Question

Suppose \[f : R^{n} \rightarrow R\]is differentiable and ||f(t)|| = 1 for all t. Show that the dot product of Df(t) and f(t) is always 0. Describe what this means geometrically.

Answer 1

@ganeshie8

Answer 2

any ideas?

Answer 3

Lets think about geometry first

Answer 4

when n=2, \(\|f(t)\|=1\) gives us circular motion :

Answer 5

so in Rn it would be a ball?

Answer 6

yeah we can say the motion is \(n\) dimensional ball

Answer 7

then how can we calculate the Df and f dot product?

Answer 8

any leads?

Answer 9

see if you can show it is true for n = 3, then try to generalize it

Answer 10

woops, use f(t) instead of r(t)

Answer 11

how do i put these into english words tho. I suck at english. lol

Answer 12

$$

\Large || A ||^2 =A ~. A\\

\\ \therefore \\
\Large ||f(t)||^2 = f(t)~.f(t)

$$

Answer 13

what about Df(t) ? it's Rn - R so it's gonna be a 1xN matrix

Answer 14

hmm

Answer 15

still not getting it

Answer 16

the left side is zero, because its a derivative of a constant
$$



$$

Answer 17

since f(t) and f'(t) the the line and it's perpendicular line, shouldn't be always have a dot product of zero? Since A dot B = |A||B| cos , and the cos is always gonna be pi/2

Answer 18

$$
\Large || \vec{a} ||^2 =\vec{a} ~. \vec{a}\\

\\ \therefore \\
\Large ||f(t)||^2 = f(t)~.f(t)\\
\Large ||f(t)|| = 1 \\
\\ \therefore\\
\Large 0 =\frac{d}{dt}(1^2) = \frac{d}{dt}(||f(t)||^2) = (f(t)~.f(t)) ~'
\\ \Large = { f'(t) . f(t) + f(t) . f ' (t)\\
= 2 f ' (t) . f(t)
\\ \therefore
\\
\Large 0 = 2 f ' (t) . f(t)
\\ \therefore
\\ \Large 0 = f ' (t) . f(t)
}
$$

Answer 19

yes but how do you know that f(t) and f ' (t) are perpendicular

Answer 20

" since f(t) and f'(t) `the the line` and it's perpendicular line, shouldn't be always have a dot product of zero? Since A dot B = |A||B| cos , and the cos is always gonna be pi/2"

what did you mean here?

Answer 21

oh ok. I might confused it with some sort of lemma that can be used directly.

Answer 22

I meant " is the line and ......"

Answer 23

I drew r ' (t) perpindicular to r(t) in the picture, but that is what we want to prove.
it is true that r ' (t) is tangent to the circle , but its not obvious that r ' (t) is perpindicular to r (t) .

Answer 24

got you. Thank you ! You are my favourite

Answer 25

This result is consistent with elementary geometry, but it is better not to use elementary geometry. Elementary geometry will only tell us about at most 2 or 3 dimensions, one dimension it is trivially true.
Vector algebra goes to n dimensions , so it is more general and more powerful.


Now there might be a way to prove this with the cos theta definition of dot product, but I don't see a simple proof.

Answer 26

yea, anyway, thank you for all the help and your time.XD

Answer 27

Note that this is a vector valued function, so we used a shorthand:
$$

\Large f(t)=(f_1(t),f_2(t),\ldots,f_n(t))

$$

Answer 28

otherwise it looks like a 1 valued function