Explanation Given: The parametric curve is ( 1 + ln t , 1 t ) . Concept used: The derivative d y d x in the parametric form is, d y d x = d y d t d x d t Here, d y d t and d x d t are the derivatives with respect to t . Calculation: The given parametric curve is ( 1 + ln t , 1 t ) . Therefore, the value of x in terms of t is, x = 1 + ln t The value of y in terms of t is, y = 1 t To eliminate the parameter t , y is solved as function of x . First, express the parameter t in terms of x as, x = 1 + ln t ln t = x − 1 e ln t = e x − 1 t = e x − 1 Now, substitute e x − 1 for t in the function y as, y = 1 e x − 1 Therefore, the equation y = f ( x ) is, y = 1 e x − 1 Here, the function f ( x ) is, f ( x ) = 1 e x − 1 The derivative of y with respect to t is, d y d t = d d t ( 1 t ) = d d t ( t − 1 ) = − ( t − 1 − 1

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 12.1, Problem 64E

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To find: An equation y=f(x) and dydx for the parametric curve.

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Chapter 12.1, Problem 63E

Chapter 12.1, Problem 65E

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Ch. 12.1 - Prob. 1PQ Ch. 12.1 - Prob. 2PQ Ch. 12.1 - Prob. 3PQ Ch. 12.1 - Prob. 4PQ Ch. 12.1 - Prob. 5PQ Ch. 12.1 - Prob. 6PQ Ch. 12.1 - Prob. 7PQ Ch. 12.1 - Prob. 1E Ch. 12.1 - Prob. 2E Ch. 12.1 - Prob. 3E

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An equation y = f ( x ) and d y d x for the parametric curve.

Solution Summary: The author calculates the parametric curve (1+mathrmlnt,

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To find: An equation y=f(x) and dydx for the parametric curve.

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Chapter 12 Solutions