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Anonymous Sun 25 Oct 2015 16:06:51 No.7614999 View Reply Report Delete

Quoted By: >>7615024 >>7615071 >>7615097 >>7615106 >>7615107

How the fuck do I prove this /sci/
a(t) is a vector-valued function

Anonymous

Anonymous Sun 25 Oct 2015 16:09:56 No.7615012 Report Delete

Quoted By:

Start by solving an example, say, a(t) = [t, -t^2, t^3]. You'll see why when you work it out, and then the proof is easy.

Anonymous

Anonymous Sun 25 Oct 2015 16:10:44 No.7615016 Report Delete

Quoted By:

If you don't assume anything about a, then this is obviously false (consider a(t) = (cos t, sin t))

Anonymous

Anonymous Sun 25 Oct 2015 16:12:25 No.7615024 Report Delete

Quoted By: >>7615034

>>7614999
I can think of examples where this isn't true though.
take v(t) = cos(t)i + sin(t)j
then v'(t) = -sin(t)i + cos(t)j
Their dot product is zero, but the product of their magnitudes is 1.

Anonymous

Anonymous Sun 25 Oct 2015 16:14:34 No.7615034 Report Delete

Quoted By: >>7615040

>>7615024
So I have to prove shit that's not even true
Gg professor, gg

Anonymous

Anonymous Sun 25 Oct 2015 16:17:03 No.7615040 Report Delete

Quoted By: >>7615069

>>7615034
Just give the counter example.
There's simply no way to prove this without knowing more about the function.

Anonymous

Anonymous Sun 25 Oct 2015 16:24:08 No.7615069 Report Delete

Quoted By:

>>7615040
K I'll do
Thanks

Anonymous

Anonymous Sun 25 Oct 2015 16:24:45 No.7615071 Report Delete

Quoted By:

>>7614999
it's only true if they are both parallel and in the same direction
this can be easily shown if you know what the dot product is

Anonymous

Anonymous Sun 25 Oct 2015 16:31:09 No.7615097 Report Delete

Quoted By:

>>7614999
You probably should have been told by now that the derivative of a single variable vector function is it's tangent vector, so this is obviously false by inspection.

Anonymous

Anonymous Sun 25 Oct 2015 16:33:45 No.7615106 Report Delete

Quoted By:

>>7614999
is there any more information about a you didn't give here?

Anonymous

Anonymous Sun 25 Oct 2015 16:34:11 No.7615107 Report Delete

Quoted By: >>7615115 >>7615139

>>7614999
Maybe they meant to write |a(t)|d/dt|a(t)|

Anonymous

Anonymous Sun 25 Oct 2015 16:36:20 No.7615115 Report Delete

Quoted By: >>7615133 >>7615139

>>7615107
is that true? i don't care to check it right now

Anonymous

Anonymous Sun 25 Oct 2015 16:39:50 No.7615133 Report Delete

Quoted By:

>>7615115
Yeah.

Anonymous

Anonymous Sun 25 Oct 2015 16:42:22 No.7615139 Report Delete

Quoted By: >>7615200

>>7615107
>>7615115
According to a quick back-of-the-envelope calculation, this seems to be true.

Anonymous

Anonymous Sun 25 Oct 2015 16:57:51 No.7615200 Report Delete

Quoted By: >>7615222

>>7615139
i don't see how it can be true. i get |a| * 1/sqrt(a d/dt a)
but then agian, i put close to no effort in it, so i might as well be wrong

Anonymous

Anonymous Sun 25 Oct 2015 17:03:21 No.7615222 Report Delete

Quoted By: >>7615235

>>7615200
$a\cdot a'=(a_1a_1'+\cdots+a_na_n')=\frac{1}{2}((a_1^2)'+\cdots+(a_n^2)')=\frac{1}{2}(|a|^2)'=|a||a|'$

Anonymous

Anonymous Sun 25 Oct 2015 17:05:31 No.7615235 Report Delete

Quoted By: >>7615266

>>7615222
a*a'=(a_1a_1'+...+a_na_n')=1/2((a_1^2)'+...+(a_n^2)')=1/2(|a|^2)'=|a||a|'

Anonymous

Anonymous Sun 25 Oct 2015 17:15:47 No.7615266 Report Delete

Quoted By:

>>7615235
ah, yes, i made a stupid mistake. my bad

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