Parameterised Curve Proof Part 1

[I like Serena]

Yes, well, I speculated it might be a constant.
And yay! It was.

I'm afraid in this case it isn't.
I didn't know before that you would come up with harder and harder problems.
I think this one doesn't even have a neat simple solution.

 

[bugatti79]

Yes, well, I speculated it might be a constant.
And yay! It was.

I'm afraid in this case it isn't.
I didn't know before that you would come up with harder and harder problems.
I think this one doesn't even have a neat simple solution.

Ok. One final question and I will leave this one..:-) How did you determine it was a constant in the first case and not in the second?

THanks in advance!

 

[I like Serena]

The derivative of r(t)=u+vt was r'(t)=v which is a constant.


Let's step through the process, shall we?

What you need, is:
|| r_1'(s) || = 1

So:
|| r'(t(s)) t'(s) || = 1

Assuming t'(s) is positive, that means:
|| r'(t(s))|| = 1 / t'(s) = s'(t)

This can also be written as:
s'(t) = ||r'(t)||

Integrate both sides with respect to t, and you find s(t).

In your previous problem, you had ||r'(t)||=||v||, which is constant and as such easy to integrate.

 

[bugatti79]

The derivative of r(t)=u+vt was r'(t)=v which is a constant.


Let's step through the process, shall we?

What you need, is:
|| r_1'(s) || = 1

So:
|| r'(t(s)) t'(s) || = 1

Assuming t'(s) is positive, that means:
|| r'(t(s))|| = 1 / t'(s) = s'(t)

This can also be written as:
s'(t) = ||r'(t)||

Integrate both sides with respect to t, and you find s(t).

In your previous problem, you had ||r'(t)||=||v||, which is constant and as such easy to integrate.

Ok, Thanks Alot! :-)