0
$\begingroup$

The perpendicular line of a curve at a point $p$ is the line that goes through $p$ and is perpendicular to the tangent line at $p$.

Find the tangent and the perpendicular line of the curve $\gamma (t)=(2 \cos t-\cos 2t, 2\sin t-\sin 2t)$ at the point that correponds at $t=\frac{\pi}{4}$.

I have done the following:

The tangent line of the curve at $t=\frac{\pi}{4}$ is $$l(t)=\gamma (\frac{\pi}{4})+t\gamma '(\frac{\pi}{4}) \Rightarrow l(t)=(\sqrt{2}+(2-\sqrt{2})t, \sqrt{2}-1+\sqrt{2}t)$$

Is this correct?

Could you explain to me how we can find the perpendicular line?

$\endgroup$
  • 2
    $\begingroup$ The vector $<x, y>$ is perpendicular to $<-y, x>$ in $\mathbb{R}^2$. Apply this to the direction vector of your tangent line. $\endgroup$ – ChocolateAndCheese Sep 29 '15 at 23:30
  • $\begingroup$ The direction vector of the tangent line is $\gamma '(t)$ or $\gamma'(\frac{\pi}{4})$ ? @ChocolateAndCheese $\endgroup$ – Mary Star Sep 29 '15 at 23:37
  • 1
    $\begingroup$ That is correct! $\endgroup$ – ChocolateAndCheese Sep 29 '15 at 23:38
1
$\begingroup$

These are the steps you should take:

(1) find a vector parallel to $\gamma$ at $t = \pi/4$

Since you claim to have the tangent line already, simply take the difference of two points on the tangent line.

(2) find a vector perpendicular to the above vector, call it $v$. This bears no explanation.

(3) Then your solution will be

$$ (x_0,y_0 ) + v\cdot t = L(t)$$ Where $(x_0,y_0) = (x,y)\Big|_{t=\pi/4} $.

$\endgroup$
  • $\begingroup$ The tangent line is $l(t)=\gamma (\frac{\pi}{4})+t\gamma '(\frac{\pi}{4})$. So a vector parallel to the tangent line at $t=\frac{\pi}{4}$ is $\gamma '(\frac{\pi}{4})$, right? So to find the vector $v$ of step $(2)$ we have to do the following: $$v \cdot \gamma '(\frac{\pi}{4})=0 \Rightarrow (v_1, v_2) \cdot (2-\sqrt{2}, \sqrt{2})=0 \Rightarrow (2-\sqrt{2})v_1+\sqrt{2}v_2=0$$ Is it correct so far? Or have I undersood it wrong? $\endgroup$ – Mary Star Sep 29 '15 at 23:48
  • $\begingroup$ There are infinitely many vectors that are perpendicular to a given vector. Any and all of them are acceptable (assuming there are no other constraints on the $t$ values). The easiest is: given $(a,b)$ a perpendicular vector is $(-b,a)$. $$(a,b)\cdot(-b,a) = -ab+ab =0 $$ Therefore if $(2-\sqrt{2},\sqrt{2})$ is parallel, then $(-\sqrt{2}, 2-\sqrt{2})$ is perpendicular. $\endgroup$ – Liam Sep 30 '15 at 0:13
0
$\begingroup$

Notice, we have $$x=2\cos t-\cos 2t \implies \frac{dx}{dt}=-2\sin t+2\sin 2t$$ $$y=2\sin t-\sin 2t \implies \frac{dy}{dt}=2\cos t-2\cos 2t$$

Hence, the slope of the tangent to the curve is given as $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2\cos t-2\cos 2t}{-2\sin t+2\sin 2t}=\frac{\cos t-\cos 2t}{\sin 2t-\sin t}$$ Hence, the slope of the tangent at $t=\frac{\pi}{4}$ $$=\left(\frac{dy}{dx}\right)_{t=\frac{\pi}{4}}=\frac{\cos\frac{\pi}{4} -\cos \frac{\pi}{2}}{\sin \frac{\pi}{2}-\sin \frac{\pi}{4}}=\frac{\frac{1}{\sqrt 2}-0}{1-\frac{1}{\sqrt 2}}=\sqrt 2+1$$

Now, the point of tangency $(2\cos t-\cos 2t, 2\sin t-\sin 2t)$ at $t=\frac{\pi}{4}$ is $(\sqrt 2, \sqrt 2-1)$ hence, the equation of the tangent is given by point-slope form as follows

$$\color{}{y-(\sqrt 2-1)=(\sqrt 2+1)(x-\sqrt 2)}$$ $$\color{red}{y=(\sqrt 2+1)x+2\sqrt 2+1}$$

Now, the equation of the line perpendicular passing through the point $(\sqrt 2, \sqrt 2-1)$ to the tangent is given as $$\color{}{y-(\sqrt 2-1)=\frac{-1}{(\sqrt 2+1)}(x-\sqrt 2)}$$ $$\color{}{y-(\sqrt 2-1)=-(\sqrt 2-1)(x-\sqrt 2)}$$ $$\color{red}{y=-(\sqrt 2-1)x+1}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.