# A Gem of Geometry and Analysis

Discussion in 'Physics & Math' started by rpenner, Feb 28, 2017.

1. ### rpennerFully WiredStaff Member

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Three special curves, symmetric about the x-axis, go vertically through the point (x=1,y=0). $r = \sqrt{x^2 + y^2}$

The vertical line: $x = 1$
The unit hyperbola: $x^2 - y^2 = 1$
The unit circle. $x^2 + y^2 = 1$

If we ignore parts left of the y-axis for now, we can describe each of these curve's x-values as functions of y.

The vertical line: $X_{0}(y) = \sqrt{1}$
The unit hyperbola: $X_{-1}(y) = \sqrt{1 + y^2}$
The unit circle. $X_{1}(y) = \sqrt{1 - y^2} ; \quad -1 \leq y \leq 1$

Note: $X_{\alpha}(y) = X_{\alpha}(-y)$ means these curves are symmetric about the x-axis.

Now, if for a certain value $y_t$ we want to know the area (A) in the "triangle" or "wedge" between the origin, the points $\left( X_{\alpha}(y_t), \pm y_t \right)$* and the curve we have:

The vertical line:
$A_{0}(y_t) = 2 \int_0^{y_t} X_0(y) - \frac{X_{0}(y_t)}{y_t} y \, dy \\ = 2 \int_0^{y_t} 1 dy - \frac{1}{y_t} \int_0^{y_t} 2 y dy \\ = 2 y_t - \frac{1}{y_t} y_t^2 = y_t$
The unit hyperbola:
$A_{-1}(y_t) = 2 \int_0^{y_t} X_{-1}(y) - \frac{X_{-1}(y_t)}{y_t} y \, dy \\ = 2 \int_0^{y_t} \sqrt{1 + y^2} dy - \frac{\sqrt{1 + y_t^2}}{y_t} \int_0^{y_t} 2 y dy \\ = y_t \sqrt{1 + y_t^2} + \ln \left( \sqrt{1 + y_t^2} + y_t \right) - y_t \sqrt{1 + y_t^2} \\ = \ln \left( X_{-1}(y_t) + y_t \right)$
The unit circle.
$A_{1}(y_t) = 2 \int_0^{y_t} X_{1}(y) - \frac{X_{1}(y_t)}{y_t} y \, dy \\ = 2 \int_0^{y_t} \sqrt{1 - y^2} dy - \frac{\sqrt{1 - y_t^2}}{y_t} \int_0^{y_t} 2 y dy \\ = y_t \sqrt{1 - y_t^2} - i \ln \left( \sqrt{1 - y_t^2} + i y_t \right) - y_t \sqrt{1 - y_t^2} \\ = -i \ln \left( X_{1}(y_t) + i y_t \right)$

Now if we parameterize the original curves in terms of the areas swept out, we have:

The vertical line: $t = A_{0}(y) = y \Rightarrow x(t) = 1, y(t) = t$
The unit hyperbola: $\rho = A_{-1}(y) = \ln( X(y) + y ) = \ln ( x + y ) \\ \Rightarrow x(\rho) = \frac{1}{2} \left(e^{\rho} + e^{-\rho} \right), y(\rho) = \frac{1}{2} \left(e^{\rho} - e^{-\rho} \right)$
The unit circle: $\theta = A_{1}(y) = -i \ln( X(y) + i y ) = \frac{1}{i} \ln ( x + i y ) \\ \Rightarrow x(\theta) = \frac{1}{2} \left(e^{i \theta} + e^{-i \theta} \right), y(\theta) = \frac{1}{2 i} \left(e^{i \theta} - e^{-i \theta} \right)$

So that's the motivation behind:
$\cosh \rho = \frac{1}{2} \left(e^{\rho} + e^{-\rho} \right) \\ \sinh \rho = \frac{1}{2} \left(e^{\rho} - e^{-\rho} \right) \\ \sinh^{-1} y = \ln \left( y + \sqrt{1 + y^2} \right) \\ e^{\rho} = \cosh \rho + \sinh \rho \\ \cos \theta = \frac{1}{2} \left(e^{i \theta} + e^{-i \theta} \right) \\ \sin \theta = \frac{1}{2 i} \left(e^{i \theta} - e^{-i \theta} \right) \\ \sin^{-1} y = -i \ln \left( iy + \sqrt{1 - y^2} \right) \\ e^{i \theta} = \cos \theta + i \sin \theta$

Which follows from the geometry of the curves and $\ln x = \int_1^{x} \frac{1}{t} dt$ being the inverse function of $e^x$.

* // Edit this has been fixed.

