7.2 Calculus of Parametric Curves

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Instructor: Number two, find the equation of the tangent line to the curve defined by these equations. And specifically, we're given down here when t equals 2. So the equation of the tangent line is going to be the same as it always has been. First we need to find our derivative. Following the same pattern here, the derivative of y is 2. Derivative of x is 2t. So our slope is-- our slope function in terms of t, is 1 over t. So the slope of this line is going to be dy dx evaluated at t equals 2, which would make our slope 1/2. 

Now what is our point at t equals 2? Well, our point with t equals 2, is going to be 1 comma. Plugging in 1 there, we get-- that can't be right. Nope, that's a 3. There we go. So 1, 3. I plugged in 1 there. OK. 

So we have the point 1, 3, and a slope of 1/2. So y minus y1 equals mx minus x1. Use our point slope form there. 1/2 half times x minus 1. Distributing there. 1/2 x minus one half, which means y equals one half x plus 2 and 1/2. So 5/2. And there is the equation of our tangent line with this curve. 

Example three. Calculate the second derivative. Now this is actually using the same function.

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