# Change in area of a triangle formed by position vectors

1. Sep 14, 2009

### MissMCHP

1. The problem statement, all variables and given/known data
Consider two particles located at positions $$\vec{r_{1}}(t)$$ and $$\vec{r_{2}}(t)$$, respectively, with corresponding velocities given respectively by $$\vec{v_{1}}(t)$$ and $$\vec{v_{2}}(t)$$. Consider the triangle formed by the origin and the position vectors of the two particles. Show that the rate of change of the area A(t) of this triangle is given by the expression
$$\frac{dA}{dt} = \frac{\left|r_{2}\right|^{2}(v_{1}\bullet r_{1})+\left|r_{1}\right|^{2}(v_{2}\bullet r_{2})-(r_{2}\bullet r_{1})(v_{1}\bullet r_{2}+v_{2}\bullet r_{1})}{2\left|r_{2}\times r_{1}\right|}$$

2. Relevant equations
$$\left|a\times b\right|=|a||b|sin\theta$$ maybe?

3. The attempt at a solution
Originally, I thought I might get the formula for the area of the triangle, then simply differentiate it in respect to t. However, I only got as far as $$A=|r_{1}\times r_{2}|/2$$before realizing I don't know how to differentiate the magnitude of a cross product. However, I can't think of any other way to express the area of the triangle in terms of the two vectors...

Any hints would be greatly appreciated.

2. Sep 15, 2009

### lanedance

before jumping into the cross product, trying differntiating a magnitude of a vector may help:

magnitude of a vector
$$|\textbf{u}|= (\textbf{u} \bullet \textbf{u})^{\frac{1}{2}}$$

differentiating
$$\frac{d}{dt}|\textbf{u}| = \frac{d}{dt}|(\textbf{u} \bullet \textbf{u})^{\frac{1}{2}} = \frac{ \frac{d}{dt}(\textbf{u} \bullet \textbf{u})}{2|\textbf{u}|} = \frac{ (\textbf{u}' \bullet \textbf{u}+\textbf{u} \bullet \textbf{u}')}{2|\textbf{u}|} = \frac{ \textbf{u}' \bullet \textbf{u} }{|\textbf{u}|} = \textbf{u}' \bullet \widehat{\textbf{u}}$$