Question

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.

Answer 1

Hint: In this question we have to find the tangents of the given parametric equations, first we will derive the two equations then we will find ${\displaystyle \frac{dy}{dx}}$, now we will get two values for $t$ by substituting the origin in the two equations as the two equations passes through the origin, now substitute the two values in ${\displaystyle \frac{dy}{dx}}$ to get the slopes of the required tangents, now by using slope intercept form i.e., $y=mx+b$, we will get the required tangent lines.

Complete step by step solution:
Given parametric equations are $x=\sin t$, $y=\sin \left(t+\sin t\right)$,
Now we will derive the first equation i.e.,
$\Rightarrow {\displaystyle \frac{dx}{dt}}={\displaystyle \frac{d}{dt}}\sin t$,
Now simplifying we get,
$\Rightarrow {\displaystyle \frac{dx}{dt}}=\cos t$,
Now we will derive the second equation we get,
$\Rightarrow {\displaystyle \frac{d}{dt}}y={\displaystyle \frac{d}{dt}}\sin \left(t+\sin t\right)$,
Now simplifying by applying derivatives we get,
$\Rightarrow {\displaystyle \frac{dy}{dt}}=\cos \left(t+\sin t\right)\left({\displaystyle \frac{d}{dt}}\left(t+\sin t\right)\right)$,
Further simplifying we get,
$\Rightarrow {\displaystyle \frac{dy}{dt}}=\cos \left(t+\sin t\right)\left(1+\cos t\right)$,
Now we divide both the equations we get,
$$\Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}}={\displaystyle \frac{\cos \left(t+\sin t\right)\left(1+\cos t\right)}{\cos t}}$$,
Now simplifying we get,
$$\Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{\cos \left(t+\sin t\right)\left(1+\cos t\right)}{\cos t}}$$,
Now we have to find the tangent at the origin i.e., $\left(0,0\right)$, so here substitute $x=0$, we get,
$\Rightarrow 0=\sin t$,
Now by trigonometric ratios we get,
$\Rightarrow t=0$or $t=\pi$,
Now substitute these values of $t$ into ${\displaystyle \frac{dy}{dx}}$ to find the equation of tangent line,
First substitute $t=0$, we get,
$\Rightarrow {{\displaystyle \frac{dy}{dx}}}_{t=0}={\displaystyle \frac{\cos \left(0+\sin 0\right)\left(1+\cos 0\right)}{\cos 0}}$,
Now simplifying we get,
$\Rightarrow {{\displaystyle \frac{dy}{dx}}}_{t=0}={\displaystyle \frac{\cos \left(0\right)\left(1+\cos 0\right)}{\cos 0}}$,
Further simplifying we get,
$\Rightarrow {{\displaystyle \frac{dy}{dx}}}_{t=0}={\displaystyle \frac{1\left(1+1\right)}{1}}={\displaystyle \frac{2}{1}}=2$,
Next substitute $t=\pi$, we get,
$\Rightarrow {{\displaystyle \frac{dy}{dx}}}_{t=\pi }={\displaystyle \frac{\cos \left(\pi +\sin \pi \right)\left(1+\cos \pi \right)}{\cos \pi }}$,
Now simplifying we get,
$\Rightarrow {{\displaystyle \frac{dy}{dx}}}_{t=\pi }={\displaystyle \frac{\cos \left(\pi +0\right)\left(1+\cos \pi \right)}{\cos \pi }}$,
Further simplifying we get,
$\Rightarrow {{\displaystyle \frac{dy}{dx}}}_{t=\pi }={\displaystyle \frac{\cos \left(\pi \right)\left(1+\cos \pi \right)}{\cos \pi }}$,
Now simplifying we get,
$\Rightarrow {{\displaystyle \frac{dy}{dx}}}_{t=\pi }={\displaystyle \frac{\left(-1\right)\left(1-1\right)}{-1}}=0$,
So we have the gradients to our two tangent lines and they are $m_{\tan 0}=2$ and $m_{\tan 1}=0$, and we know that the tangent passes through the origin i.e.,$\left(0,0\right)$,
So, for the equation of a straight line $y=mx+b$ which passes through the origin, we get,
$\Rightarrow y=mx+0$.
Now substituting the two value of $m$in the slope intercept form, then our two tangents will be:
First tangent is when $m=2$, by substituting we get,
$\Rightarrow y_1=\left(2\right)x+0=2x$,
Now second tangent is when $m=0$, by substituting value we get,
$\Rightarrow y_2=\left(0\right)x+0=0$,
So, the two tangents will be $y=0$ and $y=2x$, now we will graph all the tangents, and the given curves then the graph will be

Note: Parametric equation, a type of equation that employs an independent variable called a parameter often denoted by $t$ and in which dependent variables are defined as continuous functions of the parameter and are not dependent on another existing variable.