Last edited: Feb 28, 2017
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3. ### The GodValued Senior Member

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This new thread again begs the question, what is required to justify the name of site as sciforums.

News articles from inferno stopped, cause poor response; alternative theories..generally no discussion, only the pusher is abused by this forum piranhas.

So such tutorials? I doubt as majority here lack intellect even to decipher the basics of what rpenner is talking about.

This forum is more or less gripped by Trump Mania and US politics. It appears to me that one of the Mods (Kittamaru) has forgotten her brief and making this as antitrumpforum rather than science forum.

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5. ### Confused2Registered Senior Member

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A student at the back of the remedial class has started to make progress after swapping all x's and y's.

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7. ### arfa branecall me arfValued Senior Member

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I might comment that the OP is a nice refresher, me having done something similar long ago, but I think it was with the asymptotes of the hyperbola.

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What rpenner said, without the circle which would intersect the hyperbola at {-1,1}.

Last edited: Mar 1, 2017
8. ### arfa branecall me arfValued Senior Member

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Animation of rpenner's "areas swept out". The parameter t is the position of the intersections of the red and blue lines, with the vertical line at x = 1.

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Last edited: Mar 1, 2017
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9. ### rpennerFully WiredStaff Member

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Imagine my disappointment to learn these are not arfa brane's original works. https://en.wikipedia.org/wiki/Hyperbolic_angle#Comparison_with_circular_angle

The differential equation $y' + a y = 0$ has general solution
$y(x) = \left\{ { \begin{array}{lll} k_0 e^{-a x} & \quad & ; \; a \neq 0 \\ k_1 & & ; \; a = 0 \end{array} } \right.$
, while the differential equation $y'' + b y' + c y = 0$ has general solution:
$y(x) = \left\{ \begin{array}{lll} k_2 e^{ \frac{-b + \sqrt{b^2 - 4c}}{2} \, x} + k_3 e^{ \frac{-b - \sqrt{b^2 - 4c}}{2} \, x} ; b^2 \neq 2c & \quad & ; \; b^2 \neq 4c \\ k_4 e^{-\frac{b}{2} x} + k_5 \, x \, e^{-\frac{b}{2} x} & & ; \; b \neq 0, \, b^2 = 4c \\ k_6 + k_7 x & & ; \; b = c = 0 \end{array} \right.$

This first solution can also be written as:
$y(x) = e^{- \frac{b}{2} x} \left( k_2 e^{\sqrt{ \left(\frac{b}{2}\right)^2 - c} \, x} + k_3 e^{-\sqrt{ \left(\frac{b}{2}\right)^2 - c} \, x} \right) \\ = e^{- \frac{b}{2} x} \left( (k_2 + k_3 ) \cosh \left( \sqrt{ \left(\frac{b}{2}\right)^2 - c} \, x \right) + (k_2 - k_3) \sinh \left( \sqrt{ \left(\frac{b}{2}\right)^2 - c} \, x \right) \right) \\ = e^{- \frac{b}{2} x} \left( (k_2 + k_3 ) \cos \left( \sqrt{ c - \left(\frac{b}{2}\right)^2 } \, x \right) + i (k_2 - k_3) \sin \left( -i \sqrt{ \left(\frac{b}{2}\right)^2 - c} \, x \right) \right)$
and if $b^2 - 4c$ is real,
$y(x) = e^{- \frac{b}{2} x} \left( (k_2 + k_3 ) \cos \left( \sqrt{ c - \left(\frac{b}{2}\right)^2 } \, x \right) + i (k_2 - k_3) \sin \left( \sqrt{ c - \left( \frac{b}{2}\right)^2 } \, x \right) \right)$

Although these differential equations seem rather abstract, they arise naturally in modeling simple physical systems, including analog electric circuits and partially account for the ubiquitousness of sinusoidal signals showing up on equipment for a certain class of science fiction. Thank you, ITV.

Last edited: Mar 9, 2017

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11. ### sweetpeaRegistered Senior Member

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Rpenner, I don't get that last remark about ITV. Were you thinking of Doctor Who? That's BBC.

12. ### rpennerFully WiredStaff Member

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My mistake, it's ITC, not ITV. IMDB says it aired on ITV in the UK, but I wasn't there and can't actually name from memory the network or local channel which brought it to us in the US.

http://www.imdb.com/title/tt0072564/

See http://catacombs.space1999.net/main/cguide/umndabow.html "Close up of dome 3 screen."

Ben Carson might be wrong about the way human memory works, but we do let a fascinating amount of trivia pile up over the years.

Last edited: Mar 9, 2017