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. Each orientation can define a coordinate system, since each unit vector can point in the positive direction of an axis. Two coordinate systems related by less than a quarter of a rotation are considered equivalent in contrast to unique, to avoid counting a numerous amount of coordinate systems formed by rotations. Finally, which of the resulting orientations of unit vectors form …

### 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 express $\vec{\nabla}$ in terms of spherical coordinates? I will find out. This post lists nine derivatives relating spherical coordinates to Cartesian coordinates. As suggested in Reference [1], these nine derivatives are useful for expressing del in terms of spherical coordinates. I return to the definition of …

### 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 coordinate system? How the Spherical Coordinate System is Different The first step is to convert $f(x,y,z)$ into a function $g(r,\phi,\theta)$ of the spherical coordinates, $r$, $\phi$, and $\theta$. Here, $\phi$ is the azimuthal angle in the $x-y$ plane, and $\theta$ is the polar angle. The …

### 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 post. This matrix equation can be rewritten by substituting the nine derivatives listed at the beginning of the current post. Replacing each $a$ with the corresponding independent variable, the result is: This matrix equation includes nine equations relating dummy variables in Cartesian coordinates to dummy …

### 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 to apply the product rule to the right side. $\frac{dx}{d\phi}\big|_{\phi^+} = r \sin\theta \frac{d \cos \phi}{d\phi}\big|_{\phi^+} + \cos \phi \frac{dr \sin\theta}{d\phi}\big|_{\phi^+}$. Using this derivative of cosine, the previous equation becomes $\frac{dx}{d\phi}\big|_{\phi^+} = -r \sin\theta \sin\phi + \cos \phi \frac{d r \sin\theta}{d\phi}\big|_{\phi^+}$. From this post about the derivative of …

### 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 $x$, replace $f$ with $\cos$: $\frac{d\cos(x)}{dx}\bigg|_{a^+} = \lim_{\Delta x \rightarrow 0^+} \frac{ \cos(a + \Delta x) – \cos(a) }{\Delta x} $. Recall the angle addition identities. Specifically, $ \cos(A+B) = \cos A \cos B – \sin A \sin B$. Using this identity, the numerator of the previous …

### 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 as a function of another variable $x$. Then $y$ can be written as a function of $x$: $y(x)$. What is $\frac{dy(x)}{dx}\bigg|_{a^+}$ if only $y(u)$ and $u(x)$ are known? Solution: Start by writing $\frac{dy(x)}{dx}\bigg|_{a^+} = \frac{dy(x)}{dx}\bigg|_{a^+}$. According to this post, the $|_{a^+}$ symbols can be moved as …

### 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 $\theta$, $r \sin \phi $ is a constant function with respect to $\theta$. From this post, it follows that $\frac{dr \sin \phi }{d\theta} \bigg|_{\theta^+} = 0$, so $ \frac{dy(\theta)}{d\theta} \bigg|_{\theta^+} = r \sin \phi \frac{d \sin\theta }{d\theta}\big|_{\theta^+} $ In this post, it was shown that $ \frac{d \sin\theta}{d\theta} \bigg|_{\theta^+} = …

### 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 …

### 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 $\phi$, $ r \sin \theta $ is a constant function with respect to $\phi$. From this post, it follows that $\frac{d r \sin \theta }{d\phi} \bigg|_{\phi^+} = 0$, so $ \frac{dy(\phi)}{d\phi} \bigg|_{\phi^+} = r \sin \theta \frac{d \sin \phi }{d\phi}\big|_{\phi^+} $ In this post, it was shown that $ \frac{d \sin\phi}{d\phi} \bigg|_{\phi^+} = …

### 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} \bigg|_{\phi^+} = 0 }$.

### 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 $r$, $ \cos\theta $ is a constant function with respect to $r$. From this post, it follows that $\frac{d ( \cos\theta ) }{dr} \bigg|_{r^+} = 0$, so $ \frac{dz(r)}{dr} \bigg|_{r^+} = \cos\theta \frac{dr}{dr}\big|_{r^+} $. In this post, it was shown that $ \frac{dr}{dr} \big|_{r^+} = 1 $, so $\boxed{ \frac{dz(r)}{dr} …

### Derivative No. 2

Using the methods in this post, I would like to evaluate $\frac{dy(r)}{dr}\bigg|_{r^+}$ with $y(r)=r \sin\theta\sin\phi$. This method is not the traditional method taught in a Calculus course because I only want to use properties and rules that I have derived myself on this website. Substituting, the expression to evaluate is $ \frac{d ( r \sin \theta \sin \phi )}{dr} \bigg|_{r^+}$. From the product rule, $ \frac{dy(r)}{dr} \bigg|_{r^+} = r \frac{d(\sin \theta \sin \phi)}{dr}\big|_{r^+} + \sin \theta \sin \phi \frac{dr}{dr}\big|_{r^+}$ Each $x$ in the product rule has been replaced by $r$. Since $ \sin \theta \sin \phi $ does not depend on $r$, $ …

### Derivative No. 1

I would like to evaluate $ \frac{dx(r)}{dr} \bigg|_{x^+}$ with $x(r) = r \sin\theta \cos\phi$. Substituting, the expression to evaluate is $ \frac{d(r \sin\theta \cos\phi )}{dr} \bigg|_{x^+}$ From the product rule, $ \frac{dx(r)}{dr} \bigg|_{r^+} = \sin\theta \cos\phi \frac{dr}{dr} \bigg|_{r^+} + r \frac{d ( \sin\theta \cos\phi)}{dr} \bigg|_{r^+}$. Each $x$ in the product rule has been replaced by $r$. Since $ \sin\theta \cos\phi $ does not depend on $r$, $ \sin\theta \cos\phi $ is a constant function with respect to $r$. Differentiating the constant function, $\frac{d ( \sin\theta \cos\phi)}{dr} \bigg|_{r^+} = 0$, so $ \frac{dx(r)}{dr} \bigg|_{r^+} = \sin\theta \cos\phi \frac{dr}{dr} \bigg|_{r^+} $. From the Calculus …

### The Derivative of a Constant Function

From this post, one definition of a derivative is $\lim_{\Delta x\rightarrow0^+}\frac{f(a+\Delta x)-f(a)}{\Delta x}\equiv\frac{d f(x)}{dx}\big|_{a^+}$. In this case, the values of $\Delta x$ are restricted to positive values due to the $+$ in $0^+$ written in the limit. A function that does not vary with respect to an independent variable is called a constant function. On a graph with perpendicular $x$ and $y$ axes, a constant function looks like a horizontal line. The slope, or $\frac{\Delta y}{\Delta x}$, of a constant function $f(x) = C \in \mathbb{R}$ is equal to $0$ because $\Delta y=0$ and $\Delta x \ne 0$. A slope of …