Vertical tangents will occur where the derivative is

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Vertical tangents will occur where the derivative is not defined and so we’ll get vertical tangents at values of t for which we have, Vertical Tangent for Parametric Equations 0, provided 0 dx dy dt dt = Let’s take a quick look at an example of this. Example 2 Determine the x-y coordinates of the points where the following parametric equations will have horizontal or vertical tangents. 3 2 3 3 9 x t t y t = = Solution We’ll first need the derivatives of the parametric equations. ( ) 2 2 3 3 3 1 6 dx dy t t t dt dt = = = Horizontal Tangents We’ll have horizontal tangents where, 6 0 0 t t = = Now, this is the value of t which gives the horizontal tangents and we were asked to find the x-y coordinates of the point. To get these we just need to plug t into the parametric equations. Therefore, the only horizontal tangent will occur at the point (0,-9). Vertical Tangents In this case we need to solve, ( ) 2 3 1 0 1 t t = = ± The two vertical tangents will occur at the points (2,-6) and (-2,-6). For the sake of completeness and at least partial verification here is the sketch of the parametric curve.
Calculus II © 2007 Paul Dawkins 29 The final topic that we need to discuss in this section really isn’t related to tangent lines, but does fit in nicely with the derivation of the derivative that we needed to get the slope of the tangent line. Before moving into the new topic let’s first remind ourselves of the formula for the first derivative and in the process rewrite it slightly. ( ) ( ) d y dy d dt y dx dx dx dt = = Written in this way we can see that the formula actually tells us how to differentiate a function y (as a function of t ) with respect to x (when x is also a function of t ) when we are using parametric equations. Now let’s move onto the final topic of this section. We would also like to know how to get the second derivative of y with respect to x . 2 2 d y dx Getting a formula for this is fairly simple if we remember the rewritten formula for the first derivative above. Second Derivative for Parametric Equations 2 2 d dy d y d dy dt dx dx dx dx dx dt = =
Calculus II Note that, 2 2 2 2 2 2 d y d y dt d x dx dt Let’s work a quick example. Example 3 Find the second derivative for the following set of parametric equations. 5 3 2 4 x t t y t = =
© 2007 Paul Dawkins 30
Calculus II

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