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Tutor profile: Jackson D.

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Jackson D.
Science, Engineering, and Mathematics Tutor
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Questions

Subject: Calculus

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Question:

Find the formula for the volume of a sphere of radius $$r$$ using integration with spherical coordinates.

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Jackson D.
Answer:

Set up a triple integral containing the Jacobian for a transformation into spherical coordinates: $$\iiint_V f(\rho,\phi,\theta)\, \rho^2sin(\phi)\, d\rho d\phi d\theta$$ Because we are finding a volume, there is no function $$f(\rho,\phi,\theta)$$, so disregard that term. Define the limits: $$\rho = [0,r]$$ $$\phi = [0,\pi]$$ $$\theta = [0,2\pi]$$ Integrate: $$\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{r} \rho^2sin(\phi)\, d\rho d\phi d\theta$$ $$=\int_{0}^{2\pi}\int_{0}^{\pi}\Big[_{0}^{r} \frac{\rho^3}{3}\Big]sin(\phi)\, d\phi d\theta$$ $$=\frac{r^3}{3}\int_{0}^{2\pi}\int_{0}^{\pi}sin(\phi)\, d\phi d\theta$$ $$=\frac{r^3}{3}\int_{0}^{2\pi}\Big[_{0}^{\pi}-cos(\phi)\Big] d\theta$$ $$=\frac{r^3}{3}\Big[-cos(\pi)--cos(0)\Big]\int_{0}^{2\pi}d\theta$$ $$=\frac{2r^3}{3}\int_{0}^{2\pi}d\theta=\frac{2r^3}{3}\Big[_{0}^{2\pi}\theta = \frac{2r^3}{3}*2\pi$$ $$V = \frac{4}{3}\pi\,r^3$$

Subject: Civil Engineering

TutorMe
Question:

What is the deflection of a fixed cantilevered beam of length $$L$$ and flexural rigidity $$EI$$ under a very small axial compressive load $$P$$ and bending moment $$M_{0}$$ applied at the free end? Hint: Use the Taylor expansion for $$cos(x)$$.

Inactive
Jackson D.
Answer:

Taylor expansion: $$cos(x) = \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!} = 1-\frac{x^2}{2}+\,\,...$$ When $$x$$ is very small (near zero), $$sec(x) = cos(x)$$ Therefore: $$sec(x) = cos(x) = \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!} = 1-\frac{x^2}{2}+\,\,...$$ Deflection (from the secant formula): $$\delta = \frac{M_{0}}{P}(sec\bigg(L\sqrt{\frac{P}{EI}}\bigg)-1)$$ $$\delta = \frac{M_{0}}{P}(1-\frac{\bigg(L\sqrt{\frac{P}{EI}}\bigg)^2}{2}-1)$$ $$\delta = \frac{M_{0}}{P}(1-\frac{L^2*(\sqrt{P})^2*(\sqrt{EI})^{-2}}{2}-1)$$ $$1-1 = 0$$ $$(\sqrt{P})^2 = P$$ $$(\sqrt{EI})^{-2} = \frac{1}{EI}$$ Plug in and cancel. Thus: $$\delta = \frac{M_{0}L^2}{2EI}$$

Subject: Physics

TutorMe
Question:

What is the speed of light, $$c$$ in terms of $$\epsilon_{0}$$ and $$\mu_{0}$$?

Inactive
Jackson D.
Answer:

Take a plane of current oriented out of the page. Imagine a rectangular wire loop of arbitrary length $$L$$ along a side of the plane and width extending past the furthest propagation of magnetic field. There will be a circulation of $$\vec{B}$$ (magnetic field) on one side of the plane equal to $$B*L$$. $$\oint_C\vec{B}\,dl = B*L$$ The integral form of Maxwell's version of Ampere's Law, pictured below, applies in this situation. There is no current induced in our imaginary loop, by virtue of XXX, but there is a flux of $$\vec{E}$$ equal to $$\vec{E}*area$$ where the $$area$$ is equal to the length of the loop times the propagation of the electric field into it, given by the velocity of the electromagnetic wave times time. $$\oint_C\vec{B}\,dl = \mu_{0}I+\mu_{0}\epsilon_{0}\frac{d\phi_{E}}{dt}, I = 0$$ $$\oint_C\vec{B}\,dl = \mu_{0}\epsilon_{0}\frac{d\phi_{E}}{dt}$$ $$\phi_{E} = E(L*vt)$$ Thus: $$\frac{d\phi_{E}}{dt} = ELv$$ We find that $$\vec{B} = \frac{\vec{E}}{v}$$ by employing Maxwell's version of Faraday's Law and finding the flux of $$\vec{B}$$ through another, perpendicular arbitrary loop. $$\oint_C\vec{E}\,dl = E*L = \frac{-d\phi_{B}}{dt}$$ $$\phi_{B} = B(L*vt)$$ $$\frac{d\phi_{E}}{dt} = BLv$$ Thus: $$\oint_C\vec{E}\,dl = E*L = BLv$$ Cancel $$L$$ and rearrange. $$\vec{B} = \frac{\vec{E}}{v}$$ Finally, using some algebra, we get the following. $$\oint_C\vec{B}\,dl = B*L = \mu_{0}\epsilon_{0}\frac{d\phi_{E}}{dt} = \mu_{0}\epsilon_{0}ELv$$ Cancel $$L$$ and rearrange. Input the expression relating $$\vec{E}, \vec{B},$$ and $$v$$ $$\frac{B}{E} = \frac{\frac{E}{v}}{E} = \mu_{0}\epsilon_{0}v$$ Cancel $$\vec{E}$$ and rearrange $$v^2 = \frac{1}{\mu_{0}\epsilon_{0}}$$ $$v = \frac{1}{\sqrt{\mu_{0}\epsilon_{0}}} = c$$, speed of light

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