guys how are you doing today? Can you take some time to look over this equation of a curve defined in terms of $\theta $ and suggest how I can find the equation of the normal at the point $(a, \frac{a\pi}{2} - a)? $ In other words, finding the normal when $\theta = \frac{\pi}{2}\,$.
$$ x(\theta) = a(\theta - sin(\theta)) \\y(\theta) = a(1 - cos(\theta)) $$
I'm trying to rewrite the parametric equations in terms of $\theta$ so that I can equate the two equations making a Cartesian equation with just x and y. Then I can take derivative with respect to x.
$$ y(\theta) = a(1 - cos(\theta))\\\frac{y}{a} = 1 - cos(\theta)\\1-\frac{y}{a} = cos(\theta) \\ \theta = arccos(1-\frac{y}{a})$$
Rewriting the parametric equation $x(\theta) = a(\theta-sin(\theta))$ is very hard to do.
I need to find the slope of the tangent at $\theta = \frac{\pi}{2}$ which is the slope at $(a, \frac{a\pi}{2} - a)$ then I can find the normal, right?