Day: February 8, 2021

Derivative No. 7

Using the methods in this post, I would like to evaluate $\frac{dy(\theta)}{d\theta}\bigg|_{\theta^+}$ with $y(\theta)=r\sin\theta\sin\phi$ Substituting, the expression to evaluate is $ \frac{d \sin \theta r \sin \phi  }{d\theta} \bigg|_{\theta^+}$. From the product rule, $ \frac{dy(\theta)}{d\theta} \bigg|_{\theta^+} = \sin\theta \frac{d r \sin \phi  }{d\theta}\big|_{\theta^+} + r \sin \phi \frac{d \sin\theta }{d\theta}\big|_{\theta^+}$ Since $r \sin \phi $ does not depend on …

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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 $\theta$, $ …

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Derivative No. 5

Using the methods in this post, I would like to evaluate $\frac{dy(\phi)}{d\phi}\bigg|_{\phi^+}$ with $y(\phi)=r\sin\theta\sin\phi$. Substituting, the expression to evaluate is $ \frac{d r \sin \theta \sin \phi }{d\phi} \bigg|_{\phi^+}$. From the product rule, $ \frac{dy(\phi)}{d\phi} \bigg|_{\phi^+} = r \sin \theta \frac{d \sin \phi }{d\phi}\big|_{\phi^+} + \sin \phi \frac{d  r \sin \theta  }{d\phi}\big|_{\phi^+}$ Since $ r \sin \theta $ does not depend on $\phi$, $ …

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Derivative No. 4

Using the methods in this post, I would like to evaluate $\frac{dz}{d\phi}\bigg|_{\phi^+}$ with $z=r\cos\theta$. Substituting, the expression to evaluate is $ \frac{d (r \cos\theta )}{d\phi} \bigg|_{\phi^+}$. Since $ r \cos\theta $ does not depend on $\phi$, $ r \cos\theta $ is a constant function with respect to $\phi$. From this post, it follows that $ \boxed { \frac{dz}{d\phi} …

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Derivative No. 3

Using the methods in this post, I would like to evaluate $\frac{dz(r)}{dr}\bigg|_{r^+}$ with $z(r)=r\cos\theta$. Substituting, the expression to evaluate is $ \frac{d (r \cos\theta)}{dr} \bigg|_{r^+}$. From the product rule, $ \frac{dz(r)}{dr} \bigg|_{r^+} = r \frac{d( \cos\theta )}{dr}\big|_{r^+} + \cos\theta \frac{dr}{dr}\big|_{r^+}$ Each $x$ in the product rule has been replaced by $r$. Since $ \cos\theta $ does not depend on …

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