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7

$dx$ is called a differential. The notation is due to Leibniz and while the original meaning might have changed from being an infinitesimal quantity to a modern style differential, one still keeps the notation just because it allows to do some basic math on the differentials (like you wrote) and get reasonable results. Like the chain rule $$\frac{df}{dx} = ... 6 This is a nice question, and I don't have a comprehensive answer for you, but perhaps this will help a bit. For conservative forces, it's generally convenient to define the scalar potential V. The force will satisfy \vec{F} = \vec{\nabla}V, and your potential V = r (yay, system is conservative). Since the kinetic + potential energy is constant, and ... 4 The technique being used here is called separation of variables. It can be done without separating differentials by using the chain rule. Specifically, you have (simplifying the notation a little bit)$$\frac{dv}{dt} = \frac{mg - cv}{m}.$$By dividing both sides by the expression on the right, you get$$\frac{dv}{dt} \frac{m}{mg-cv} = 1.$$Now integrate ... 4 Recall where the negative sign comes from in physics -- it is simply due to your coordinate system and point of view. The difference is analogous to the difference between work done by gravity and work done on gravity. 4 Consider the retarted propagator of a free right moving quantum mechanical particle with linear dispersion$$ G(\omega,k,\delta)=\frac{1}{\omega-vk+i \delta} $$(the +i \delta is there for causality reasons and have set to be 0 in the end (in a limiting procedure)) Calculating the corresponding density of states requires the calculation of ... 3 I recommend Div, Grad, Curl, and All That: An Informal Text on Vector Calculus by H. M. Schey. 2 Since$$ \frac{dw_1}{dt}w_1+\frac{dw_2}{dt}w_2=\frac12\frac{d}{dt}(w_1^2+w_2^2), $$your differential equation is equivalent to:$$\tag{1} \frac{du}{dt}=\alpha u, $$with$$ u=w_1^2+w_2^2,\quad \alpha=-2\frac{c}{I}. $$Solving (1) we get:$$ w_1^2(t)+w_2^2(t)=\beta\exp\left(-2\frac{ct}{I}\right), $$where$$ \beta=w_1^2(0)+w_2^2(0). $$2 This question was already answered by Gil Bor, but I just wanted to add some remarks. Consider quaternionic projective space \mathbb{H}P^n for instance. It is quaternion-Kahler, but cannot be Kahler for topological reasons, since it has b_2 = 0. On the other hand, a Kahler manifold has by definition a closed Kahler form, which is not exact, since its ... 2 We will show that one need not calculate the energy stored in the electrostatic field by integration. To that end, we proceed. We start with the electrostatic energy W stored as given by$$W=\frac{\epsilon_0}{2}\int_V \vec E\cdot \vec E \,dV \tag 1$$where V is all of space. We now assume that the field is induced by a charge distribution ... 2 Consider an orbit$$\gamma: \quad t\mapsto z(t)=r(t)e^{i\phi(t)}$$of such a particle. We are interested in the polar representation$$\phi\mapsto r(\phi)\tag{1}$$of the resulting curve \hat\gamma\subset {\mathbb C}. Denoting the differentiation with respect to t by a \cdot and the differentiation with respect to \phi by a ' we have \dot ... 2 Let I_i be the intensity at 50m and I_f be the intensity at 100m. The intensity is proportional to the inverse of the square, but not equal to it. At 50m, we have L=10log(\frac{I_i}{I_0})=130. From that, I_i=10. Because it is proportional to the inverse of the square of the distance, \frac{I_f}{I_i} = \frac{d_i^2}{d_f^2} = \frac{1}{4}. Then, ... 2 By using Earth's orbital parameters, you are calculating the sun's standard gravitational parameter. You need to use the orbital parameters for a satellite of Earth to get the value for Earth. You also need to keep the units consistent, so the radius should be in km. For the sun, it comes out close at 1.31E11 For the Earth, I know GEO is 42164km and ... 2 It is actually possible to define these kind of concepts only when every statement is meaningful: the example you reported lacks of sense (or many passages are in the best case implied) and rigour although it is a very common way to present introductory physics. Just for the case you mentioned: given the position of a material point with respect to time as ... 2 To be precise, we have for an electrostatic field$$\phi(\vec r_1)-\phi(\vec r_2)=\int_{C_{12}}\vec E(\vec r')\cdot d\vec r'$$where C_{12} is any contour that begins at \vec r_1 and ends at \vec r_2. For \vec E(\vec r) =\hat r E_r(r)=\hat r\frac{\lambda}{2\pi \epsilon_0\,r}, we have$$\phi(\vec r)-\phi(\vec r_0)=\int_{\vec r}^{\vec r_0} \hat ...

2

From the get-go we should be able to tell the equation (2) is incorrect, since it equals $$\left(\int_{-\infty}^\infty p(x_1)dx_1\right)\cdots\left(\int_{-\infty}^\infty p(x_{n-2})dx_{n-2}\right)\int_{-\infty}^\infty p(x_{n-1})p(x-x_{n-1})dx_{n-1} \tag{2-bad}$$ which is $1\cdots 1\cdot P_2(x)$ instead of the $P_n(x)$ it's supposed to be. The correct ...

