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Differentiation

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^+} = …

Differentiation

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 …

Differentiation

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^+} = …

Differentiation

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 }$.

Differentiation

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} …

Differentiation

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 $r$, $ …

Differentiation

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 …

Coordinate Systems Differentiation Integration

The Crux of Calculus

Define $\Delta x \equiv x_2 – x_1$, to be consistent with this post. Similarly, define $\Delta y \equiv y_2 – y_1$ and $\Delta z \equiv z_2 – z_1$. The Cartesian coordinates are $x$, $y$, & $z$. In contrast, the spherical coordinates are $r$, $\theta$, & $\phi$. Here, $\phi$ is the azimuthal angle in the $xy$-plane. Next, use this post to obtain the equations relating Cartesian coordinates to spherical coordinates. In particular: $x = r \sin\theta \cos \phi $ $y = r \sin \theta \sin \phi $ $z = r \cos\theta $ Note that $x$ changes if $r$ changes, $\theta$ changes, and/or $\phi$ changes. …

Differentiation

Definition of a Derivative

I copy the definitions of three different types of derivatives from [1]: $ \lim_{\Delta x \rightarrow 0} \frac{f(a+\Delta x) – f(a)}{\Delta x} \equiv \frac{d f(x)}{dx}\big|_{a}$ $ \lim_{\Delta x \rightarrow 0^+} \frac{f(a+\Delta x) – f(a)}{\Delta x} \equiv \frac{d f(x)}{dx}\big|_{a^+}$ $ \lim_{\Delta x \rightarrow 0^-} \frac{f(a+\Delta x) – f(a)}{\Delta x} \equiv \frac{d f(x)}{dx}\big|_{a^-}$ These definitions are best understood geometrically in terms of secant and tangent lines. Note that $ \Delta x$ never equals zero because otherwise the fraction would be undefined. The tangent line intersects the function at only one point. [1] David V. Widder. Advanced Calculus. Dover 1989.