Calculus Notes 11.2

# Calculus Notes 11.2 - Calculus- Stewart Dr. Berg Summer...

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Unformatted text preview: Calculus- Stewart Dr. Berg Summer ‘09 Page 1 11.2 11.2 Calculus With Parametric Curves Tangents Suppose a curve is defined by the equation y = F ( x ) and can also be expressed as the parametric equations x = f ( t ) and y = g ( t ) where F , f , and g are all differentiable. Then when we differentiate y = g ( t ) = F ( x ) = F ( f ( t )) with respect to t and apply the chain rule, we get ′ g ( t ) = ′ F ( f ( t )) ′ f ( t ) = ′ F ( x ) ′ f ( t ) so that ′ F ( x ) = ′ g ( t ) ′ f ( t ) if ′ f ( t ) ≠ . Using Leibniz notation yields: Proposition dy dx = dy dt dx dt if dx dt ≠ . Note that a curve has a horizontal tangent when dy dt = (provided that dx dt ≠ ) and has a vertical tangent when dx dt = (provided that dy dt ≠ ). It can also be useful to consider d 2 y dx 2 . Corollary d 2 y dx 2 = d dt dy dx dx dt . Example A Find the x- y coordinates of the places where the curve given by x = t 3 − 3 t and y = 3 t 2 − 9 has vertical or horizontal tangents. Indeed dy dx = D t 3 t 2 − 9 ( ) D t t 3 − 3 t ( ) = 6 t 3 t 2 − 3 = 2 t ( t − 1)( t + 1) . Thus, we get a horizontal tangent when t = at (0, –9), and there are vertical tangents when t = ± 1 at (–2, –6) and (2, –6). Here is the graph of the curve: Calculus- Stewart Dr. Berg Summer ‘09 Page 2 11.2 Notice that the curve crosses itself at the origin. This corresponds to two values of Notice that the curve crosses itself at the origin....
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## This note was uploaded on 06/06/2010 for the course M 408 taught by Professor Hodges during the Spring '08 term at University of Texas.

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Calculus Notes 11.2 - Calculus- Stewart Dr. Berg Summer...

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