37-Calculus-of-ParametricEq

# 3 3 2 3 2 2 5 12 2 1 5 12 8 2 1 5 12 8 t t t dy

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• hain2005
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3 0 3 2 3 2 2 5 12 2 1 5 12 8 2 1 5 12 8 t t t dy undefined dx t t dy dx t t dy dx t t    The graph is shown and it is clear that you can rule out t = 0 . There is no vertical tangent at the point (0,4) . Only the slopes of 1 8 are possible. And from which, you find that the 2 possibles tangent lines are: 4 8 x y In other words, there are 2 tangent lines for this parametric curve something that you had not seen in Cal I.

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Arc Length of Parametric Equation The familiar arc length equation is: 2 2 L dx dy Multiply the quantity under the radical sign 2 dt dt then take one of the 2 dt outside and we will have: 2 2 2 2 2 2 2 2 dt dx dy L dx dy dx dy dt dt dt dt dt dt dy dt dx L t t 2 1 2 2 Note that the integration is performed with respect to “dt” and the limits of integration are in terms of the parameter “t”. Find the length for 3 2 2 x t y t for t = 0 to t = 1 . 2 2 2 1 1 4 2 2 1 0 0 9 16 9 16 t t dx dy L dt t t dt t t dt dt dt You can integrate via u-substitution by letting 2 9 16 18 u t du t dt Find the length for cos sin x t y t for t = 0 to t = 2π 2 2 2 2 2 2 2 1 0 0 sin cos 1 2 t t dx dy L dt t t dt dt dt dt The parametric equation is a circle with a radius of 1 and what you found is the circumference of the circle. Note how easy it is to work with the parametric form . Using the length integral on the original form 2 2 1 x y can be tricky!
Find the length for 3 3 cos sin x t y t for t = 0 to t = 2 π The graph of this curve is shown below and as you can see, you will have issue with the sharp points because the derivative does not exist. If you can find the length of just one segment and times 4, you can get the total length. And ¼ of the graph is generated for t = 0 to t = π /2. 2 2 2 /2 /2 4 2 4 2 2 2 1 0 0 9cos sin 9sin cos 9cos sin t t dx dy L dt t t t t dt t t dt dt dt Therefore: /2 /2 2 2 0 0 9cos sin 3 cos sin L t t dt t t dt Using the u-substitution with sin cos u t du tdt You will find that /2 /2 2 0 0 3 3 cos sin sin 1.5 2 L t t dt t The total length is 4 times this: (4)(1.5) = 6 units.

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The circle circumference revisited. Find the length for cos sin x t y t for t = 0 to t = This is nearly identical to the example that you previously saw. The difference is that one of the integration limits extends to . Let’s see what we get: 2 2 2 6 6 2 2 1 0 0 sin cos 1 6 t t dx dy L dt t t dt dt dt dt
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