Explanation Given: Equation of curve y = x 2 and point x = 1 Formula used: O Q → = r → ( t 0 ) + 1 k ( t 0 ) N ∧ ( t 0 ) where O Q → is the center of circle T ∧ ( t ) = r → ( t ) ‖ r → ( t ) ‖ N ∧ ( t ) = T ∧ ' ( t ) ‖ T ∧ ' ( t ) ‖ K ( t ) = ‖ r → ' ( t ) × r → " ( t ) ‖ ‖ r → ' ( t ) ‖ 3 Equation of circle C → ( t ) = O Q → + 1 K ( t 0 ) 〈 cos t , sin t 〉 Calculation: Equation of space curve y = x 2 . this equation can be written as parametric form. r → ( x ) = 〈 x , f ( x ) 〉 r → ( x ) = 〈 x , x 2 〉 r → ' ( x ) = 〈 1 , 2 x 〉 T ∧ ( x ) = 〈 1 , 2 x 〉 ‖ 〈 1 , 2 x 〉 ‖ = 〈 1 , 2 x 〉 1 + 4 x 2 N ∧ ( x ) = T ' ∧ ( x ) ‖ T ∧ ' ( x ) ‖ T ∧ ' ( x ) = d d x ( 〈 1 , 2 x 〉 1 + 4 x 2 ) = d d x 〈 1 1 + 4 x 2 , 2 x 1 + 4 x 2 〉 = 〈 d d x ( 1 1 + 4 x 2 ) , d d x ( 2 x 1 + 4 x 2 ) 〉 〈 − 1 2 × 8 x ( 1 + 4 x 2 ) 3 2 , 1 + 4 x 2 .2 − 2 x × 1 2 × 8 x 1 + 4 x 2 ( 1 + 4 x 2 ) 〉 ( apply chain rule of differentiation ) = 〈 − 4 x ( 1 + 4 x 2 ) 3 2 , 2 ( 1 + 4 x 2 ) − 8 x 2 ( 1 + 4 x 2 ) 3 2 〉 Simplification = 〈 − 4 x ( 1 + 4 x 2 ) 3 2 , 2 ( 1 + 4 x 2 ) 3 2 〉 N ∧ ( x ) = T ∧ ' ( x ) ‖ T ∧ ' ( x ) ‖ = 〈 − 4 x ( 1 + 4 x 2 ) 3 2 , 2 ( 1 + 4 x 2 ) 3 2 〉 ( − 4 x 1 + 4 x 2 ) 3 2 + ...

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Author: Rogawski

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Chapter 14.4, Problem 59E

To determine

The parametrization of osculating circle at the point indicated

$y = x^{2}, x = 1$

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Chapter 14.4, Problem 58E

Chapter 14.4, Problem 60E

Chapter 14 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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 parametrization of osculating circle at the point indicated y = x 2 , x = 1

Solution Summary: The author explains the parametrization of osculating circle at the point indicated y = x 2. The equation can be written as parametric form.

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The parametrization of osculating circle at the point indicated

$y = x^{2}, x = 1$

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Chapter 14 Solutions

The parametrization of osculating circle at the point indicated y = x 2 , x = 1

Solution Summary: The author explains the parametrization of osculating circle at the point indicated y = x 2. The equation can be written as parametric form.

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The parametrization of osculating circle at the point indicated y=x2,x=1

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Chapter 14 Solutions

The parametrization of osculating circle at the point indicated

$y = x^{2}, x = 1$