Explanation Given: Equation of curve in polar coordinate. r → ( θ ) = 〈 f ( θ ) . cos θ . f ( θ ) sin θ 〉 Formula used: K ( θ ) = | f ( θ ) 2 + 2 f ' ( θ ) 2 − f ( θ ) . f " ( θ ) | ( f ( θ ) 2 + f ' ( θ ) 2 ) 3 / 2 Calculation: r → ( θ ) = 〈 f ( θ ) . cos θ , f ( θ ) sin θ 〉 r → ' ( θ ) = d d θ ( r ( θ ) ) = d d θ ( 〈 f ( θ ) . cos θ , f ( θ ) . sin θ 〉 ) Apply product rule of differentiation = 〈 f ( θ ) ( − sin θ ) + cos θ f ' ( θ ) , f ( θ ) cos θ + sin θ f ' ( θ ) 〉 r → ' ( θ ) = 〈 − f ( θ ) . sin θ + f ' ( θ ) cos θ , f ( θ ) cos θ + sin θ f ' ( θ ) 〉 r → " ( θ ) = d d θ ( r ' ( θ ) ) = d d θ 〈 − f ( θ ) . sin θ + f ' ( θ ) cos θ , f ( θ ) cos θ + sin f ' ( θ ) 〉 Distribute the differentiation = 〈 d d θ ( − f ( θ ) . sin θ + f ' ( θ ) . cos θ ) . d d θ ( f ( θ ) cos θ + sin θ f ' ( θ ) ) 〉 d d θ ( − f ( θ ) . sin + f ' ( θ ) . cos θ ) (Apply product rule of differentiation) = − f ' ( θ ) . sin θ − f ( θ ) cos θ + f " ( θ ) cos θ − f ' ( θ ) sin θ d d θ ( f ( θ ) cos θ + sin θ f ' ( θ ) ) (Apply product rule of differentiation) = − f ( θ ) sin θ + f ' ( θ ) cos θ + sin θ f " ( θ ) + f " ( θ ) . cos θ r → " ( θ ) = 〈 − 2 t ' ( θ ) . sin θ − f ( θ ) cos θ + f " ( θ ) cos θ , − f ( θ ) . sin θ + 2 f ' ( θ ) . cos θ + sin θ f " ( θ ) 〉 r → " ( θ ) = 〈 − 2 t ' ( θ ) . sin θ − f ( θ ) cos θ + f " ( θ ) cos θ , − f ( θ ) . sin θ + 2 f ' ( θ ) . cos θ + sin θ f " ( θ ) 〉 r → ' ( θ ) × r → " ( θ ) = 〈 − f ( θ ) . sin θ + f ' ( θ ) cos θ , f ( θ ) cos θ + sin θ f ' ( θ ) 〉 × 〈 − 2 t ' ( θ ) . sin θ − f ( θ ) cos θ + f " ( θ ) cos θ , − f ( θ ) sin θ + 2 f ' ( θ ) ...

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Chapter 14.4, Problem 74E

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To prove:

Show that a curve $\vec{r} (θ) = 〈 f (θ) \cos θ, f (θ) \sin θ 〉$ in polar coordinate has curvature $K (θ) = \frac{| f {(θ)}^{2} + 2 f ' {(θ)}^{2} - f (θ) . f " (θ) |}{{(f {(θ)}^{2} + f ' {(θ)}^{2})}^{3 / 2}}$

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Chapter 14.4, Problem 73E

Chapter 14.4, Problem 75E

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Chapter 14 Solutions

EBK CALCULUS

Ch. 14.1 - Prob. 1PQ Ch. 14.1 - Prob. 2PQ Ch. 14.1 - Prob. 3PQ Ch. 14.1 - Prob. 4PQ Ch. 14.1 - Prob. 5PQ Ch. 14.1 - Prob. 6PQ Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E

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To prove: Show that a curve r → ( θ ) = 〈 f ( θ ) cos θ , f ( θ ) sin θ 〉 in polar coordinate has curvature K ( θ ) = | f ( θ ) 2 + 2 f ' ( θ ) 2 − f ( θ ) . f " ( θ ) | ( f ( θ ) 2 + f ' ( θ ) 2 ) 3 / 2

Solution Summary: The author explains that a curve in polar coordinate has curvature K ( ) = 2 + 2 f.

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To prove:

Show that a curve $\vec{r} (θ) = 〈 f (θ) \cos θ, f (θ) \sin θ 〉$ in polar coordinate has curvature $K (θ) = \frac{| f {(θ)}^{2} + 2 f ' {(θ)}^{2} - f (θ) . f " (θ) |}{{(f {(θ)}^{2} + f ' {(θ)}^{2})}^{3 / 2}}$

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Chapter 14 Solutions

To prove: Show that a curve r → ( θ ) = 〈 f ( θ ) cos θ , f ( θ ) sin θ 〉 in polar coordinate has curvature K ( θ ) = | f ( θ ) 2 + 2 f ' ( θ ) 2 − f ( θ ) . f " ( θ ) | ( f ( θ ) 2 + f ' ( θ ) 2 ) 3 / 2

Solution Summary: The author explains that a curve in polar coordinate has curvature K ( ) = 2 + 2 f.

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To prove: Show that a curve r→(θ)=〈 f(θ)cosθ,f(θ)sinθ 〉 in polar coordinate has curvature K(θ)=| f(θ)2+2f '(θ)2−f (θ). f "(θ) |(f(θ)2+f ' (θ)2)3/2

Want to see the full answer?

Chapter 14 Solutions

To prove:

Show that a curve $\vec{r} (θ) = 〈 f (θ) \cos θ, f (θ) \sin θ 〉$ in polar coordinate has curvature $K (θ) = \frac{| f {(θ)}^{2} + 2 f ' {(θ)}^{2} - f (θ) . f " (θ) |}{{(f {(θ)}^{2} + f ' {(θ)}^{2})}^{3 / 2}}$