Explanation Given: y = e x , x = 0 Calculation: y = e x y ' = e x y ' ' = e x If R is radius curvature will be, k = 1 R Where, k = | y ' ' | ( 1 + ( y ' ) 2 ) 3 2 = ( e x ) ( 1 + ( e x ) 2 ) 3 2 When x = 0 k = 1 ( 1 + 1 ) 3 2 = 1 2 2 So, R can be determined k = 1 R k = 1 2 2 R = 2 2 The equation of the circle can be described as, ( x − a ) 2 + ( y − b ) 2 = R 2 Where, ( a , b ) - Coordinates of the center of circle and R - its radius If above equation is denoted as below, f = ( x − a ) 2 + ( y − b ) 2 − R 2 Then derivatives of f with respect to x will as follow, f ' = 2 ( x − a ) + 2 ( y − b ) y ′ f ' ' = 2 + 2 ( y ′ ) 2 + 2 ( y − b ) y ′ ′ Assuming that the curves osculate at the point ( x 0 , y 0 ) the equations of the osculating circle ( a

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Chapter 14.4, Problem 63E

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Findthe equation of the osculating circle at the point indicated or indicate that none exists.

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Chapter 14.4, Problem 62E

Chapter 14.4, Problem 64E

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Chapter 14 Solutions

CALCULUS (CLOTH)

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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Findthe equation of the osculating circle at the point indicated or indicate that none exists.

Solution Summary: The author explains that the equation of the osculating circle is (x+2)2+(y-3).

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Findthe equation of the osculating circle at the point indicated or indicate that none exists.

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Chapter 14 Solutions