Is Acceleration a Scalar Multiple of Arc Length?

[jaejoon89]

Homework Statement

Show that d^2 R / dt^2 is NOT a scalar which is a multiple of d^2 R / ds^2 where R is a vector, s is arc length

Homework Equations

and

The Attempt at a Solution

I was thinking maybe it has something to do with the fact k = |d^2 R / ds^2|
a = d^2 R / dt^2 = d|v|/dt * T + k |v|^2 N
so perhaps intuitively it can't be a multiple? I'm not really sure how to go about this...

 

[slider142]

Parameterize R as a function of s and note that s is a function of t. Then apply the chain rule.

 

[jaejoon89]

How does parameterizing show that for the second derivative?

Also, how would you do it when you aren't given a function?

s = int[dR/dt]dt from 0 to t ... but I don't see how you can really go from there

 

[slider142]

Let R be a function of s and s be a function of t. Then the second derivative of R with respect to s is $\frac{d^2R}{ds^2}$. What is the second derivative of R with respect to t (Just use the chain rule)? Is it a scalar multiple of $\frac{d^2R}{ds^2}$?

 

[jaejoon89]

That's what I'm asking. I don't know how to properly show it.

s = int[dR/dt]dt from 0 to t
Thus, s(t)

Given d^2 R / ds^2
= d^2 R / ds^2 * ds^2 / dt^2 = d^2 R / dt^2 by the chain rule
Is that what you mean?

 

[slider142]

Given d^2 R / ds^2
= d^2 R / ds^2 * ds^2 / dt^2 = d^2 R / dt^2 by the chain rule
Is that what you mean?

That formula is not true. The chain rule applies to a single derivative only, not to second derivatives. You can, however, apply the chain rule to
$$\frac{d^2 R}{dt^2} = \frac{d}{dt}\left(\frac{dR}{dt}\right)$$
to find it in terms of dR/ds. Remember R is a function of s and s is a function of t, so you have R(s(t)).

 

[jaejoon89]

Doesn't that imply that it is a multiple, when the whole point is to show that it is not?

That is why I am still confused. I understand using the chain rule to show:
d/ds (dR/ds) = d^2 R / ds^2
or
d/dt (dR/dt) = d^2 R / dt^2
But not how it is used to show that the second isn't a multiple of the first.

 

[slider142]

Doesn't that imply that it is a multiple, when the whole point is to show that it is not?

That is why I am still confused. I understand using the chain rule to show:
d/ds (dR/ds) = d^2 R / ds^2
or
d/dt (dR/dt) = d^2 R / dt^2
But not how it is used to show that the second isn't a multiple of the first.

d/dt is not a number, it is a differential operator. Note that dR/dt = (dR/ds)(ds/dt) by the chain rule. You then have to find (d/dt)[(dR/ds)(ds/dt)] for which you will have to use the product rule.

 

[jaejoon89]

Thank you. It makes A LOT more sense now.