Find the equation of the tangent line to $W$ at the point where $t=\dfrac{\pi}{6}$.
Let $W$ be the curve with parametric equations $\begin{cases} u(t) = \cos(3t) \\ v(t) = \cos(2t) \end{cases}$. Find the equation of the tangent line to $W$ at the point where $t=\dfrac{\pi}{6}$.
Enter your answer in the form $\verb#y=mx+b#$.
I don't have any specific formula in mind.
$\endgroup$ 03 Answers
$\begingroup$Alright, so you remember how we define slope as $\frac{dv}{du}$?
So $\frac{dv}{du}$ is equal to $\displaystyle \frac{\frac{dv}{dt}}{\frac{du}{dt}}$.
$u(\frac{π}{6})=0$ and $v(\frac{π}{6})=\frac{1}{2}$.
When $t=\frac{π}{6}$, $\frac{du}{dt}=3$ and $\frac{dv}{dt}=\sqrt{3}$.
So, $\frac{dv}{du}=\frac{\sqrt{3}}{3}$.
The tangent line is $\boxed{y=\frac{\sqrt{3}}{3}x+\frac{1}{2}}$.
Have a nice weekend!
$\endgroup$ $\begingroup$just find the gradient of the function $f : \mathbb{R} \longrightarrow \mathbb{R}^2, t \mapsto \binom{u(t)}{v(t)}$ at that point and then you know that the tangent to the curve is orthogonal to the gradient vector.
$\endgroup$ $\begingroup$If your curve is $\varphi(t)=\begin{pmatrix}u(t)\\v(t)\end{pmatrix}$, then you get the parametric form of your tangent by $T(t)=\varphi'\left(\frac{\pi}6\right)t+\varphi\left(\frac{\pi}6\right)$.
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