Vector calculus and parametrisation

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Homework Help Overview

The discussion revolves around a physics assignment involving a parametric curve defined by the equation x(t) = t + t^(1/3). Participants are tasked with sketching the curve, calculating its derivative, and finding a new parametrization that ensures the tangent vector is well-defined at all points, particularly at the origin.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss how to interpret the parametric equations and relate them to Cartesian coordinates. There are attempts to derive the equations for x and y from the parametric form and questions about the existence of the tangent vector at specific points.

Discussion Status

Some participants have made progress on parts a) and b), while others express uncertainty about part c). There is a recognition that the tangent vector is undefined at the origin, and multiple interpretations of how to achieve a valid parametrization are being explored.

Contextual Notes

Participants note the challenge of finding a parametrization that maintains a well-defined tangent vector across the curve, particularly at points where the original parametrization leads to undefined behavior.

Metalor
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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
 
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Metalor said:
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.

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.
 
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?
 
##\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.
 
Metalor said:
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?

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

Yes this is right.
 
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 :)
 
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?
 
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
Metalor said:
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?
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 .)
 

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