(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: Graphical, Numerical, Algebraic

3rd Edition

ISBN: 9780133688399

Author: Finney, Ross L.

Publisher: Pearson/Prentice Hall

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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: Graphical, Numerical, Algebraic

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

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

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

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