Physics Teacher's Notes

. Related post:  [GENPHY2] Right Hand Rule   At this point, let us consider one example of the right hand rule in nature. The magnetic force \(\vec{F}_m\) experienced by a moving charge \(q\) with velocity \(\vec{v}\) in a region with magnetic field \(\vec{B}\) is given by:  \[\begin{equation}\label{eqn:fm} q \vec{v} \times \vec{B} = \vec{F}_m\end{equation}\]  Just as in the previous post , let us use color codes for the vectors involved in the cross product:  The first vector \(q\vec{v}\) is red .  The second vector \(\vec{B}\) is green .  The third vector (the result of the cross product) \(\vec{F}_m\) is blue .  Before we proceed, notice that I specifically included the \(q\) (which is a scalar quantity) in the first vector. This has an important implication. The direction of the first vector is not simply the direction of the velocity; instead, it should include the effect of the sign  of the charge \(q\)....

. The exponential function is a special kind of function because it is the solution to the following differential equation:  \[\begin{equation}\label{eqn:diffeqn}\frac{dx}{dt} = rx\end{equation}\] i.e. the derivative (here shown as a rate of change) in the value of \(x\) is proportional to itself. The constant \(r\) is the exponential rate: if \(r > 0\), we will find an exponential growth, while if \(r < 0\), we will have an exponential decay.  Before solving Eqn (\(\ref{eqn:diffeqn}\)), let us discuss why this equation is important. Many processes in the real world starts out with trends comparable to the one modeled by this equation. For example, consider the case of the spread of infectious diseases, just like the one we are having right now. Let \(x(t)\) represent the number of infected individuals at a particular time \(t\). These infected individuals are the ones who can transmit the infection to the other individuals; more other people can be infecte...

. The right hand rule (RHR) is important for understanding the directions of the vectors involved in a cross product. Let me teach you some techniques and tricks that work. 😊  RHR for the Coordinate Directions  Coordinate Systems are Right-Handed. The three-dimensional coordinate systems that we use in mathematics actually follow the RHR. For example, the rectangular (Cartesian) coordinate system \(xyz\) follows this relationship for their unit vectors:  \[\begin{equation} \label{eqn:rhrxyz} \hat{i} \times \hat{j} = \hat{k}\end{equation}\]  Therefore, in understanding the RHR for cross products, we can refer to the right-handedness of coordinate systems. The Cartesian coordinate system is illustrated in Figure 1 below using the common directions we associate with them: \(+x\) is rightward, \(+y\) is upward, and \(+z\) is outward (by the way, the circle with a dot inside means outward, while the circle with an "X" inside means inward). For our ...

. 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...