Derivative No. 6

Using the methods in this post, I would like to evaluate

$\frac{dx(\theta)}{d\theta}\bigg|_{\theta^+}$

with

$x(\theta)=r\sin\theta\cos\phi$.


Substituting, the expression to evaluate is

$ \frac{d r \sin\theta \cos \phi }{d\theta} \bigg|_{\theta^+}$.

From the product rule,

$ \frac{dx(\theta)}{d\theta} \bigg|_{\theta^+} = \sin\theta \frac{d r \cos \phi }{d\theta}\big|_{\theta^+} + r \cos \phi \frac{d \sin\theta }{d\theta}\big|_{\theta^+}$

Since $ r \cos \phi $ does not depend on $\theta$, $ r \cos \phi $ is a constant function with respect to $\theta$. From this post, it follows that

$\frac{d r \cos \phi }{d\theta} \bigg|_{\theta^+} = 0$, so

$ \frac{dx(\theta)}{d\theta} \bigg|_{\theta^+} = r \cos \phi \frac{d \sin\theta }{d\theta}\big|_{\theta^+} $

In this post, it was shown that

$ \frac{d \sin\theta}{d\theta} \bigg|_{\theta^+} = \cos\theta $.

In the current post, the independent variable is $\theta$ instead of $x$.

Therefore,

$ \boxed{ \frac{dx(\theta)}{d\theta} \bigg|_{\theta^+} = \cos\theta r \cos \phi   = z(r,\theta,\phi) \cos \phi  } $.

Written by

I am a research assistant in theoretical physics. This website, derive-it.com, serves to organize my ongoing learning and research as well as to provide a resource to other learners around the world.

Leave a Reply