Unit Vectors for Ellipse: How Do You Find the Tangential and Normal Vectors?

[psid]

Homework Statement

Find the tangential and normal unit vectors for an ellipse with major axis of length a in the x-direction and minor axis of length b in the y-direction.

Homework Equations

For a circle, the unit vectors are defined as
$$\hat{r}=\cos{\theta}\hat{i}+\sin{\theta}\hat{j}$$
$$\hat{\theta}=-\sin{\theta}\hat{i}+\cos{\theta}\hat{j}$$

The Attempt at a Solution

For the circular case the derivation is easy, one just takes the derivate of
$$\vec{r}=r\cos{\theta}\hat{i}+r\sin{\theta}\hat{j}$$
with respect to $$\theta$$ and r. Now one can take the derivative with respect to $$\theta$$ and hope that this gives the vector I'm looking for, but I'm not sure. How about the vector normal to the tangent?

 

[HallsofIvy]

Why did you not use the appropriate form for homework problems? In particular, why have you not shown any attempt to do this yourself?

(In order that anyone be able to do this, you will need to specify the directions for the major and minor axes.)

 

[gabbagabbahey]

First step is to parameterize the ellipse into the form $\vec{r}=\vec{r}(t)=x(t)\hat{i}+y(t)\hat{j}+z(t)\hat{k}$ (I used $t$ as the parameter, but you could just as well use $\theta$ or any other variable). For example, a circle in the xy-plane of radius $R$ centered at the origin can be parameterized as $\vec{r}(t)=R\cos(t)\hat{i}+R\sin(t)\hat{j}$.

Once you've parameterized your ellipse, the unit tangent $\hat{T}$ and unit normal $\hat{N}$ can be easily found by taking derivatives with respect to your chosen parameter:

$$\hat{T}=\frac{\vec{v}}{v}=\frac{\frac{d\vec{r}}{dt}}{\left| \frac{d\vec{r}}{dt} \right|}$$

$$\hat{N}=\frac{\frac{d\hat{T}}{dt}}{\left| \frac{d\hat{T}}{dt} \right|}$$

You should have similar definitions for these unit vectors given in whatever calculus text you are using...do you not?

 

[psid]

Thanks for the reply. The above expression yields tangent vector rather easily:

$$\hat{T}=\frac{-a\sin{\theta}\hat{i}+b\cos{\theta}\hat{j}}{\sqrt{a^{2}\sin^{2}{\theta}+b^{2}\cos^{2}{\theta}}}.$$

But taking the derivative of this yields multiple pages of output in Mathemathica and is not very useful.

Looking the tangent vector one can see that vector defined as

$$\hat{N}=\frac{b\cos{\theta}\hat{i}+a\sin{\theta}\hat{j}}{\sqrt{a^{2}\sin^{2}{\theta}+b^{2}\cos^{2}{\theta}}}$$
should be orthogonal with $$\hat{T}$$ for all $$\theta$$.

So perhaps this could be used as the normal vector. Am I correct?

 

[gabbagabbahey]

That looks fine to me; just be sure that you are being asked for the outward normal, and not the inward normal (Your problem statement doesn't say what definition you are using for the unit normal) which differs from your result by a negative sign.