User talk:Reedbeta

Derivation of the equation for simple harmonic motion
There must be an easier way to derive this.

The desired closed-form solution is: $$x(t) = A \cos \left ( \omega t + \phi \right )$$ for some constants $$A$$, $$\phi$$.

A differential equation for a 1-dimensional harmonic oscillator is:


 * $$\ F = -kx$$

Since force is mass times acceleration,


 * $$\frac {d^2 x} {dt^2} = - \frac{k}{m} x$$

To solve this second-order differential equation, we introduce a new variable v, reducing it to a system of linear first-order differential equations. Also writing $$\omega^2 = k / m$$ for convenience:


 * $$\frac{dx}{dt} = v$$


 * $$\frac{dv}{dt} = - \omega^2 x$$

Rewrite this system using matrices:


 * $$\frac{d}{dt} \begin{bmatrix} x \\ v \end{bmatrix}

= \begin{bmatrix} 0 && 1 \\ -\omega^2 && 0 \end{bmatrix} \begin{bmatrix} x \\ v \end{bmatrix}$$

The coefficient matrix has characteristic polynomial


 * $$\ \det (A - \lambda I) = \lambda^2 + \omega^2$$

so its complex eigenvalues are $$\pm \omega i$$. A complex eigenbasis is then:


 * $$\mathcal{B} = \left \{ \begin{bmatrix} - \frac {i}{\omega} \\ 1 \end{bmatrix}, \begin{bmatrix} \frac {i}{\omega} \\ 1 \end{bmatrix} \right \}$$,

which leads to the general solution of our system,


 * $$x(t) = c_1 e^{\omega i t} \frac{-i}{\omega} + c_2 e^{- \omega i t} \frac{i}{\omega}$$

Here, $$c_1$$ and $$c_2$$ are the (complex) coordinates of $$[x_0, v_0]^T$$ (that is, the initial state vector of the system) with respect to the complex eigenbasis $$\mathcal{B}$$.

Applying Euler's formula:


 * $$x(t) = \frac{-ic_1}{\omega} \left ( \cos \omega t + i \sin \omega t \right ) + \frac{ic_2}{\omega} \left ( \cos - \omega t + i \sin - \omega t \right )$$

Using the trigonometric identities $$\ \cos a = \cos -a$$ and $$\ \sin a = - \sin -a$$, this simplifies to:


 * $$x(t) = \frac{i(c_2 - c_1)}{\omega} \cos \omega t + \frac{c_1 + c_2}{\omega} \sin \omega t$$

Now, let us return to the coordinates $$c_1$$ and $$c_2$$. We know that


 * $$\begin{bmatrix} x_0 \\ v_0 \end{bmatrix} = \begin{bmatrix} - \frac{i}{\omega} && \frac{i}{\omega} \\ 1 && 1 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}$$

Inverting the matrix, we have


 * $$\begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} \omega i && 1 \\ - \omega i && 1 \end{bmatrix} \begin{bmatrix} x_0 \\ v_0 \end{bmatrix}$$ where $$x_0$$ and $$v_0$$ are both real.

Substituting back into the equation for $$x(t)$$, we have


 * $$x(t) = x_0 \cos \omega t + \frac{v_0}{\omega} \sin \omega t$$

Another trig identity gives:


 * $$x(t) = \sqrt{x_0^2 + \frac{v_0^2}{\omega^2}} \sin (\omega t + \theta)$$, where $$\theta = \arctan \frac{x_0 \omega}{v_0}$$.

If we then let


 * $$A = \sqrt{x_0^2 + \frac{v_0^2}{\omega^2}} = \sqrt{x_0^2 + \frac{mv_0^2}{k}}$$ and


 * $$\phi = \theta - \frac{\pi}{2} = \arctan \frac{x_0 \omega}{v_0} - \frac{\pi}{2}$$,

then we finally end up at the desired solution


 * $$\ x(t) = A \cos (\omega t + \phi)$$.

Gradient of an ellipsoid
The basic ellipsoid equation is
 * $$f(x,y,z) = {x^2 \over a^2}+{y^2 \over b^2}+{z^2 \over c^2} - 1$$

So the gradient is
 * $$\nabla f(x,y,z) = \left\langle \frac{2x}{a^2}, \frac{2y}{b^2}, \frac{2z}{c^2} \right\rangle$$

ArbCom elections are now open!
MediaWiki message delivery (talk) 13:01, 23 November 2015 (UTC)