Parameterised Curve Proof Part 1

[I like Serena]

First things first. You seem to have miscalculated when getting the length of r'(t).
You're not supposed to just drop i, j, and k.

What you get, should satisfy ||r_1'(s)|| = 1.
Does it?

 

[bugatti79]

First things first. You seem to have miscalculated when getting the length of r'(t).
You're not supposed to just drop i, j, and k.

What you get, should satisfy ||r_1'(s)|| = 1.
Does it?

hmmm...why not? I dropped it based on the definition (in general terms) that if the vector v=(ai+bj+ck) then its magnitude is ||v||=(a^2+b^2+c^2)^0.5...?

 

[I like Serena]

hmmm...why not? I dropped it based on the definition (in general terms) that if the vector v=(ai+bj+ck) then its magnitude is ||v||=(a^2+b^2+c^2)^0.5...?

This is correct, but that's not what you did.

 

[bugatti79]

First things first. You seem to have miscalculated when getting the length of r'(t).
You're not supposed to just drop i, j, and k.

What you get, should satisfy ||r_1'(s)|| = 1.
Does it?

Ok, I shouldn't have added like terms, they need to be kept separated?

||r'(t)||=||-2sin(t)-2sin(t)+1||=(-2sin^2(t) -2sin^2(t)+1)^(1/2).


It can only satisfy ||r_1'(s)|| = 1 if we make c=(-2sin^2(t) -2sin^2(t)+1)^(1/2) if we let s(t)=ct...?

 

[I like Serena]

When taking the derivative you should not drop the i, j, or k.
That happens later, when you calculate the length.

When calculating the squares, you need to add parentheses to indicate you square the entire term and not just the sine.

 

[bugatti79]

Uummm... s(t) depends on a parameter t that is not in r_1(s)...
But I think what you mean is probably right.

I'd say it something like this:

s is the length of the curve r_1 from 0 to s.

With s(t) = ||v||t, the curve r(t(s)) has this property.




Yes.



Yes.

Ok, I see the error.

||r'(t)||=||-2sin(t)i-2sin(t)j+1k||=(8sin^2(t)+1)^(1/2).

If we let s(t)=ct then using the same idea we must have c=8sin^2(t)+1)^(1/2) in order for ||r_1'(s)||=1...?


Is this the same as saying ||r_1'(s)||=1 is satisfied?

 

[I like Serena]

Ok, I see the error.

||r'(t)||=||-2sin(t)i-2sin(t)j+1k||=(8sin^2(t)+1)^(1/2).

Good!

If we let s(t)=ct then using the same idea we must have c=8sin^2(t)+1)^(1/2) in order for ||r_1'(s)||=1...?


Is this the same as saying ||r_1'(s)||=1 is satisfied?

No.
If you let s(t)=ct, then r_1(s)=r(t(s))=r(s/c).
Can you calculate r_1'(s) for this case?
Please substitute and write out the entire formula.

 

[bugatti79]

Good!




No.
If you let s(t)=ct, then r_1(s)=r(t(s))=r(s/c).
Can you calculate r_1'(s) for this case?
Please substitute and write out the entire formula.

r_1'(s)=(dr/dt)(dt/ds) = (-2sin(t)i-2sin(t)j+1k)(1/c)...?

 

[I like Serena]

r_1'(s)=(dr/dt)(dt/ds) = (-2sin(t)i-2sin(t)j+1k)(1/c)...?

This is only true if c is a constant.
With your solution c is not a constant but a function of t.

What you need to do is define (dt/ds) such that it yields ||r_1'(s)||=1, and then integrate it.

 

[bugatti79]

Yes. You really need to start making distinctions between vectors, scalars, infinitesimal vectors, and infinitesimal scalars.

Anyway, your arc speed is ||r'(t)||=||v||.

For your reparameterisation you want it to be 1.

So perhaps s(t) is something like s(t)=c t for some constant c.
What will the arc speed become with this parameterisation?

This is only true if c is a constant.
With your solution c is not a constant but a function of t.

What you need to do is define (dt/ds) such that it yields ||r_1'(s)||=1, and then integrate it.

but in the previous case you had the same eqn s(t)=ct and you treated c as a constant, is it not a function of t here also...?

 

[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 a lot! :-)