Question

a curve is defined parametrically by the equations x=2sintheta y=cos^2(theta) find the equation of the tangent to this curve at (2;0)

Answer 1

x = $2sin\theta$  and y = $cos^2\theta$   at (2, 0)

The first thing that we should do is find the derivative so we can get the slope of the tangent line.

             $\frac{dy}{dx} = \large\frac{\large\frac{dy}{d\theta}}{\large\frac{dx}{d\theta}}$ $= \large\frac{2cos\theta}{-2sin\theta cos\theta}$ $=\large -\frac{1}{sin\theta}$

At this point we’ve got a small problem. The derivative is in terms of $\theta$ and all we’ve got is an x-y­ coordinate pair. The next step then is to determine the value(s) of $\theta$ which will give this  point. We find these by plugging the x and y values into the parametric equations and solving for $\theta$

           2 = 2$sin\theta$    $\to$    $\theta = \large\frac{\pi}{2} + 2\pi k$

           0 = $cos^2\theta$     $\to$    $\theta$ $= \large\frac{\pi}{2} + \pi k$

Since we already know the x and y coordinates of the point all that we need to do is find the slope of the tangent line.

             $m = \large\frac{dy}{dx} \large|_{x = -\large\frac{\pi}{2}} =$ 1

$y - y_0 = m (x - x_o)$

y - 0 = 1* (x - 2)

y = x - 2