Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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Question

Chapter 3.1, Problem 19E

(a)

To determine

The line that is tangent to the curve at a point.

(a)

Expert Solution

Answer to Problem 19E

The required equation of the tangent line is y=3x−2 .

Explanation of Solution

Given information:

The curve: y=x3

The point = (1,1)

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=x3 .

Write the given definition using the above function as:

f′(a)=limx→ax3−a3x−a

Substitute 1 for a in the above limit and simplify.

f′(1)=limx→1x3−(1)3x−(1)=limx→1x3−1x−1=limx→1(x−1)(x2+x+1)(x−1)=limx→1(x2+x+1)

Simplify the above limit further.

limx→1(x2+x+1)=(1)2+(1)+1=1+1+1=3

So, the slope of the tangent line is 3 and the point is (1,1) .

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

Substitute 3 for m , 1 for y1 and 1 for x1 in the above equation and simplify.

(y−1)=3(x−1)y−1=3x−3y−1+1=3x−3+1y=3x−2

Hence, the required equation of the tangent line is y=3x−2 .

(b)

To determine

The line that is normal to the curve at a point.

(b)

Expert Solution

Answer to Problem 19E

The required equation of the normal line is y=−13x+43 .

Explanation of Solution

Given information:

The curve: y=x3

The point =(1,1)

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=x3 .

Write the given definition using the above function as:

f′(a)=limx→ax3−a3x−a

Substitute 1 for a in the above limit and simplify.

f′(1)=limx→1x3−(1)3x−(1)=limx→1x3−1x−1=limx→1(x−1)(x2+x+1)(x−1)=limx→1(x2+x+1)

Simplify the above limit further.

limx→1(x2+x+1)=(1)2+(1)+1=1+1+1=3

So, the slope of the tangent line is 3 and the point is (1,1) .

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

So, the slope of the required normal is, m=−13

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

Substitute −13 for m , 1 for y1 and 1 for x1 in the above equation and simplify.

(y−1)=−13(x−1)y−1=−13x+13y−1+1=−13x+13+1y=−13x+43

Hence, the required equation of the normal line is y=−13x+43 .

Chapter 3.1, Problem 18E

Chapter 3.1, Problem 20E

Chapter 3 Solutions

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

Ch. 3.1 - Prob. 1QR Ch. 3.1 - Prob. 2QR Ch. 3.1 - Prob. 3QR Ch. 3.1 - Prob. 4QR Ch. 3.1 - Prob. 5QR Ch. 3.1 - Prob. 6QR Ch. 3.1 - Prob. 7QR Ch. 3.1 - Prob. 8QR Ch. 3.1 - Prob. 9QR Ch. 3.1 - Prob. 10QR

Ch. 3.1 - Prob. 1E Ch. 3.1 - Prob. 2E Ch. 3.1 - Prob. 3E Ch. 3.1 - Prob. 4E Ch. 3.1 - Prob. 5E Ch. 3.1 - Prob. 6E Ch. 3.1 - Prob. 7E Ch. 3.1 - Prob. 8E Ch. 3.1 - Prob. 9E Ch. 3.1 - Prob. 10E Ch. 3.1 - Prob. 11E Ch. 3.1 - Prob. 12E Ch. 3.1 - Prob. 13E Ch. 3.1 - Prob. 14E Ch. 3.1 - Prob. 15E Ch. 3.1 - Prob. 16E Ch. 3.1 - Prob. 17E Ch. 3.1 - Prob. 18E Ch. 3.1 - Prob. 19E Ch. 3.1 - Prob. 20E Ch. 3.1 - Prob. 21E Ch. 3.1 - Prob. 22E Ch. 3.1 - Prob. 23E Ch. 3.1 - Prob. 24E Ch. 3.1 - Prob. 25E Ch. 3.1 - Prob. 26E Ch. 3.1 - Prob. 27E Ch. 3.1 - Prob. 28E Ch. 3.1 - Prob. 29E Ch. 3.1 - Prob. 30E Ch. 3.1 - Prob. 31E Ch. 3.1 - Prob. 32E Ch. 3.1 - Prob. 33E Ch. 3.1 - Prob. 34E Ch. 3.1 - Prob. 35E Ch. 3.1 - Prob. 36E Ch. 3.1 - Prob. 37E Ch. 3.1 - Prob. 38E Ch. 3.1 - Prob. 39E Ch. 3.1 - Prob. 40E Ch. 3.1 - Prob. 41E Ch. 3.1 - Prob. 42E Ch. 3.1 - Prob. 43E Ch. 3.1 - Prob. 44E Ch. 3.1 - Prob. 45E Ch. 3.2 - Prob. 1QR Ch. 3.2 - Prob. 2QR Ch. 3.2 - Prob. 3QR Ch. 3.2 - Prob. 4QR Ch. 3.2 - Prob. 5QR Ch. 3.2 - 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