Question

Given that r(t)•r’(t)=0 for all values of t, prove that ||r|| = c (constant)

Given that r(t)•r’(t)=0 for all values of t, prove that ||r|| = c (constant)

@Mathematics

Answer 1

r(t)•r'(t) = aa'+bb'+cc'

Answer 2

Not sure how that proves ||r|| = c must be constant though

Answer 3

Consider the function f(t) = r(t)•r(t).

Then f'(t) = r'(t)•r(t) + r(t)•r'(t) = 2 r(t)•r'(t)

Now if r(t)•r'(t) = 0, then f'(t) = 0 therefore .... can you fill in the rest now?

Answer 4

hmm

Answer 5

Remember that
\[ ||r||^2 = r . r \]

Answer 6

Ah yea

Answer 7

Ok, that explains it, just gotta think it over a bit. thanks :)

Answer 8

We know now that \( f(t) = constant \). And by definition \( f(t) = ||r||^2 \) Hence ...

Answer 9

Is letting f(t) = r(t)•r(t) the only way to do it?

Answer 10

Hence ||r||= f(t)^2 = c^2

Answer 11

square roots

Answer 12

square roots?

Answer 13

Not squared. f(t) = ||r||^2 = c
implies ||r|| = sqrt(c) not c^2

Answer 14

hmm

Answer 15

But if f(t) = c

Answer 16

and f(t) = ||r||^2

Answer 17

ok, it should be f(t) = ||r||^2 = c?

Answer 18

Yes

Answer 19

And therefore \( ||r|| = \sqrt{c} \), a constant.

Answer 20

I wouldnt have thought to set f(t) = r(t)•r(t), is that the only way to prove this?

Answer 21

I just introduced f(t) to make it a bit easier to handle. But you do the same thing this way:

\[ (||r||^2)' = (r.r)' = r.r' + r'.r = 2r.r' = 0 \ \ \implies \ \ ||r||^2 = c \ \ \implies \ ||r|| = \sqrt{c} \]

Answer 22

As to whether there is another way to prove it, nothing comes to mind right now.

Nonetheless, this is a short and elegant proof; why not use it?

Answer 23

Yea, thanks for the help

Answer 24

I am having trouble with this step: (r.r)′=r.r′+r′.r

Answer 25

is that r * r or r dot r?

Answer 26

I see the product rule being used so thats whats making me think it is r times r

Answer 27

definitely the dot product between r and r: r . r.

The expression r * r, r times r, is meaningless as these are vectors.

Answer 28

so the derivative of the dot product is the same as the derivative of a product?

Answer 29

r.r′+r′.r cause isnt that the product rule?

Answer 30

It's a lemma/basic result in vector calculus that if u(t) and v(t) are vector functions, that the product rule holds:

\[ \frac{d \ }{dt} u(t) . v(t) = u'(t) . v(t) + u(t) . v'(t) \]

Answer 31

You should prove that for yourself by writing u and v in components, multiplying it out and differentiating.

Answer 32

hmm, ok, I wont take anymore of your time with this problem, ill analyze what you've written for a better understanding, thanks :)

Answer 33

ok