Function parameters and cartesian curves

Question

Given

$$x = \cos t + \cos 2t \,,\; y = \sin t + \sin 2t ,$$

find the tangent line for the parameter at point $(-1, 1),$ and draw a graph of the curve.

To find the point you could simply do $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ and find $\dfrac{dy}{dx}$ by taking the quotient, so you get $$ \dfrac{dy}{dx} =\dfrac{\cos t+2 \cos 2t}{-\sin t-2 \sin 2t}$$

How exactly would you define parameter $t$ in this case as a value to find the equation of the tangent? And when trying to draw the graph, you would need the Cartesian curve, which can be fetched by eliminating the parameter by solving for $t$. Is there a trigonometric identity that would make it possible to solve for $t$ using $X$ and $Y$ equations?

Answer 1

Firstly, you forgot to use the chain rule. Remember that $\frac d{dt}\sin 2t=2\cos 2t$, etc. So instead we have $$\frac {dy}{dx}=\frac{\cos t+2\cos t}{-\sin t-2\sin 2t}.$$

As you can see this is a functino of $t$. So to find the tangent line at $(-1,1)$, we need to know for what value of $t$ our parametric curve passes through $(-1,1)$. So we solve $$\begin{cases}\cos t+\cos 2t=-1\\ \sin t+\sin 2t =1.\end{cases}$$ Now \begin{gather} \cos t+\cos 2t = -1\\ \cos t+\cos^2t-\sin^2 t=-\sin^2 t-\cos^2t\\ \cos t+2\cos^2t=0\\ \cos t(1+2\cos t)=0\\ \cos t=0\vee 2\cos t=-1\\ t=\frac \pi2+k\pi\vee t=\frac{\pm2\pi}3+2k\pi \end{gather}

Now take $\tilde t_k=\frac{\pm2\pi}3+2k\pi$. We can simply calculate that $\sin \tilde t_k +\sin 2\tilde t_k=0$, so non of these is a solution to our system of equations.

Now we check the other possibilities for $t$ and see that $$\sin \left(\frac\pi2+k\pi\right)+\sin (\pi+2k\pi)=\begin{cases}1&\text{for $k$ odd}\\ -1& \text{for $k$ even}\end{cases}. $$

So we find that our curve passes through $(-1,1)$ whenever $t=\frac\pi2 +(2m+1)\pi$, for some integer $m$.

Now just substitute this $t$ into the expression we found for $\frac {dy}{dx}$ and you'll be able to find the tangent line at $(-1,1)$.

Answer 2

You are on the parametric form, it is not necessary to know the geometric significance of $t$,

$$x = \cos t + \cos 2t \,,\; y = \sin t + \sin 2t ,$$

The given point is generated for parameter $t = \pi/2 .$ Using standard form and slope computed, evaluate for this point and it is done.

$$ \dfrac{y-(\sin t + \sin 2t)}{x-(\cos t + \cos 2t) } = \dfrac{\cos t+2 \cos 2t}{-\sin t-2 \sin 2t}. $$

Answer 3

Another way by finding implicit equation \begin{align*} x^{2} &= \cos^{2} t+2\cos t \cos 2t+ \cos^{2} 2t \\ y^{2} &= \sin^{2} t+2\sin t \sin 2t+ \sin^{2} 2t \\ x^{2}+y^{2} &= 2+2(\cos t \cos 2t+\sin t \sin 2t) \\ &= 2+2\cos t \\ \cos t &= \frac{x^{2}+y^{2}}{2}-1 \\ x &= \cos t+2\cos^{2} t-1 \\ &=(\cos t+1)(2\cos t-1) \\ &= \frac{(x^{2}+y^{2})}{2} (x^{2}+y^{2}-3) \\ 0 &= (x^{2}+y^{2})(x^{2}+y^{2}-3)-2x \\ 0 &= x^{4}+2x^{2}y^{2}+y^{4}-3x^{2}-3y^{2}-2x \\ \end{align*}

In principle, the tangent can be found by implicit differentiation and plug in the point $(-1,1)$. The rest of work is left as an exercise, but showing the graph instead:

P.S.: The curve is actually a shifted limaçon of Pascal, $r=2\cos \theta+1$.