Physics Teacher's Notes

. Integrate the following functions. 1. \(y(x) = x^2 + 5x + 9\). SOLUTION. The integral \(Y(x)\) can be written as three separate indefinite integrations,  \(Y(x) = \int x^2 dx + 5 \int x dx + 9 \int dx = \boxed{\frac{1}{3}x^3 + \frac{5}{2}x^2 + 9x + C }\textrm{   } \blacksquare\). 2. \(R(\theta) = \sin^2 \theta \cos \theta\). SOLUTION. Let \(u = \sin \theta\) such that \(du = \cos \theta \textrm{ } d\theta\). Thus,  \(\int \sin^2\theta \cos \theta \textrm{ } d\theta = \int u^2 du = \frac{1}{3}u^3 + C = \boxed{\frac{1}{3}\sin^3 \theta + C} \textrm{   } \blacksquare\). 3. \(Z(x) = 5x^2 \sec^2\left(x^3\right) \tan\left(x^3\right) \). SOLUTION. Let \(u = \tan\left(x^3\right)\), such that \(du = 3x^2 \sec^2 \left(x^3\right) dx\).  \(\int 5x^2 \sec^2\left(x^3\right) \tan\left(x^3\right) dx = \frac{5}{3} \int u du = \boxed{\frac{5}{6}\tan^2 \left(x^3\right) + C} \textrm{   } \blacksquare\).  4. \(A(y)...

. Given a square wave,  \[\begin{equation}f(t) = 1 \textrm{, for } 0 \leq t \leq t_0\end{equation}\] The Fourier Transform is given by:  \[\begin{equation}F(\xi) = \int_{-\infty}^{\infty} f(t) e^{-i 2\pi \xi t} dt\end{equation}\] Substituting the first equation to the second one yields: \[\begin{equation}F(\xi) = \int_{0}^{t_0} e^{-i 2\pi \xi t} dt\end{equation}\] \[\begin{equation}F(\xi) = -\frac{1}{i 2\pi \xi}\int_{0}^{t_0} e^{-i 2\pi \xi t} \cdot (-i 2\pi \xi) dt\end{equation}\] \[\begin{equation}F(\xi) = -\frac{1}{i 2\pi \xi} \left[e^{-i 2\pi\xi t}\right]_0^{t_0}\end{equation}\] \[\begin{equation}F(\xi) = -\frac{1}{i 2\pi \xi} \left[e^{-i 2\pi\xi t_0} - 1\right]\end{equation}\] \[\begin{equation}F(\xi) = \frac{1}{i 2\pi \xi} \left[1 - e^{-i 2\pi\xi t_0} \right]\end{equation}\] We can write this as: \[\begin{equation}F(\xi) = \frac{1}{i 2\pi \xi}\frac{e^{i 2\pi\xi (t_0/2)}}{e^{i 2\pi\xi (t_0/2)}} \left[1 - e^{-i 2\pi\xi t_0} \right]\end{equa...

. Part 1. Solutions and Solution Sets 1. The width (shorter dimension) of the A0 paper is \(x\). When the paper is folded crosswise along the middle of its length (longer dimension) \(y\), the resulting halves of size A1 would have the same aspect ratio (width/length) as the original paper. Repeating the process of preserving the aspect ratio by folding results in the smaller sizes A2 (half of A1 ), A3 (half of A2 ), A4 (half of A3 ), and so on. a. Using ratio and proportion, solve for the equation relating \(y\) and \(x\). SOLUTION. Note that the aspect ratios of A0 and A1 are equal. Thus, \[\begin{equation}\frac{x}{y} = \frac{y/2}{x}\end{equation}\] By cross multiplication and taking the square root, we get: \[\begin{equation}\boxed{y = \sqrt{2}x}\end{equation}\] b. Solve for the aspect ratios (length \(\times\) width) in terms of \(x\) for paper sizes A0 to A4 . [i.e. A0 is (\(\alpha_0 x \times x\)), A1 is (\(\alpha_1 x \times \beta_1 x\)), A...