Explanation Definition Used: For a parametric equation x = f ( t ) , y = g ( t ) the derivative of y with respect to x in terms of t is given by d y d x = d g d t d f d t . Calculation: Note that, for a curve to have a horizontal tangent the slope of the tangent must be zero and for a slope to have a vertical tangent the slope of the tangent must be ∞ . The slope of a tangent is nothing but the value of derivative d y d x of the curve. Use definition and calculate the value d y d x as shown below. d y d x = d d t ( 2 sin t ) d d t ( sin 2 t ) = 2 cos t 2 cos 2 t = cos t cos 2 t To obtain the horizontal tangent, equate d y d x to 0. d y d x = 0 cos t cos 2 t = 0 cos t = 0 t = cos − 1 ( 0 ) The given interval is 0 ≤ t ≤ 2 π . In the given interval t = cos − 1 ( 0 ) has two solutions namely t = π 2 and t = 3 π 2 . Observe that t = π 2 corresponds to the point ( sin 2 π 2 , 2 sin π 2 ) = ( 0 , 2 ) and t = 3 π 2 corresponds to the point ( sin 6 π 2 , 2 sin 3 π 2 ) = ( 0 , − 2 ) . Therefore, the curve x = sin 2 t , y = 2 sin t ; 0 ≤ t ≤ 2 π has the horizntal tangent at ( 0 , 2 ) and ( 0 , − 2 ) . To obtain the vertical tangent equate d y d x to ∞ . d y d x = ∞ cos t cos 2 t = ∞ cos 2 t = 0 t = cos − 1 ( 0 ) 2 The given interval is 0 ≤ t ≤ 2 π

MyLab Math with Pearson eText -- 24 Month Access -- for Calculus with Integrated Review

3rd Edition

ISBN: 9780135243435

Author: Bill Briggs

Publisher: Pearson Education (US)

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Chapter 12.1, Problem 101E

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To estimate: The points at which the curve x=sin2t,y=2sint;0≤t≤2π has the vertical and horizontal tangent; sketch the graph.

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Chapter 12.1, Problem 100E

Chapter 12.1, Problem 102E

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Chapter 12 Solutions

MyLab Math with Pearson eText -- 24 Month Access -- for Calculus with Integrated Review

Ch. 12.1 - Identify the graph generated by the parametric...Ch. 12.1 - Prob. 2QC Ch. 12.1 - Prob. 3QC Ch. 12.1 - Prob. 4QC Ch. 12.1 - Prob. 5QC Ch. 12.1 - Prob. 6QC Ch. 12.1 - Explain how a pair of parametric equations...Ch. 12.1 - Prob. 2E Ch. 12.1 - Prob. 3E Ch. 12.1 - Prob. 4E

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To estimate: The points at which the curve x = sin 2 t , y = 2 sin t ; 0 ≤ t ≤ 2 π has the vertical and horizontal tangent; sketch the graph.

Solution Summary: The author illustrates how the slope of a tangent is nothing but the value of derivative dx of the curve.

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To estimate: The points at which the curve x=sin2t,y=2sint;0≤t≤2π has the vertical and horizontal tangent; sketch the graph.

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Chapter 12 Solutions