Parameterised Curve Proof Part 1

[I like Serena]

Hmm, your problem statement said nothing about m/s.
So how did you assume it would be m/s?

Anyway, even assuming the speed is in m/s, what's stopping us from defining a new curve that has a unit-less speed of 1?

 

[bugatti79]

Hmm, your problem statement said nothing about m/s.
So how did you assume it would be m/s?

Anyway, even assuming the speed is in m/s, what's stopping us from defining a new curve that has a unit-less speed of 1?

There was nothing said of m/s, I can't think of velocity being anything other than m/s!

I don't know how you would have a unit less speed. I give up :-)

 

[I like Serena]

Well, let's make a side trip into physics.
Did you know that meters and seconds are actually the same unit with a conversion factor (speed of light in vacuum) between them?
In other words, in a sense m/s is also dimensionless!

This is however not really relevant to your problem.
But if you want to give up, that's up to you of course!

 

[bugatti79]

Hmm, your problem statement said nothing about m/s.
So how did you assume it would be m/s?

Anyway, even assuming the speed is in m/s, what's stopping us from defining a new curve that has a unit-less speed of 1?

c would have to be 1 to make ||r_1(t)||=1 if v is dimensionless.

 

[I like Serena]

You wrote before that ||r_1'(s)|| = ||v/c||
So if c = 1, then ||r_1'(s)|| = ||v||, which is presumably not 1.

Btw, you seem to have dropped a quote that should have been there to indicate the derivative.
And you also seem to have replaced the parameter s of r_1(s) by t.
Really!

 

[bugatti79]

You wrote before that ||r_1'(s)|| = ||v/c||
So if c = 1, then ||r_1'(s)|| = ||v||, which is presumably not 1.

Btw, you seem to have dropped a quote that should have been there to indicate the derivative.
And you also seem to have replaced the parameter s of r_1(s) by t.
Really!

Well, I think I have exhausted all my possible answers based on my understanding. I don't know h0w ||r_1'(s)||=1 if c is not 1.

 

[I like Serena]

Well, if v is a vector and c is a scalar, what if c=||v||?

And if you want, you can let v keep a unit of [m/s] and have c be dimensionless.
That should satisfy your sense of what speed should be.

 

[bugatti79]

Well, if v is a vector and c is a scalar, what if c=||v||?

And if you want, you can let v keep a unit of [m/s] and have c be dimensionless.
That should satisfy your sense of what speed should be.

hmm... so if s(t) is the distance along the curve r_1(s) we would have s(t)=||v||t?

So in a nutshell we were finding a suitable expression for the distance s(t) such that we can maintain the requirement that ||r_1'(s)||=1 for all s?


Is it a similar approach for this last exercise?

reparameterise the curve r(t)=2cos(t)i+2cos(t)j+tk by it arc length. start at t=0

 

[I like Serena]

hmm... so if s(t) is the distance along the curve r_1(s) we would have s(t)=||v||t?

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.

So in a nutshell we were finding a suitable expression for the distance s(t) such that we can maintain the requirement that ||r_1'(s)||=1 for all s?

Yes.

Is it a similar approach for this last exercise?

reparameterise the curve r(t)=2cos(t)i+2cos(t)j+tk by it arc length. start at t=0

Yes.

 

[bugatti79]

Is it a similar approach for this last exercise?

reparameterise the curve r(t)=2cos(t)i+2cos(t)j+tk by it arc length. start at t=0

||r'(t)||=||-2sin(t)-2sin(t)+1||=(-4sin(t)+1)^(1/2).

If we let s(t)=ct then using the same idea we must have c=(-4sin(t)+1)^0.5 in order for ||r_1'(s)||=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 Alot! :-)