Reasoning about Left and Right Handed Coordinate Systems

Arranging Unit Vectors How many unique orientations can three unit vectors be arranged, if all three unit vectors must be perpendicular? Here, a unique orientation is one that cannot be rotated by less than 90 degrees into another orientation–admittedly, this is a somewhat arbitrary definition. Anyway, each orientation can define a coordinate system, since each …

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 …

How to Integrate in a Spherical Coordinate System

Review of Integration Integration with Cartesian coordinates is simple. The general form is $\int\int\int f(x,y,z)dxdydz$ in which $f(x,y,z)$ is an arbitrary function of the Cartesian coordinates. However, there may be cases in which integrating with spherical coordinates is more convenient. Given the above, general form for integration with Cartesian coordinates, how can one integrate in a spherical …

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 …

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 …

Differentiating the Cosine Function

In this post, the derivative of the cosine function is found. To do this, the steps in reference 1 are followed. Start with a definition of a derivative, from this post: $\frac{df(x)}{dx}\bigg|_{a^+} \equiv \lim_{\Delta x \rightarrow 0^+} \frac{ f(a + \Delta x) – f(a) }{\Delta x} $. Since $f(x)$ and $\cos(x)$ are both functions of $x$, …

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 …

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 …

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 $\theta$, $ …

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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