# Vector calculus and parametrisation

1. Apr 28, 2013

### Metalor

Ok, I just found out I have a physics assignment due tomorrow and I have no idea how to do it so I came here for help as none of the maths assistants at Uni could help me. I'm having trouble with:

1. Consider the parametric curve given by the equation x(t)= t<i> + t^(1/3)<j> - <> denotes a vector

a) Sketch x(t) , b) calculate x'(t). does this vector exist at t=0? c) Find a new parametrisation of the curve for which the tangent vector is well defined at all point. What is the value of the vector at the origin?

I don't understand what the equation is describing and I THINK 'i' and 'j' are vectors (handout read t* i[hat] + t^1/3j[hat], so I don't have a clue how to sketch this. I think I've got question b) and x'(t) = i + j/3*t^(2/3) which is undefined when t = 0, so the vector does not exist.

Help!! :S

2. Apr 28, 2013

### Sunil Simha

I think I can help you with part a). Assume x(t) to be a position vector in the xy plane. Now the x component of the position vector is x = t right? Similarly, the y component is y = t1/3. So all you need to do is relate y and x i.e. write y as some function of x and plot it on the xy plane.

3. Apr 28, 2013

### Metalor

I've just done a bit of reading online, and have this so far:

x(t) = ti + (t^(1/3))j gives two equations x = t and y = x^1/3 with gives the equation y= x^(1/3) and then sketch from there?

Is this correct or am I way off?

4. Apr 28, 2013

### DrewD

$\hat{i}$ is the unit vector in the $x$ direction and $\hat{j}$ is the unit vector in the $y$ direction. You can also think of this as the point set $(t,t^{1/3})$ if that helps you visualize the curve. Personally, I never like the i,j notation, but a lot of people do. It shouldn't be difficult to write this in the familiar form $y=f(x)$ instead of the parametric way it is given. This may make part c) easier.

5. Apr 28, 2013

### DrewD

Looks okay so far.
Edit: actually you have what looks like a typo, but I'm pretty sure you understand it.

6. Apr 28, 2013

### Sunil Simha

Yes this is right.

7. Apr 28, 2013

### Metalor

Oh wow, you guys are fast!! I wasn't expecting a response for a few hours but you beat me to it. Thanks very much :)

8. Apr 29, 2013

### Metalor

Hey guys,

So I've got a) and b) done but I"m still not sure about c)

It's asking me to find a new parametrisation of the curve for which the tangent vector is well defined at all points. And what is the value of the vector at origin?

Does that just mean write it as y = x^1/3? But at the origin, the tangent vector is undefined?

9. Apr 29, 2013

### Sunil Simha

When you look at the curve of y=x1/3 you can see that the slope at x=0 is infinite. So I'm not sure how to create a new parametrisation to fulfill the requirements. I hope DrewD or someone else can help you there.

10. Apr 29, 2013

### SammyS

Staff Emeritus
A vector tangent to $\displaystyle \ y=x^{1/3} \$ would have an x component of zero, and a non-zero y component.

You need a parametrization such that x'(t) = 0, when x(t) = 0 .

There is a fairly obvious one available.

(Solve $\displaystyle \ y=x^{1/3} \$ for x .)