2
$\begingroup$

As the title shows this question concerns nothing but chain rule. We now have:

$\frac{d\theta}{dt}=\frac{x}{r^{2}}\frac{dy}{dt}-\frac{y}{r^{2}}\frac{dx}{dt}$

I am assuming by chain rule we have $\frac{d\theta}{dt}=\frac{d\theta}{dx}\frac{dx}{dt}+\frac{d\theta}{dy}\frac{dy}{dt}$

But we have $\theta=\arccos[\frac{x}{r}]=\arcsin[\frac{y}{r}]$

Thus taking the derivative we should assume $\frac{d\theta}{dx}=-\frac{1}{y},\frac{d\theta}{dy}=\frac{1}{x}$ because $\frac{d}{dx}\arccos[\frac{x}{r}]=-\frac{1}{r\sqrt{1-\frac{x^{2}}{r^{2}}}}=-\frac{1}{\sqrt{r^{2}-x^{2}}}=-\frac{1}{y}$

However we know $-\frac{y}{r^{2}}\not=-\frac{1}{y}$ I computed this a few times but do not know where I got wrong. The relationship in the title is in Berkeley Problems in Mathematics.

  • 0
    @DonAntonio: The relationship is not incorrect, but it is incorrect to treat $r$ as a constant, which is why I commented earlier.2012-07-22

1 Answers 1

2

After the clarification (polar coordinates) and summarizing the answers given in the comments above: $x=r\cos\theta\,\,,\,\,y=r\sin\theta\,\,\,,\,\,r\geq0\,,\,\,\theta\in [0,2\pi]$ I'm assuming the radius is always non-negative, though not all do this.

From the above, $\,\theta=\arctan\frac{y}{x}\,\,,\,x\neq 0$ (the case $\,x=0\,$ is an easy particular case depending on the sign of $\,y\,$), so if both rectangular coordinates are derivable functions of some parameter $\,t\,$, we'd get: $\frac{d\theta}{dt}=\frac{d\theta}{dx}\frac{dx}{dt}+\frac{d\theta}{dy}\frac{dy}{dt}$ $\frac{d\theta}{dx}=-\frac{y}{x^2}\frac{1}{1+\left(\frac{y}{x}\right)^2}=-\frac{y}{x^2+y^2}$ $\frac{d\theta}{dy}=\frac{1}{x}\frac{1}{1+\left(\frac{y}{x}\right)^2}=\frac{x}{x^2+y^2}$

Observe that writing the expressions for $\,x,y\,$ from the beginning you get two differential equations. I'll leave this here as I'm not completely sure whether this already answers your question.

  • 0
    thanks a lot, but no need to do that. Gerry's answer is enough.2012-07-22