0
$\begingroup$

I am trying to derive an expression for some partial derivatives in two different ways, but they seem to lead to inconsistent results.

I have got the Cartesian components of a two-dimensional position vector $x(t)$ and $y(t)$ and I need to evaluate their sensitivity with respect to the angle $\varphi(t)$, which is defined as $$\varphi(t) = \tan^{-1} \left( \frac{y(t)}{x(t)} \right)$$

According to the answers to this question, by using the chain rule, I can compute the partial derivative of $x(t)$ with respect to $\varphi(t)$ as $$ \frac{\partial x}{\partial \varphi} = \frac{\partial x / \partial x}{\partial \varphi / \partial x} = \frac{1}{\partial \varphi / \partial x} = \left( \frac{\partial}{\partial x} \left( \tan^{-1} \left( \frac{y(t)}{x(t)} \right) \right) \right)^{-1} = -\frac{x(t)^2 + y(t)^2}{y(t)} $$

However, by expressing the position in polar coordinates $$ x(t) = r(t) \cos(\varphi(t)) $$ $$ y(t) = r(t) \sin(\varphi(t)) $$ the derivative of $x(t)$ with respect to $\varphi(t)$ appears to be $$ \frac{\partial x}{\partial \varphi} = -r(t) \sin(\varphi(t)) = -r(t) \frac{y(t)}{r(t)} = -y(t) $$ This is a completely different result from the one obtained using the chain rule. So, which one is correct and why? Also, what am I doing wrong?

$\endgroup$
2

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.