What is the significance of taking the derivative of y=sin^-1(x)?

[JDude13]

I was mucking around with graphmatica and decided to take the derivitive of
$$y=\sin^{-1}(x)$$
and got
$$y=\frac{1}{\sqrt{1-x^2}}$$
which is the lorentz factor with
$$\beta=x$$
and
$$\gamma=y$$

Does this have any significance?

 

[Meir Achuz]

No significance for relativity.
It just comes from sin^2 +cos^2=1.

 

[HallsofIvy]

If $y= sin^{-1}(x)$ then x= sin(y). dx/dy= cos(y) so that $dy/dx= 1/cos(y)= 1/cos(sin^{1}(x))$. Since, as Meir Achuz said, $sin^2(x)+ cos^2(x)= 1$, $cos(x)= \sqrt{1- sin^2(z)}$ so that $cos(sin^{-1}(x)= \sqrt{1- sin^2(sin^{-1}(x)}= \sqrt{1- x^2}$.

That is, $dy/dx= 1/\sqrt{1- x^2}$.

(I assume you meant $y'= 1/\sqrt{1- x^2}$ not $y= 1/\sqrt{1- x^2}$.)

 

[JDude13]

If $y= sin^{-1}(x)$ then x= sin(y). dx/dy= cos(y) so that $dy/dx= 1/cos(y)= 1/cos(sin^{1}(x))$. Since, as Meir Achuz said, $sin^2(x)+ cos^2(x)= 1$, $cos(x)= \sqrt{1- sin^2(z)}$ so that $cos(sin^{-1}(x)= \sqrt{1- sin^2(sin^{-1}(x)}= \sqrt{1- x^2}$.

That is, $dy/dx= 1/\sqrt{1- x^2}$.

(I assume you meant $y'= 1/\sqrt{1- x^2}$ not $y= 1/\sqrt{1- x^2}$.)

dunno. Like i said. I was mucking around with graphmatica.

 

[romsofia]

Why WOULD it have any significance?

 

[JDude13]

Why WOULD it have any significance?

Why WOULDN'T it have any significance?

 

[romsofia]

Why WOULDN'T it have any significance?

I'm asking you, I'm wondering why you would think it would have significance. Do you know how to differentiate/integrate inverse trig functions (not with the program you mentioned)?

 

[JDude13]

I'm asking you, I'm wondering why you would think it would have significance. Do you know how to differentiate/integrate inverse trig functions (not with the program you mentioned)?

no... I was just a little confused when it happened.

 

[GrayGhost]

I was mucking around with graphmatica and decided to take the derivitive of
$$y=\sin^{-1}(x)$$
and got
$$y=\frac{1}{\sqrt{1-x^2}}$$
which is the lorentz factor with
$$\beta=x$$
and
$$\gamma=y$$

Does this have any significance?

As HallsofIvy pointed out, it should be $$dy/dx=\frac{1}{\sqrt{1-x^2}}$$

I do see the significance myself, or maybe better to say "I see the relation". I'd agree with your assessment. It's just a matter of understanding how the abstract math applies to the physical model of relativity theory. If x = v/c, then dy/dx = gamma (not y = gamma) where c=1. We then have the gamma factor of relativity, and the angular orientation differential (in 4-space) between the 2 frames of interest ... y = theta = sin^-1(v/c).

GrayGhost

 

[ZapperZ]

This topic is a bit strange. There are MANY expressions in physics, chemistry, engineering, etc. that have the SAME mathematical form as this. You are fishing to put more "meaning" into this than there really is.

Zz.