# Find all points (if any) of horizontal and vertical tangency to the portion of the curve shown....

## Question:

Find all points (if any) of horizontal and vertical tangency to the portion of the curve shown. (Enter your answers as a comma-separated list.)

Involute of a circle:

{eq}x=cos\theta +\theta sin\theta {/eq}

{eq}y=sin\theta - \theta cos \theta {/eq}

{eq}0 < \theta \le 2\pi {/eq}

## Finding a Tangent Line to a Parametric Curve

The slope of a tangent line to a function {eq}y=f(x) {/eq} is given by the derivative of the function with respect to x, {eq}\dfrac{dy}{dx} = f'(x) {/eq}. If instead the function is defined parametrically as {eq}x=f(t) {/eq} , {eq}y=g(t) {/eq}, the slope of the tangent line to the curve is still given by {eq}\dfrac{dy}{dx} {/eq} where {eq}\dfrac{dy}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}} = \dfrac{g'(t)}{f'(t)} {/eq}

A tangent line is horizontal when its slope is zero and is vertical when it has undefined slope. To find all horizontal tangent lines, find all points where the derivative is equal to zero. To find all vertical tangent lines, find all points where the derivative is undefined. Finding the derivative, we have

{eq}\dfrac{dy}{dx} = \dfrac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\\ \dfrac{dy}{dx} = \dfrac{\cos(\theta) + \theta\sin(\theta) - \cos(\theta)}{-\sin(\theta) + \theta\cos(\theta) + \sin(\theta)}\\ \dfrac{dy}{dx} = \dfrac{\theta\sin(\theta)}{\theta\cos(\theta)}\\ \dfrac{dy}{dx} = \dfrac{\sin(\theta)}{\cos(\theta)} {/eq}

The tangent line is horizontal when the numerator is zero but the denominator is not. We have {eq}\sin(\theta) = 0\\ \theta = \pi, 2\pi {/eq}

To find the points where the tangent line is horizontal, substitute the values into the equations for x and y.

{eq}x=\cos(\pi) +\pi\sin(\pi) = -1\\ y=\sin(\pi) - \pi\cos(\pi) = \pi\\ {/eq}

and

{eq}x=\cos(2\pi) +2\pi\sin(2\pi) = 1\\ y=\sin(2\pi) - 2\pi\cos(2\pi) = -2\pi\\ {/eq}

So the tangent lines are horizontal at the points {eq}(-1, \pi) {/eq} , {eq}(1, -2\pi) {/eq}.

The tangent lines are vertical when the denominator of the derivative is zero but the numerator is not zero. We have

{eq}\cos(\theta) = 0\\ \theta = \frac{\pi}{2} , \frac{3\pi}{2} {/eq} To find the points where the tangent line is vertical, substitute the values into the equations for x and y.

{eq}x=\cos(\pi/2) +\frac{\pi}{2}\sin(\pi/2) = \frac{\pi}{2}\\ y=\sin(\pi/2) - \frac{\pi}{2}\cos(\pi/2) = 1\\ {/eq}

and

{eq}x=\cos(3\pi/2) +\frac{3\pi}{2}\sin(3\pi/2) = -\frac{3\pi}{2}\\ y=\sin(3\pi/2) - \frac{3\pi}{2}\cos(3\pi/2) = -1\\ {/eq}

So the tangent lines are vertical at the points {eq}\left(\frac{\pi}{2}, 1\right) {/eq} , {eq}\left(-\frac{3\pi}{2}, -1\right) {/eq}.