# Tangent to a curve

1. May 24, 2015

### skrat

1. The problem statement, all variables and given/known data
Calculate the tangent vector to a curve $$r=R(1-\varepsilon \sin ^2 \varphi)$$ as a function of $\varphi$

2. Relevant equations

3. The attempt at a solution
Ok, I tried like this:

I defined a vector $$(f(\varphi),\varphi)=(R(1-\varepsilon \sin ^2 \varphi),\varphi)$$ and than I calculated the derivative $$\frac{\partial }{\partial \varphi}$$ therefore my result is $$(-\varepsilon R \sin(2\varphi),1).$$

But I have a friend claiming that the result is $$(-\varepsilon R \sin(2\varphi),R(1-\varepsilon \sin ^2 \varphi)).$$ So, my question to you is... which is right? :/

2. May 24, 2015

### hunt_mat

Looks like you've computed the normal. You need to find a vector such that your vector dotted with it give zero. I think both of you are wrong.

3. May 24, 2015

### Orodruin

Staff Emeritus
First of all: What basis are you using to write down your vector? Unless you specify a basis, your two numbers mean nothing. What basis is your friend using?

Edit: Second, how do you define the tangent vector of a curve?

4. May 24, 2015

### skrat

It's polar basis $r$ being the distance from the origin and $\varphi$ the polar angle.

The function $$R(1-\varepsilon \sin ^2\varphi)$$ turns out to be an ellipse. $R$ and $\varepsilon$ are two constants, where $\varepsilon \ll 1$.

5. May 24, 2015

### Orodruin

Staff Emeritus
And what is the definition of the tangent vector of a curve?

6. May 25, 2015

### skrat

Yeah that could be the reason why my result is wrong.

If I am not mistaken, the vector is $(f(\varphi),\varphi)$ and normalized tangent vector should be $$t=\frac{(\frac{\partial f(\varphi)}{\partial \varphi},\frac{\partial }{\partial \varphi}\varphi)}{|(f(\varphi),\varphi)|}$$

7. May 25, 2015

### Orodruin

Staff Emeritus
Your big problem is that you are trying to only deal with the components of a vector, and even the components you quote are wrong. For any curve $\vec r(t)$, the tangent vector is given by $\vec t = d\vec r/dt$. In polar coordinates, the position vector $\vec r$ is given by $\vec r = r \vec e_r$, where $\vec e_r$ is the radial unit vector. There is no component in the angular direction.

However, you do need to note that $\vec e_r$ is not a constant vector and in order to differentiate this with respect to time, you will also need to consider the time derivative of $\vec e_r$. This is a big disadvantage of trying to write everything as a list of numbers only. The first step is to stop referring to vectors as a list of numbers without any reference as to what basis is being used.

8. May 25, 2015

### skrat

Ok, lets write $$\vec r=R(1-\varepsilon \sin ^2 \varphi)\vec e_r=s(\varphi)\vec e_r$$ than tangent vector $\vec t$ is $$\vec t=\frac{\partial }{\partial \varphi }s(\varphi) \vec e_r+s\frac{\partial }{\partial \varphi }\vec e_r.$$ The second term is $$\frac{\partial }{\partial \varphi }e_r=\frac{\partial }{\partial \varphi }(\cos\varphi, \sin \varphi)=(-\sin\varphi,\cos\varphi)$$ but IF (and only if) I am not mistaken, than by definition $$\vec e_\varphi =s(-\sin\varphi,\cos\varphi).$$ Meaning the tangent vector is $$t=\frac{\partial s(\varphi)}{\partial \varphi }\vec e_r + \vec e_\varphi .$$

9. May 25, 2015

### Orodruin

Staff Emeritus
There is an $s$ too much in front here. Remember that $\vec e_\varphi$ is a unit vector.

And one too few in front of the $\vec e_\varphi$ here. Otherwise you are doing well now.