Explanation Given information : f is a differential function of θ . Formula used: The relation between the polar coordinates and the Cartesian coordinates is given by x = r c o s θ y = r s i n θ Proof: The relation between the polar coordinates and the Cartesian coordinates is given by x = r c o s θ y = r s i n θ Since r = f ( θ ) , x = f ( θ ) c o s θ y = f ( θ ) s i n θ } ......(1) Differentiating (1), with respect to ‘ θ ’, we have d x d θ = f ' ( θ ) c o s θ − f ( θ ) s i n θ d y d θ = f ' ( θ ) sin θ + f (<

7th Edition

ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Question

Chapter 6.3, Problem 57PS

To determine

To prove:The tangent line to the polar curve r=f(θ) at the point P(r,θ) has slope

m=f(θ)cosθ+f'(θ)sinθ−f(θ)sinθ+f'(θ)cosθ , whenever the denominator is not zero.

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Chapter 6.3, Problem 56PS

Chapter 6.3, Problem 58PS

Chapter 6 Solutions

Calculus

Ch. 6.1 - Prob. 1PS Ch. 6.1 - Prob. 2PS Ch. 6.1 - Prob. 3PS Ch. 6.1 - Prob. 4PS Ch. 6.1 - Prob. 5PS Ch. 6.1 - Prob. 6PS Ch. 6.1 - Prob. 7PS Ch. 6.1 - Prob. 8PS Ch. 6.1 - Prob. 9PS Ch. 6.1 - Prob. 10PS

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Prob. 51PS Ch. 6.6 - Prob. 52PS Ch. 6.6 - Prob. 53PS Ch. 6.6 - Prob. 54PS Ch. 6.6 - Prob. 55PS Ch. 6.6 - Prob. 56PS Ch. 6.6 - Prob. 57PS Ch. 6.6 - Prob. 58PS Ch. 6.6 - Prob. 59PS Ch. 6.6 - Prob. 60PS Ch. 6 - Prob. 1PE Ch. 6 - Prob. 2PE Ch. 6 - Prob. 3PE Ch. 6 - Prob. 4PE Ch. 6 - Prob. 5PE Ch. 6 - Prob. 6PE Ch. 6 - Prob. 7PE Ch. 6 - Prob. 8PE Ch. 6 - Prob. 9PE Ch. 6 - Prob. 10PE Ch. 6 - Prob. 11PE Ch. 6 - Prob. 12PE Ch. 6 - Prob. 13PE Ch. 6 - Prob. 14PE Ch. 6 - Prob. 15PE Ch. 6 - Prob. 16PE Ch. 6 - Prob. 17PE Ch. 6 - Prob. 18PE Ch. 6 - Prob. 19PE Ch. 6 - Prob. 20PE Ch. 6 - Prob. 21PE Ch. 6 - Prob. 22PE Ch. 6 - Prob. 23PE Ch. 6 - Prob. 24PE Ch. 6 - Prob. 25PE Ch. 6 - Prob. 26PE Ch. 6 - Prob. 27PE Ch. 6 - Prob. 28PE Ch. 6 - Prob. 29PE Ch. 6 - Prob. 30PE Ch. 6 - Prob. 1SP Ch. 6 - Prob. 2SP Ch. 6 - Prob. 3SP Ch. 6 - Prob. 4SP Ch. 6 - Prob. 5SP Ch. 6 - Prob. 6SP Ch. 6 - Prob. 7SP Ch. 6 - Prob. 8SP Ch. 6 - Prob. 9SP Ch. 6 - Prob. 10SP Ch. 6 - Prob. 11SP Ch. 6 - Prob. 12SP Ch. 6 - Prob. 13SP Ch. 6 - 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To prove: The tangent line to the polar curve r = f ( θ ) at the point P ( r , θ ) has slope m = f ( θ ) c o s θ + f ' ( θ ) s i n θ − f ( θ ) sin θ + f ' ( θ ) c o s θ , whenever the denominator is not zero.

Solution Summary: The author explains that the tangent line r=f(theta ) has a slope and the relation between polar and Cartesian coordinates is given.

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To prove:The tangent line to the polar curve r=f(θ) at the point P(r,θ) has slope

m=f(θ)cosθ+f'(θ)sinθ−f(θ)sinθ+f'(θ)cosθ , whenever the denominator is not zero.

Want to see the full answer?

Chapter 6 Solutions

To prove: The tangent line to the polar curve r = f ( θ ) at the point P ( r , θ ) has slope m = f ( θ ) c o s θ + f ' ( θ ) s i n θ − f ( θ ) sin θ + f ' ( θ ) c o s θ , whenever the denominator is not zero.

Solution Summary: The author explains that the tangent line r=f(theta ) has a slope and the relation between polar and Cartesian coordinates is given.

Videos

To prove:The tangent line to the polar curve r=f(θ) at the point P(r,θ) has slope m=f(θ)cosθ+f'(θ)sinθ−f(θ)sinθ+f'(θ)cosθ , whenever the denominator is not zero.

Want to see the full answer?

Chapter 6 Solutions

To prove:The tangent line to the polar curve r=f(θ) at the point P(r,θ) has slope

m=f(θ)cosθ+f'(θ)sinθ−f(θ)sinθ+f'(θ)cosθ , whenever the denominator is not zero.