Explanation Given: The function r ( t ) = 〈 cosh t , sinh t 〉 and the point t = 0 . Calculation: Find the first and the second derivative of the function r ( t ) = 〈 cosh t , sinh t 〉 . The first and the second derivative of the given function are r ′ ( t ) = 〈 sinh t , cosh t 〉 and r ″ ( t ) = 〈 cosh t , sinh t 〉 respectively. So, r ( 0 ) = 〈 cosh ( 0 ) , sinh ( 0 ) 〉 = 〈 1 , 0 〉 = 〈 1 , 0 , 0 〉 , r ′ ( 0 ) = 〈 sinh 0 , cosh 0 〉 = 〈 0 , 1 〉 = 〈 0 , 1 , 0 〉 , r ″ ( 0 ) = 〈 cosh 0 , sinh 0 〉 = 〈 1 , 0 〉 = 〈 1 , 0 , 0 〉 . And, ‖ r ′ ( 0 ) ‖ = 0 + 1 2 + 0 = 1 . Now, r ′ ( 0 ) × r ″ ( 0 ) = 〈 0 , 1 , 0 〉 × 〈 1 , 0 , 0 〉 = 〈 0 , 0 , − 1 〉 So, ‖ r ′ ( 0 ) × r ″ ( 0 ) ‖ = 0 + 0 + ( − 1 ) 2 = 1 . So, the curvature at t = 0 is, k ( 0 ) = ‖ r ′ ( 0 ) × r ″ ( 0 ) ‖ ‖ r ′ ( 0 ) ‖ 3 = 1 1 3 = 1 . Thus, the radius of the osculating circle at t = 0 is r ( 0 ) = 1 k ( 0 ) = 1 1 = 1

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ISBN: 9781319050740

Author: Jon Rogawski, Colin Adams, Robert Franzosa

Publisher: W. H. Freeman

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Question

Chapter 13.4, Problem 68E

To determine

The equation of the osculating circle of the curve r(t)=〈cosht,sinht〉 at the point t=0 or state that the osculating circle does not exist.

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Chapter 13.4, Problem 67E

Chapter 13.4, Problem 69E

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The equation of the osculating circle of the curve r ( t ) = 〈 cosh t , sinh t 〉 at the point t = 0 or state that the osculating circle does not exist.

Solution Summary: The author states that the osculating circle of the curve r(t)=langle mathrmcosh

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The equation of the osculating circle of the curve r(t)=〈cosht,sinht〉 at the point t=0 or state that the osculating circle does not exist.

Want to see the full answer?

Chapter 13 Solutions