# How we can find the perpendicular line?

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?

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

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

• 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? – Mary Star Sep 29 '15 at 23:48
• 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. – Liam Sep 30 '15 at 0:13

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