Parametric curve /derivative problems

[ParoxysmX]

Homework Statement

Consider the parametric curve given by the equation

x(t) = ti + t^(1/3)j1. Calculate x'(t). Does the vector exist at $t=0$?

2. Find a new parametrisation of the curve for which the tangent vector is well
defined at all points. What is the value of the vector at the origin?

The Attempt at a Solution

For q.1, is it as simple as finding dx/dt? which would be $1 + \frac{1}{3\sqrt[3]{t^{2}}}$?

As for q.2, I'm completely stumped.

Any help would be greatly appreciated.

 

[HallsofIvy]

Yes, it really is as simple as finding dx/dt. No, what you have is NOT dx/dt. x(t) is a vector and dx/dt is a vector: dx/dt= i+ $\frac{1}{3}t^{-2/3}$j.

As for (2), $e^s$ exists for all s so try $e^s$i+ $e^{s/3}$j.
Do you see why that is also a parameterization for the curve?

 

[ParoxysmX]

Yes, it really is as simple as finding dx/dt. No, what you have is NOT dx/dt. x(t) is a vector and dx/dt is a vector: dx/dt= i+ $\frac{1}{3}t^{-2/3}$j.

Ah yes, my mistake. So this should mean that dx/dt does not exist at t=0, since $\frac{1}{3}t^{-2/3}$j becomes 1/0, giving i + ∞j.

As for (2), $e^s$ exists for all s so try $e^s$i+ $e^{s/3}$j.
Do you see why that is also a parameterization for the curve?

Yeah. dx/ds in this case would be defined at s=0, which is what we want. Is this sort of like 'moving' the origin?