Find the normal vector to the given curve at given point

Question

Consider a curve whose parametrization is given by $\gamma(t)=(2\cos t-\cos(2t),2\sin t-\sin(2t))$. Find the normal line to the curve at the point $t=\dfrac{\pi}{4}$.

Now I know that the normal line is perpendicular to the tangent line at the point on the curve. So if my normal line is $N(t)=(N_1(t),N_2(t))$ then $\nabla {N(t)}\cdot\nabla\gamma(t)=0$ at the point $t$ where $N(t)$ lies on the curve.

Now I get that $\dfrac{dN_1(t)}{dt}=\dfrac{\cos(2t)-\cos t}{\sin(2t)-\sin t}\dfrac{dN_2(t)}{dt}$.

What to do next? I put $t=\pi/4$ to get a relation between $dN_1(t)/dt$ and $dN_2(t)/dt$, but that's it.

Answer 1

It is way easier to computationally just use what you've already said. You have $\gamma'(\pi/4) \in \mathbb{R}^2$ and it is very easy to get perpendicular vectors in the plane. Now let $\vec{n}$ be perpendicular to $\gamma'(\pi/4)$ then your normal line is given by $l: \vec{\gamma(\pi/4)} + t \vec{n}$.

Answer 2

Although it’s very easy to find a vector orthogonal to a given one in $\mathbb R^2$, you don’t need to do that explicitly here. One form of equation for a line is $$\mathbf n\cdot\mathbf x=\text{const}$$ where $\mathbf n$ is a vector normal to the line. You’ve already observed that the tangent and normal to a curve are orthogonal, so we can use $\nabla\gamma(\pi/4)$ as this normal. Thus, an equation of the normal line to the given curve at $t=\pi/4$ is $$\nabla\gamma\left(\frac\pi4\right)\cdot\mathbf x=\nabla\gamma\left(\frac\pi4\right)\cdot\gamma\left(\frac\pi4\right).$$ You should be able to take it from here.

Answer 3

The normal line to this curve does not need differential operators to be found; it can be obtained directly from the parametric representation. $$\gamma(t)=(x,y)=(2\cos t-\cos 2t,2\sin t-\sin 2t)$$ $$\gamma(\tfrac\pi4)=(\sqrt2,\sqrt2-1)$$ $$\gamma'(t)=(\tfrac{dx}{dt},\tfrac{dy}{dt})=(-2\sin t+2\sin 2t,2\cos t-2\cos 2t)$$ $$\gamma'(\tfrac\pi4)=(-\sqrt2+2,\sqrt2)$$ Since the normal to a curve at a point on the curve has slope $-\frac{dx}{dy}$: $$-\frac{dx}{dy}=-\frac{\tfrac{dx}{dt}}{\tfrac{dy}{dt}}= \frac{\sqrt2-2}{\sqrt2}=1-\sqrt2$$ The normal line to the curve at $t=\frac\pi4$ has slope $1-\sqrt2$ and passes through $(\sqrt2,\sqrt2-1)$. $$\sqrt2-1=(1-\sqrt2)\sqrt2+c;\ c=1$$ The equation of this normal line is thus $$y=(1-\sqrt2)x+1$$