Explanation Given: A curve C in polar form r = f ( θ ) is parameterized by r ( θ ) = ( f ( θ ) cos θ , f ( θ ) sin θ ) because the x-and y-coordinates are given by x = r cos θ and y = r sin θ . Proof: a. The parameterized curve in the polar form: r ( θ ) = ( f ( θ ) cos θ , f ( θ ) sin θ ) .... ( 1 ) We have to evaluate the derivative of r ( θ ) . For that differentiate the equation (1) on both sides with respect to θ and we get: r ' ( θ ) = ( f ' ( θ ) cos θ − f ( θ ) sin θ , f ' ( θ ) sin θ + f ( θ ) cos θ ) Now, ‖ r ' ( θ ) ‖ = ( f ' ( θ ) cos θ − f ( θ ) sin θ ) 2 + ( f ' ( θ ) sin θ + f ( θ ) cos θ ) 2 = f ' ( θ ) 2 ( cos 2 θ + sin 2 θ ) + f ( θ ) 2 ( cos 2 θ + sin 2 θ ) = f ( θ ) 2 + f ' ( θ ) 2 Thus, it is proved that ‖ r ' ( θ ) ‖ = f ( θ ) 2 + f ' ( θ ) 2 .... ( 2 ) . b. The polar equation of the semicircle is given by: r = 2 cos θ , 0 ≤ θ ≤ π 2 From the part (a), we have r = f ( θ ) so, we can say that f ( θ ) = 2 cos θ .... ( 3 ) . Now, differentiate the equation (2) on both sides with respect to θ and we get: f ' ( θ ) = − 2 sin θ .... ( 4 ) Now, put the values of equation (3) and equation (4) in the equation (2) and we get: ‖ r ' ( θ ) ‖ = f ( θ ) 2 + f ' ( θ ) 2 = ( 2 cos ( θ ) ) 2 + ( − 2 sin ( θ ) ) 2 = 4 cos 2 θ + 4 sin 2 θ = 2 Now, the x-and y-coordinates are given by x = r cos θ and y = r sin θ so, the line integral is evaluated as: ∫ C ( x − y ) 2 d s = ∫ 0 π 2 ( r cos θ ...

4th Edition

ISBN: 9781319434526

Author: Rogawski

Publisher: MAC HIGHER

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Vector calculus is an important branch of mathematics and it relates two important branches of mathematics namely vector and calculus.

Question

Chapter 17.2, Problem 60E

To determine

To Prove:

a. $‖ r' (θ) ‖ = \sqrt{f {(θ)}^{2} + f' {(θ)}^{2}}$ .

b. Evaluate $\int_{C} {(x - y)}^{2} d s$ , where C is the semicircle with polar equation $r = 2 \cos θ, 0 \leq θ \leq \frac{π}{2}$ .

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Chapter 17.2, Problem 59E

Chapter 17.2, Problem 61E

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Chapter 17 Solutions

CALCULUS 4E (LL) W/ ACHIEVE ACCESS

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To Prove: a. ‖ r ' ( θ ) ‖ = f ( θ ) 2 + f ' ( θ ) 2 . b. Evaluate ∫ C ( x − y ) 2 d s , where C is the semicircle with polar equation r = 2 cos θ , 0 ≤ θ ≤ π 2 .

Solution Summary: The author evaluates the derivative of r ( ) by comparing the x- and y-coordinates.

Concept explainers

To Prove:

a. $‖ r' (θ) ‖ = \sqrt{f {(θ)}^{2} + f' {(θ)}^{2}}$ .

b. Evaluate $\int_{C} {(x - y)}^{2} d s$ , where C is the semicircle with polar equation $r = 2 \cos θ, 0 \leq θ \leq \frac{π}{2}$ .

Want to see the full answer?

Chapter 17 Solutions

To Prove: a. ‖ r ' ( θ ) ‖ = f ( θ ) 2 + f ' ( θ ) 2 . b. Evaluate ∫ C ( x − y ) 2 d s , where C is the semicircle with polar equation r = 2 cos θ , 0 ≤ θ ≤ π 2 .

Solution Summary: The author evaluates the derivative of r ( ) by comparing the x- and y-coordinates.

Concept explainers

To Prove: a. ‖r'(θ)‖=f(θ)2+f'(θ)2. b. Evaluate ∫C(x−y)2ds, where C is the semicircle with polar equation r=2cosθ,0≤θ≤π2.

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Chapter 17 Solutions

To Prove:

a. $‖ r' (θ) ‖ = \sqrt{f {(θ)}^{2} + f' {(θ)}^{2}}$ .

b. Evaluate $\int_{C} {(x - y)}^{2} d s$ , where C is the semicircle with polar equation $r = 2 \cos θ, 0 \leq θ \leq \frac{π}{2}$ .