(a) Answer The required equation of the tangent line is y = 5 x − 7 . Explanation Given information: The functions: f ( 2 ) = 3 and f ′ ( 2 ) = 5 Formula used: Point-slope form of the equation of a line: ( y − y 1 ) = m ( x − x 1 ) It is given that x = 2 and f ( 2 ) = 3 . Also at x = 2 ; f ′ is 5. This means the slope of the tangent line is 5 and the line passes through the point ( 2 , 3 ) . To find the equation of the tangent, use point-slope formula of an equation of a line ( y − y 1 ) = m ( x − x 1 ) . Substitute 5 for m , 3 for y 1 and 2 for x 1 in the above equation and simplify. ( y − 3 ) = 5 ( x − 2 ) y − 3 = 5 x − 10 y − 3 + 3 = 5 x − 10 + 3 y = 5 x − 7 Hence, the required equation of the tangent line is y = 5 x − 7 . (b) To determine The equation of the normal line to the graph of y = f ( x ) at the point where x = 2 . Answer The required equation of the normal line is y = − 1 5 x + 17 5 . Explanation Given information: The functions: f ( 2 ) = 3 and f ′ ( 2 ) = 5 Formula used: Point-slope form of the equation of a line: ( y − y 1 ) = m ( x − x 1 ) It is given that x = 2 and f ( 2 ) = 3 . Also at x = 2 ; f ′ is 5. This means the slope of the tangent line is 5 and a point on the normal line is ( 2 , 3 ) . It is known that the normal line is perpendicular to the tangent line at the point of tangency. So, the slope of the normal line is the negative reciprocal of the slope of the tangent line. This means slope of the normal line is − 1 5 . To find the equation of the normal line, use the point-slope form of the equation of a line ( y − y 1 ) = m ( x − x 1 ) . Substitute − 1 5 for m , 3 for y 1 and 2 for x 1 in the above equation and simplify. ( y − 3 ) = − 1 5 ( x − 2 ) y − 3 = − 1 5 x + 2 5 y − 3 + 3 = − 1 5 x + 2 5 + 3 y = − 1 5 x + 17 5 Hence, the required equation of the normal line is y = − 1 5 x + 17 5 .

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 17E

(a)

To determine

The equation of the tangent line to the graph of y=f(x) at the point where x=2 .

(a)

Expert Solution

Answer to Problem 17E

The required equation of the tangent line is y=5x−7 .

Explanation of Solution

Given information:

The functions: f(2)=3 and f′(2)=5

Formula used:

Point-slope form of the equation of a line:

(y−y1)=m(x−x1)

It is given that x=2 and f(2)=3 .

Also at x=2 ; f′ is 5.

This means the slope of the tangent line is 5 and the line passes through the point (2,3) .

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

Substitute 5 for m , 3 for y1 and 2 for x1 in the above equation and simplify.

(y−3)=5(x−2)y−3=5x−10y−3+3=5x−10+3y=5x−7

Hence, the required equation of the tangent line is y=5x−7 .

(b)

To determine

The equation of the normal line to the graph of y=f(x) at the point where x=2 .

(b)

Expert Solution

Answer to Problem 17E

The required equation of the normal line is y=−15x+175 .

Explanation of Solution

Given information:

The functions: f(2)=3 and f′(2)=5

Formula used:

Point-slope form of the equation of a line:

(y−y1)=m(x−x1)

It is given that x=2 and f(2)=3 .

Also at x=2 ; f′ is 5.

This means the slope of the tangent line is 5 and a point on the normal line is (2,3) .

It is known that the normal line is perpendicular to the tangent line at the point of tangency.

So, the slope of the normal line is the negative reciprocal of the slope of the tangent line.

This means slope of the normal line is −15 .

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

Substitute −15 for m , 3 for y1 and 2 for x1 in the above equation and simplify.

(y−3)=−15(x−2)y−3=−15x+25y−3+3=−15x+25+3y=−15x+175

Hence, the required equation of the normal line is y=−15x+175 .

Chapter 3.1, Problem 16E

Chapter 3.1, Problem 18E

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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Prob. 4RE Ch. 3 - Prob. 5RE Ch. 3 - Prob. 6RE Ch. 3 - Prob. 7RE Ch. 3 - Prob. 8RE Ch. 3 - Prob. 9RE Ch. 3 - Prob. 10RE Ch. 3 - Prob. 11RE Ch. 3 - Prob. 12RE Ch. 3 - Prob. 13RE Ch. 3 - Prob. 14RE Ch. 3 - Prob. 15RE Ch. 3 - Prob. 16RE Ch. 3 - Prob. 17RE Ch. 3 - Prob. 18RE Ch. 3 - Prob. 19RE Ch. 3 - Prob. 20RE Ch. 3 - Prob. 21RE Ch. 3 - Prob. 22RE Ch. 3 - Prob. 23RE Ch. 3 - Prob. 24RE Ch. 3 - Prob. 25RE Ch. 3 - Prob. 26RE Ch. 3 - Prob. 27RE Ch. 3 - Prob. 28RE Ch. 3 - Prob. 29RE Ch. 3 - Prob. 30RE Ch. 3 - Prob. 31RE Ch. 3 - Prob. 32RE Ch. 3 - Prob. 33RE Ch. 3 - Prob. 34RE Ch. 3 - Prob. 35RE Ch. 3 - Prob. 36RE Ch. 3 - Prob. 37RE Ch. 3 - Prob. 38RE Ch. 3 - Prob. 39RE Ch. 3 - Prob. 40RE Ch. 3 - Prob. 41RE Ch. 3 - Prob. 42RE Ch. 3 - Prob. 43RE Ch. 3 - Prob. 44RE Ch. 3 - Prob. 45RE Ch. 3 - Prob. 46RE Ch. 3 - Prob. 47RE Ch. 3 - Prob. 48RE Ch. 3 - Prob. 49RE Ch. 3 - Prob. 50RE Ch. 3 - Prob. 51RE Ch. 3 - Prob. 52RE Ch. 3 - Prob. 53RE Ch. 3 - Prob. 54RE Ch. 3 - Prob. 55RE Ch. 3 - Prob. 56RE Ch. 3 - 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The equation of the tangent line to the graph of y = f ( x ) at the point where x = 2 .

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Answer to Problem 17E

Explanation of Solution

The equation of the normal line to the graph of y=f(x) at the point where x=2 .

Answer to Problem 17E

Explanation of Solution

Chapter 3 Solutions

Additional Math Textbook Solutions