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Tutor profile: Thomas M.

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Thomas M.
Patient and kind Mathematics and Physics tutor with over four years experience!
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Questions

Subject: Calculus

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Question:

Find the area of the open cylinder of radius $r$ and height $h$

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Thomas M.
Answer:

We have; $(\textbf{r}(u,v)=r\cos(u)\textbf{e}_x + r\sin(u)\textbf{e}_y+v\textbf{e}_z$) Where we take the domain $$D$$ as $$0\leq u \leq 2\pi$$ and $$0\leq v \leq h$$. Differentiating this both with respect to $u$ and $v$ yields the following; $(\frac{\partial\textbf{r}}{\partial u} = -r \sin(u)\textbf{e}_x + rcos(u)\textbf{e}_y$) $(\frac{\partial\textbf{r}}{\partial v} = \textbf{e}_z$) So, $( \frac{\partial\textbf{r}}{\partial u} \land \frac{\partial\textbf{r}}{\partial v} = rcos(u)\textbf{e}_x + r \sin(u)\textbf{e}_y$) This is the radial vector in the $$x,y$$ plane as expected so as the normal the surface. Then we can write the following; $(\Bigg| \frac{\partial\textbf{r}}{\partial u} \land \frac{\partial\textbf{r}}{\partial v} \Bigg| dudv = rdudv$) Which will simplify the integration phase of this example. So we then integrate, $(\text{Area(S)} = \int_s dS = \int_D \Bigg| \frac{\partial\textbf{r}}{\partial u} \land \frac{\partial\textbf{r}}{\partial v} \Bigg| dudv = \int^{2\pi}_0 \int^h _0 rdvdu = 2\pi hr$) So we then find $$\text{Area(}S\text{)} = 2\pi h r$$ as expected.

Subject: Physics

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Question:

Using a Schrodinger equation in one dimension, and a de Broglie plane wave as a solution, show that when $$V=0$$ this leads to the correct nonrelativistic relationship between energy and momentum.

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Thomas M.
Answer:

Recall the Schrodinger equation; $(i\hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \Psi}{\partial x^2} + V \Psi$) With $$V=0$$ we obtain; $(i\hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \Psi}{\partial x^2}$) In general, a plane has the form; $$\psi(\textbf{r},t)= A \exp[i(\textbf{k⋅r} - \omega t)]$$ When $$\textbf{k}$$ is pointing in the positive $$x$$ direction, $$\textbf{k⋅r} = k_x x$$ or since because $$k=k_x$$ in this case, we can write; $(\psi(\textbf{r},t)= A \exp[i(kx - \omega t)]$) If we recall de Broglie's hypothesis that $$p=\frac{h}{\lambda}$$, and since $$k=\frac{2\pi}{\lambda}, p=\hslash k$$, and also recalling $$E=hv=\hslash \omega$$ We can use this to write; $(\psi(\textbf{r},t)= A \exp\Bigg[\frac{i(px - Et)}{\hslash}\Bigg]$) Putting this into Schrodinger's Equation gives us; $(i \hslash A \exp \Bigg[ \frac{i(px - Et)}{\hslash}\Bigg]\Bigg(\frac{-iE}{\hslash}\Bigg) = - \frac{\hslash^2}{2m}\frac{\partial}{\partial x} A \exp\Bigg[\frac{i(px - Et)}{\hslash}\Bigg]\Bigg(\frac{ip}{\hslash}\Bigg) $) And $(E\psi(x,t) = \frac{\hslash^2}{2m}\Bigg(\frac{ip}{\hslash}\Bigg)^2 \psi(x,t) $) Which then finally yields; $(E=\frac{p^2}{2m}$) The above is the non-relativistic formula.

Subject: Algebra

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Question:

Using de Moivre's theorem; prove that $(\cos3\theta = \cos^3 \theta - 3\cos\theta \sin^2 \theta$)

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Thomas M.
Answer:

So, if we recall de Moivre's theorem; $((\cos\theta+i\sin\theta)^n = \cos n\theta+i\sin n\theta, \qquad \forall n \in \Re^+ $) Which extends to; $([r(\cos\theta+i\sin\theta)]^n = r^n(\cos n\theta+i\sin n\theta)$) So, by de Moivre's theorem; $(\cos3\theta + i\sin3\theta = (\cos\theta+i\sin\theta)^3$) $(\cos3\theta + i\sin3\theta = cos^3\theta+3cos^2\theta(i\sin\theta)+3cos\theta(i\sin\theta)^2+(i\sin\theta)^3$) Then by simplifying we obtain; $(\cos3\theta + i\sin3\theta = cos^3\theta-3\cos\theta\sin^2\theta + i(3\cos^2\theta\sin\theta-\sin^3\theta)$) Then the equation is split into a real part and a complex part. If we focus on the real part of the equation we see that; $(\cos3\theta= cos^3\theta-3\cos\theta\sin^2\theta$) And the imaginary part; $( i\sin3\theta = i(3\cos^2\theta\sin\theta-\sin^3\theta)$) We have now shown that; $$\cos3\theta = \cos^3 \theta - 3\cos\theta \sin^2 \theta$$

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