Differentiation

Del is More than an Upside Down Triangle, Part 1

What is Del? In math, the symbol $\vec{\nabla}$ is called “del.” This symbol is defined in terms of Cartesian coordinates. $\vec{\nabla} \equiv \frac{d}{dx}\vec{e}_x + \frac{d}{dy}\vec{e}_y + \frac{d}{dz}\vec{e}_z$ The right side is a sum of unit vectors. So $\vec{\nabla}$ is a vector. This is why I write $\vec{\nabla}$ instead of just $\nabla$. Is it possible to …

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The Jacobian Matrix

Using conclusions from previous posts, the following nine derivatives have been determined. $\frac{dx(r)}{dr}\big|_{r^+}=\sin\theta\cos\phi$ $\frac{dy(r)}{dr}\big|_{r^+} = \sin\theta\sin\phi$ $\frac{dz(r)}{dr}\big|_{r^+} = \cos\theta$ $\frac{dx(\phi)}{d\phi}\big|_{\phi^+} = -r\sin\theta\sin\phi$ $\frac{dy(\phi)}{d\phi}\big|_{\phi^+} = r \sin\theta \cos\phi$ $\frac{dz(\phi)}{d\phi}\big|_{\phi^+} = 0$ $\frac{dx(\theta)}{d\theta} \big|_{\theta^+} = r\cos\theta \cos\phi$ $\frac{d y(\theta)}{d \theta} \big|_{\theta^+} = r \cos\theta \sin\phi $ $\frac{dz(\theta)}{d\theta} \big|_{\theta^+} = -r \sin\theta$ Next, recall the following result from this …

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How to Use the Product Rule

I would like to evaluate two more derivatives. They are $ \frac{dx}{d\phi}\big|_{\phi^+} $ and $ \frac{dz}{d\theta}\big|_{\theta^+} $ given $x = r \sin\theta\cos\phi$ and $z = r\cos\theta$.   Start with $ \frac{dx}{d\phi}\big|_{\phi^+} $. The first step is to substitute $x$ with $r \sin\theta \cos\phi$. $ \frac{dx}{d\phi}\big|_{\phi^+} = \frac{d r \sin\theta \cos \phi}{d\phi}\big|_{\phi^+} $. The next step is …

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Proving the Chain Rule

In this post, the chain rule is proved. This rule frequently appears in Calculus. Recall from this post that: $dx|_{a^+} \equiv \lim_{\Delta x \rightarrow 0^+} \Delta x$ and $df(x)|_{a^+} = \lim_{\Delta x \rightarrow 0^+} \Delta f (\Delta x)$. Suppose a variable $y$ can be written as a function of another variable $u$, and that $u$ can be written …

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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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