Parametric Calculus: Tangents: Example 1 Part 2 (DTube)

mes(74)

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2018-05-02 23:38

In this video I go over Part 2 of Example 1 on using calculus to determine tangents for parametric curves. In this part I show how to determine the points at which the tangent line corresponds to a horizontal or vertical asymptote line. For horizontal tangents, the derivative dy/dx is equal to zero, and this can be translated into parametric derivative form through the equation dy/dx = (dy/dt)/(dx/dt). Thus in order for dy/dx = 0, we must have the condition that dy/dt = 0 and dx/dt is not equal to zero (i.e. to not get an indefinite form 0/0). Likewise, for vertical asymptotes we need to consider when dy/dx approaches infinity, which is when dx/dt = 0 (while dy/dt is not equal to 0). Applying these equations, I show how we can solve for the parameter values, which we can then place inside the parametric equations to determine the x and y coordinates. This is a very interesting video on how to apply calculus to determine the points where a parametric curve slopes horizontally or vertically, so make sure to watch this video! Also stay tuned for Part 3 where I look for intervals of concavity and then sketch the entire curve.

Download the notes in my video: https://1drv.ms/b/s!As32ynv0LoaIhuI-182aD0bAKUDETw

View Video Notes on Steemit: https://steemit.com/mathematics/@mes/parametric-calculus-tangents-example-1-part-2