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PP 11.2 - Calculus with Parametric Curves Some curves...

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Calculus with Parametric Curves
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Some curves defined by parametric equations x = f ( t ) and y = g ( t ) can also be expressed in the form y = F ( x ). This can be accomplish by eliminating the parameter. 2 , 1 : Example 2 - = - = t y t x 2 ) 1 ( 1 1 2 - + = + = - = x y x t t x
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If we substitute x = f ( t ) and y = g ( t ) in the equation y = F ( x ), we get: g ( t ) = F ( f ( t )) If g , F , and f are differentiable, the Chain Rule gives: g’ ( t ) = F’ ( f ( t )) f’ ( t ) = F’ ( x ) f’ ( t ) 0 ) ( , ) ( ) ( ) ( = t f t f t g x F But y = F ( x ), x = f ( t ) and y = g ( t ), so have found a formula for the slope of the tangent line.
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0 , to equivalent is 0 ) ( , ) ( ) ( ) ( = = dt dx dt dx dt dy dx dy t f t f t g x F
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Tangent Lines curve. parametric a be ) ( ), ( : Let t y y t x x C = = . 0 provided is to ing correspond point the o tangent t line the of slope The a t = = = a t dt dx dt dx dt dy a t
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Comments From the formula for the slope, it can be deduced that tangent lines are horizontal when 0 = dt dy Also, tangent lines are vertical when 0 = dt dx
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Example . 1 2 , 1 3 : Let 3 2 t y t x C + = + = Find the value of the parameter that yields the point (4, 3) on the curve.
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