.

f′(4)=limx→4x−4x−4=limx→4(x−2)(x+2)(x−2)=limx→41(x+2)

Substitute 4 for x in the above limit and simplify.

limx→41(x+2)=14+2=12+2=14

Also, substitute 4 for x in the curve y=x .

y=4=2

So, the slope of the tangent line is 14 and the point through which the line passes is (4,2) .

It is known that the slope of a normal line is the negative reciprocal to the slope of the tangent line.

So, the slope of the normal line is m=−114=(−1)×41=−4

To find the equation of the tangent, use point-slope formula of an equation of a line (y−y1)=m(x−x1) .

Substitute −4 for m , 2 for y1 and 4 for x1 in the above equation and simplify.

(y−2)=−4(x−4)y−2=−4x+16y−2+2=−4x+16+2y=−4x+18

Hence, the required equation of the tangent line is y=−4x+18 .

Chapter 3.1, Problem 19E

Chapter 3.1, Problem 21E

Chapter 3 Solutions

Calculus: Graphical, Numerical, Algebraic

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The lines that are tangent to the curve at x = 4 .

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The lines that are tangent to the curve at x=4 .

Answer to Problem 20E

Explanation of Solution

The lines that are normal to the curve at x=4 .

Answer to Problem 20E

Explanation of Solution

Chapter 3 Solutions

Additional Math Textbook Solutions