2

Write a single node equation in terms of the impedances of L and C and the resistance R. $$\frac{V}{R} + \frac{V}{j \omega L} + \frac{V}{\frac{-j}{\omega C}} = I_0 \angle \phi$$ where j is i for math people. V is a phasor (complex number) like $I_0 \angle \phi$ which represents the amplitude and phase of the voltage across all 3 components which will ...

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Using mostly geometry for estimation: In other words, how much "space" does the water of the oceans occupy? Earth radius is about $r = 6400 \mbox{ km}$. Water covers about $70\%$ of the earth surface, estimation: $A_w = 4 \pi r^2 = 360,302,978 \mbox{ km}^2$ Wikipedia lists $360,570,000 \mbox{ km}^2$. Average ocean depth is $d = 3.8 \mbox{ km}$. So ...

1

A general way to derive a weak form is to multiply a test function on both sides of the equation and then integrate them. The second step is to use some kind of divergence theorems to derive the weak solution such that the solution is some what not so smooth as in the strong form. For your question here, we can derive the weak form as follows: Let the ...

1

You need to be careful with the differences between linear speed (which is the magnitude of linear velocity), linear velocity, and angular velocity. The wheel is rotating as a rigid object, so all points at the same distance from the center have the same linear speed. The linear velocity will change as the wheel goes around. There is no angular ...

1

The following should work at least approximately. The granularity of time may cause the planet not to follow a closed orbit. To improve precision you should use a smaller $\Delta t$ and/or learn something about the numerical analysis of handling this type of differential equations more accurately. Initialize variables (run tests to find combos that work for ...

1

Just apply the chain rule and use the differential equation. $$\frac{d}{dt}H(x(t),y(t)) = \partial_x H(x(t),y(t)) \frac{dx}{dt} + \partial_y H(x(t),y(t)) \frac{dy}{dt}$$ Now use the equation to get: $$\partial_x H(x(t),y(t)) \partial_y H(x(t),y(t)) - \partial_y H(x(t),y(t)) \partial_x H(x(t),y(t)) = 0$$ So the function $H$ is constant along the ...

1

Why would you need to have $||{\bf r}' (t)|| \neq 0$? this constraint is not necessary at all. You have pointed out yourself that you are using Newton's second law, which states ${\bf F} = m \frac{d {\bf v}}{dt}$, if ${\bf F}$ was to be directed in oppoiste direction to ${\bf v}$ and you started with a given value of ${\bf v} = {\bf v}_0$ then certainly ...

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You calculated $$\phi (x,y,z)=\int (x y-\sin (z)) \, dx=\text{c1}(y,z)+\frac{x^2 y}{2}-x \sin (z)$$ Now, since you are integrating with respect to $x$ only, you can have a function $c_1(y,z)$, which represents the constant of integration which usually appears. Oberve that in this case, this constant is a function, since $\frac{\partial c_1(y,z)}{\partial ... 1 First, the Fourier Transform pairs$f\leftrightarrow Fare given by \begin{align} F(p)&=\int_{-\infty}^{\infty}f(x)e^{-ipx/\hbar}dx \\\\ f(x)&=\frac{1}{2\pi\hbar}\int_{-\infty}^{\infty}F(p)e^{ipx/\hbar}dp \tag 1 \end{align} Next, we know that the Dirac Delta acts as a distribution on a test function\phi$with ... 1 I think it depends on how you organize your problem. Suppose your problem takes place in a time interval$a \le t \le b$. Some things you know are$f'(t)$and$f(b)$. Then to find$f(t)$, you may think of integrating backward from$b$. And maybe some ways to write that would involve thinking of negative$dt$. 1$9.5-u_{rest}$can be accounted for by your constant$C_0$. Since you already can see why the third option is correct (actually directly from the question by using the integrating factor), you can perform a substitution to get the first option, and then add the homogenous solution to get the second option. 1 HINT: Only two things needed: The electric field generated by a charged sphere has rotational symmetry. Gauss' law From 1. and 2. you conclude that the field outside the sphere is the same with the field generated by all the charge concentrated at the center, and that the field inside the sphere is zero. 1 The work done in assembling the entire system, i.e, building it from radius$r\rightarrow r+ dr$and the corresponding surface, to attain the full sphere, would be pretty rigorous, so that's an option. Otherwise, you could always integrate$\frac{\epsilon}{2}\int_VE^2 dV$. 1 The issue is that the physics equations often use a finite difference approximation to a derivative: $$x'=\frac{dx}{dt}\approx\frac{\Delta x}{\Delta t}$$ Whenever$x$is linear over the region of interest, that approximation is correct. In general, if the time-step ($\Delta t$) is small enough, then the linearity approximation does hold. In fact, this is the ... 1 It appears to me that they’ve simply observed that $$\lim_{\ell\to 0}\left|\mathbf{r}+\frac12\ell\mathbf{k}\right|\left|\mathbf{r}-\frac12\ell\mathbf{k}\right|=|\mathbf{r}|^2\;.$$ Thus,$\$\begin{align*} \lim_{{\ell\to ...

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