# Related Rate of Change of an Angle

• Mar 20th 2009, 12:45 PM
fxsapa
Related Rate of Change of an Angle
Hello

Out of pure maths interest, I was trying to solve the following problem:

We have: point O, anywhere on the plan; line AB, anywhere on the plan; point M, moving at speed described by func s(t) on AB, from A.
Find the rate of change of angle AOM (from time t), given that you can know everything about triangle ABO (lengths, angles...) but that this triangle is common (not iso, not rect, not equi....).

I have created a graph of this using geometric methods... with GeoGebra. It looks a bit like a Gauss curve.

When I now try to get the function for it, I tried several ways, and I couldn't. All I found was some strange graphs... (reversed U with vertical tangent on right side, graph in -1<y<1 -1<x<1; that was the latest... ?!)

I tried searching on internet but to no avail.

I know the basics of differentiation and am studying more.

If anyone has an idea...?

PS: I'll post a sketch when I've got time...
• Mar 20th 2009, 08:21 PM
Soroban
Hello, fxsapa!

Quote:

We have: point O, anywhere on the plane; line AB, anywhere on the plane.
Point M, moving on AB from A, at speed described by function s(t).

Find the rate of change of $\angle AOM$ (from time t),
given that you can know everything about $\Delta ABO$ (lengths, angles, ...)
but that this triangle is scalene.

Code:

                      O                       *                     */  *                   * /      *             b  * θ/          *               *  /              *             *    /                  *           *    /                      *         *    φ /                          *     A *  *  *  *  *  *  *  *  *  *  * B         x    M

We have: . $\Delta AOB,\:A = \angle OAB,\;\theta = \angle AOM,\;b = OA,\;x = AM$
. . . .and: . $\phi = \angle AMO \:=\:180^o - A - \theta$

In $\Delta OAM$, apply the Law of Sines:

. . $\frac{\sin\theta}{x} \:=\:\frac{\sin(180-A-\theta)}{b} \quad\Rightarrow\quad b\sin\theta \;=\;x\sin([180 - A] - \theta)
$

. . $b\sin\theta \;=\;x\bigg[\underbrace{\sin(180-A)}_{\text{This is }\sin A}\cos\theta - \underbrace{\cos(180-A)}_{\text{This is }-\cos A}\sin\theta\bigg]$

. . $b\sin\theta \;=\;x\underbrace{\bigg[\sin A\cos\theta + \cos A\sin\theta\bigg]}_{\text{This is }\sin(\theta + A)}$

. . $b\sin\theta \;=\;x\sin(\theta + A)$

Differentiate with respect to time:

. . $b\cos\theta\,\frac{d\theta}{dt} \;=\; x\cos(\theta + A)\,\frac{d\theta}{dt} + \sin(\theta + A)\,\frac{dx}{dt}$

. . $b\cos\theta\,\frac{d\theta}{dt} - x\cos(\theta + A)\,\frac{d\theta}{dt} \;=\;\sin(\theta + A)\,\frac{dx}{dt}$

. . $\bigg[b\cos\theta - x\cos(\theta+A)\bigg]\,\frac{d\theta}{dt} \;=\;\sin(\theta + A)\,\frac{dx}{dt}$

Therefore: . $\frac{d\theta}{dt} \;=\;\frac{\sin(\theta+A)\,\frac{dx}{dt}}{b\cos\th eta - x\cos(\theta +A)}$

• Mar 21st 2009, 01:59 AM
fxsapa
Thx