Explanation Given information: The curve is given as a function of x such that, y = x Formula used: The curvature of a curve is given by, κ ( x ) = | f ' ' ( x ) | ( 1 + f ' ( x ) 2 ) 3 / 2 The normal vector N , N ( x ) = T ' ( x ) ‖ T ' ( x ) ‖ where T ( x ) is the unit tangent vector. The vector for the center of the osculating circle. r ( x 0 ) + 1 κ ( x ) N Let the vector function be, r ( x ) = 〈 x , x 〉 The first derivative of y is, y ' = 1 2 x and, The second derivative od y is, y ' ' = − 1 4 x x The curvature at x = 1 can be found by the formula, κ ( x ) = | y ' ' ( x ) | ( 1 + y ' ( x ) 2 ) 3 / 2 = | − 1 4 x x | ( 1 + ( 1 2 x ) 2 ) 3 / 2 κ ( x = 4 ) = | − 1 4 ( 4 ) 4 | ( 1 + 1 4 ( 4 ) ) 3 / 2 = 1 / 32 ( 17 / 16 ) 3 / 2 = 1 32 × 16 16 17 17 κ ( 4 ) = 2 17 17 The derivative of the vector function is, r ' ( x ) = 〈 1 , 1 2 x 〉 An orthogonal vector for r ' ( x ) can be defined as,

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ISBN: 9781319434526

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 14, Problem 37CRE

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To find: The equation of the osculating circle of the given curve at x=4 .

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Chapter 14, Problem 36CRE

Chapter 14, Problem 38CRE

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Chapter 14 Solutions

CALCULUS 4E (LL) W/ ACHIEVE ACCESS

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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The equation of the osculating circle of the given curve at x = 4 .

Solution Summary: The author explains that the equation of the osculating circle at x=4 can be found by the formula.

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Chapter 14 Solutions