# How do you find the tangent lines of parametric curves?

## Show that the curve with parametric equations $x = \sin t$, $y = \sin \left(t + \sin t\right)$ has two tangent lines at the origin and find their equations. Illustrate by graphing the curve and its tangents.

Jan 16, 2018

$y = 2 x$ and $y = 0$

#### Explanation:

$x = \sin \left(t\right) , y = \sin \left(t + \sin \left(t\right)\right)$

First find the derivative:

$\frac{\mathrm{dx}}{\mathrm{dt}} = \cos t$

$\frac{\mathrm{dy}}{\mathrm{dt}} = \left(1 + \cos t\right) \cos \left(t + \sin t\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}} = \frac{\left(1 + \cos t\right) \cos \left(t + \sin t\right)}{\cos} t$

At the origin we have: $x = 0$ and $y = 0$ so from the x-coordinate:

$0 = \sin \left(t\right) \to {t}_{0} = 0 , {t}_{1} = \pi$

(There are of course more values which may satisfy this but the periodicity of the sine and cosine functions means that the other solutions such as $t = 2 \pi , 4 \pi , \ldots$ will return the same tangent as $t = 0$ and $t = 3 \pi , 5 \pi , \ldots$ will return the same tangent as $t = \pi$ so we need only consider the two solutions stated).

Checking that these values of $t$ satisfy $y = 0$

$y = \sin \left(0 + \sin \left(0\right)\right) = 0$
$y = \sin \left(\pi + \sin \left(\pi\right)\right) = \sin \left(\pi + 0\right) = 0$

So $y$ is also satisfied. Put these values of $t$ into $\frac{\mathrm{dy}}{\mathrm{dx}}$ to find the gradient of the tangent line:

For $t = 0$:
${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{0} = \frac{\left(1 + \cos \left(0\right)\right) \cos \left(0 + \sin 0\right)}{\cos} \left(0\right) = \frac{2 \cdot \cos \left(0\right)}{\cos} \left(0\right) = 2$

For $t = \pi$

${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{1} = \frac{\left(1 + \cos \left(\pi\right)\right) \cos \left(\pi + \sin \pi\right)}{\cos} \left(\pi\right)$

${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{1} = \frac{\left(1 - 1\right) \cos \left(\pi\right)}{\cos} \left(\pi\right) = 0$

So we have the gradients to our two tangent lines:

${m}_{\tan 0} = 2$ and ${m}_{\tan 1} = 0$

We also have a point at which the lines pass through, the origin: $\left(0 , 0\right)$

Obviously, for the equation of a straight line $y = m x + c$ any line which passes through the origin will simply have $c = 0.$ Therefore our two tangents will be:

${y}_{1} = 2 x$ and $y = 0$

When graphed, they look like this:

The blue line is the original parametric function and, the red line is $y = 2 x$ and the green line (lying right on the x-axis) is $y = 0$.