Critical points of a multivariable function (wave equation)

[Poetria]

Homework Statement

There is a function $f(x,y)= sin(x-v*t)$ for v>0

Which of the following best describes the critical points of the function?

Relevant Equations

$f_x(x,t) = cos(x-v*t)$
$f_t(x,t) = -v*cos(x-v*t)$

I would like to check my understanding of this problem.
There are the following possibilities:

a. Isolated points where the gradient is 0.
b. The level curves of height 0
c. The level curves of height 1.
d. The level curves of height -1.
e. None of the above.

I would choose a, c, d.

Where sin(x-v*t)=1, cos(x-v*t)=0 (the gradient is 0). Is there a pitfall here?

 

[Steve4Physics]

Homework Statement:: There is a function $f(x,y)= sin(x-v*t)$ for v>0

Which of the following best describes the critical points of the function?
Relevant Equations:: $f_x(x,t) = cos(x-v*t)$
$f_t(x,t) = -v*cos(x-v*t)$

I would like to check my understanding of this problem.
There are the following possibilities:

a. Isolated points where the gradient is 0.
b. The level curves of height 0
c. The level curves of height 1.
d. The level curves of height -1.
e. None of the above.

I would choose a, c, d.

Where sin(x-v*t)=1, cos(x-v*t)=0 (the gradient is 0). Is there a pitfall here?

Hi. You haven't explained how/why you made your choices (a, c and d).

Also you have a small typo': $f(x,y)= sin(x-v*t)$ should be $f(x,t)= sin(x-v*t)$. (However don't let the use of 't' cause any confusion. - it's just a variable name. The maths is the same as for $f(x,y)= sin(x-v*y)$).

 

[Poetria]

Hi. You haven't explained how/why you made you choices (a, c and d).

Also you have a small typo': $f(x,y)= sin(x-v*t)$ should be $f(x,t)= sin(x-v*t)$. (However don't let the use of 't' cause any confusion. - it's just a variable name. The maths is the same as for $f(x,y)= sin(x-v*y)$).

Oh, of course, f(x,t). I can't edit the post. :(

My reasoning is as follows:
"A critical point of a multivariable function is a point where the partial derivatives of first order of this function are equal to zero."
If sin(x-v*t)=1 (i.e. $x-v*t= \frac {\pi} {2}+n*\pi)$), then $f_x=cos(x-v*t)=0$

If sin(x-v*t)=-1 (i.e. $x-v*t=\frac {-\pi} {2} - n*\pi$), then $f_x= cos(x-v*t)=0$

Similarly for $f_t(x,t)=-v*cos(x-v*t)$

The gradient is a derivative, therefore if a gradient is zero at a given point, this point is a critical one.

 

[Steve4Physics]

Oh, of course, f(x,t). I can't edit the post. :(

My reasoning is as follows:
"A critical point of a multivariable function is a point where the partial derivatives of first order of this function are equal to zero."
If sin(x-v*t)=1 (i.e. $x-v*t= \frac {\pi} {2}+n*\pi)$), then $f_x=cos(x-v*t)=0$

If sin(x-v*t)=-1 (i.e. $x-v*t=\frac {-\pi} {2} - n*\pi$), then $f_x= cos(x-v*t)=0$

Similarly for $f_t(x,t)=-v*cos(x-v*t)$

The gradient is a derivative, therefore if a gradient is zero at a given point, this point is a critical one.

Your answers were:
a. Isolated points where the gradient is 0.
c. The level curves of height 1.
d. The level curves of height -1.

I think you have justified c and d - though you haven’t made the logic explicit. I would have been tempted to have added something like this:
$f_x$ and $f_y$ are both zero when $x – vt = \frac {\pi}{2}±n\pi$ $(n = 0, 1, 2, ...)$
E.g. for n = 0 this give $x = vt + \frac {\pi}{2}$ which is a straight line in the x-t plane (corresponding to f(x, t)=1).

But, more importantly, you haven’t justified your choice ‘a’.’ Can you?

 

[Poetria]

But, more importantly, you haven’t justified your choice ‘a’.’ Can you?

Well, I don't think I understand the concept 'isolated point' well. I have found a definition in the calculus book, p. 227: https://books.google.pl/books?id=1dQ7LgPsDzYC&pg=PA230&dq=isolate+points+gradient+multivariable&hl=pl&sa=X&ved=2ahUKEwiB-a6v_LryAhXTuKQKHa_kBhoQ6AEwAHoECAkQAg#v=snippet&q=isolated points &f=false

"If there is an open ball B centered at $x_0$ and contained in O with the property that for all $x \in B, x \neq x_0$, x is not a critical point of f. That is $x_0$ is the only critical point of f in B."
But it is possible to construct an open ball with more critical points in this case. Am I right?
I had to check the definition for an open ball - https://mathworld.wolfram.com/OpenBall.html

So my choice of a isn't correct.

 

[Steve4Physics]

So my choice of a isn't correct.

Agreed - choice 'a' is incorrect.

But don't get tied-up or distracted by the term 'isolated point'. It’s not meant to be that complicated. A common-sense interpretation of ‘isolated point’ is perfectly adequate here (I say that as a non-mathematician!)
___________
Diversion/Example

Consider a domain consisting of the line $y = x+ 2$ and the point (1,1).

Any point, P, on the line is not isolated. For example the point (1, 3) is not isolated. That’s because P has immediate neighbouring points – because the line is a continuum.

But the point Q(1,1) is not on the line. Q is not connected to the line. Q has no immediate neighbouring points. Q is a single unconnected dot on the xy plane. We say Q is an isolated point;

Diversion ends (and my apologies to any mathematicians reading this).
__________

Thy only way to get $f_x=0$ and $f_y= 0$ is to make $cos(x-vt) = 0$. This only occurs when $x – vt = \frac {\pi}{2}±n\pi$

Each value of n gives a unique line $x = vt + \frac {\pi}{2}±n\pi$. So $f_x=0$ and $f_y= 0$ at all points on every line and nowhere else.

Therefore there are no isolated points where $f_x=0$ and $f_y= 0$.
___________

To get a better intuition, see what the function $f(x, t)=sin(x – vt)$ looks like. I’ve used ‘y’ rather than ‘t’ and set v = 4 for illustration purposes:
https://www.wolframalpha.com/input/?i=3d+plot+sin(x-4y)
Of course this is just one region of the graph – the graph extends from -∞ to +∞ in both x and t axes.

Every point on a crest or trough is a critical point.

 

[Poetria]

Wow, that's very lucid. :) Thank you very much. :)

 

[haruspex]

But it is possible to construct an open ball with more critical points in this case.

Yes, but the test in the definition is whether you can find an open ball that does not contain any other critical points. That there are some open balls that do is irrelevant.

 

[Poetria]

Oh, ok. I got it. :) Many thanks.