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# parametrisation- curve in R^3 watch

1. (Sorry, I'm stuck again.)

Let r(t) be a parametristation of curve C in R^3. Suppose r(t) does not equal to 0, and r(t)*r'(t)=0 for all points of C. Show C must lie on the surface of a sphere centered at the origin.

2. (Original post by ellemay96)
(Sorry, I'm stuck again.)

Let r(t) be a parametristation of curve C in R^3. Suppose r(t) does not equal to 0, and r(t)*r'(t)=0 for all points of C. Show C must lie on the surface of a sphere centered at the origin.

If I understand your notation correctly, can you think of some simple function f(r) that has the property that [f(r)]' = r(t)r'(t) where the LHS is differentiated w.r.t.t (think of when you first learned the chain rule)?
3. (Original post by ellemay96)
(Sorry, I'm stuck again.)

Let r(t) be a parametristation of curve C in R^3. Suppose r(t) does not equal to 0, and r(t)*r'(t)=0 for all points of C. Show C must lie on the surface of a sphere centered at the origin.

I guess you are working with vectors here and you have ?

If so, then davos's comment still applies, but you need to find a quantity that

1) differentiates to
2) is a constant scalar (else its derivative can't be zero)
4. (Original post by atsruser)
I guess you are working with vectors here and you have ?

If so, then davos's comment still applies, but you need to find a quantity that

1) differentiates to
2) is a constant scalar (else its derivative can't be zero)
Think the logic is a little confused here.

If ,

THEN

iwe can deduce that is constant.

[Of course, you can also work the other way:

We want to show that r(t) (r = |r|) is constant.

So we want to show that dr/dt = 0.

So find an expression for r in terms of t, and differentiate. Does the given condition show that dr/dt = 0? Then we're done (hurrah!).

Of course, once you have the expression for r, you may be able to see how you can choose a function f(r) that is easier to deal with and still has the property that f(r) = constant => r = constant].
5. (Original post by DFranklin)
Think the logic is a little confused here.

If ,

THEN

iwe can deduce that is constant.
Chill out, Mr Franklin! I was merely giving a couple of hints, rather than presenting a logically unimpeachable argument. (And I'm not sure that what I wrote is inconsistent with what you said anyway ..).

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