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

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ISBN: 9781319434526

Author: Rogawski

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Chapter 12.1, Problem 63E

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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 62E

Chapter 12.1, Problem 64E

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Chapter 12 Solutions

CALCULUS 4E (LL) W/ ACHIEVE ACCESS

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 explains how the parametric curve is (1-et,t-1).

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

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Chapter 12 Solutions