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0

If you don't understand the other method, this is how you would do in the realm of real numbers : Write $$\frac{d^2\theta}{dt^2}=\frac{d\omega}{dt}=\omega\frac{d\omega}{d\theta}$$ I have used that $\omega=\frac{d\theta}{dt}$ in the last step. Now your equation is first order and you can solve it easily. $$\omega d \omega=-k^2 \theta d\theta$$ You can ...

1

The equation is a second-order linear differential equation; in standard form, $$\frac{\mathrm{d}^2 \theta(t)}{\mathrm{d}t^2} + k^2\theta = 0$$ For the case of constant coefficients, one must simply propose the ansatz $\theta(t)=e^{Rt}$, where $R$ is a constant. Plugging into the differential equation yields, $$e^{Rt}(R^2+k^2) = 0$$ This is only the case ...

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It's not as simple as that. You'll have to obtain velocity and displacement by integrating your given acceleration and using correct boundary conditions. For example: Suppose the acceleration is given by A(t) = 2t [m/s²] and the problem states that the particle starts its movement from rest and from the origin of your coordinate system, so that X(t=0)=0 ...

1

All the other answers that say the "single" integral is simply a shorthand notation are right, but it is well to remember that one can indeed construe the integral as a single integral as a Lebesgue integral (if you do nothing else, look up Lebesgue's very cute little half paragraph summary (on the Wiki page) of his idea in a letter to Paul Montel). If ...

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$\int^t_0 A x^2 dt = x_0 + A x^2 t$ is incorrect. You are assuming $x$ as a constant. $x$ is a function of time x(t). Try $\dfrac{dx}{dt}=Ax^2 \implies \dfrac{dx}{x^2}=Adt$. Now integrate both the sides in appropriate limits. $$\int_{x_0}^{x(t)}\dfrac{dx}{x^2}=\int_0^t Adt$$ $$\int_{x_0}^{x(t)}x^{-2}dx=\int_0^t Adt$$ ...

3

First problem: you say $v(t) = A x^2$, but that is a function of position, not time. Putting the definition right: $$v = \frac{dx}{dt} = A x^2$$ You can regroup terms on the same variable: $$\frac{dx}{x^2} = A dt$$ And then do the integration: $$\int \frac{dx}{x^2} = \int A dt$$ This is homework, so I will leave the integral limits and the ...

13

It is, in fact, a double integral! The first notation used $$\varPhi_E = \oint_S \vec{E} \cdot \mathrm{d}\vec{A} = \oint_S \vec{E} \cdot \hat{n} \ \mathrm{d}A$$ is simply a more compact notation. It's much easier to write $\mathrm{d} \vec{A}$ instead of, say, $r \ \mathrm{d}r \ \mathrm{d}\theta$ all the time. Furthermore, it's more general, as $\mathrm{d} ... 4 The general formula is indeed a double integral, so the most technically correct way to write it is $$\Phi_E = \iint_S \vec{E}\cdot\mathrm{d}^2\vec{A}$$ But when formulas start to involve four, five, or more integrals, it gets tedious to write them all out all the time, so there's a notational convention in which a multiple integration can be designated by ... 5 It is just a more compact notation. It is implied by the integration element$dA$that you are integrating over the surface. 1 The factor comes from the moment of inertia of the infinitesimal piece. In the disc method, each piece is a filled flat circle (a disc) of radius$r$, and the moment of inertia of a flat circle is$\frac{1}{2}mr^2$. The$\frac{1}{2}$accounts for the fact that the mass of the circle is distributed between the center and the edge. But in the shell method, ... 3 Maxwell's equations in curved spacetime are written in the form $$\begin{split}\nabla_a F^{ab} &= - 4\pi J^b,\\ \nabla_{[a} F_{bc]} &= 0,\end{split}$$ with$F$the Faraday two-form,$J^a$the current four-vector,$\nabla$the covariant derivative and$[]$denotes antisymmetrization of the indices. In terms of exterior calculus they become: $$... 1 So on in three-dimensional Euclidean space we have an isomorphism between vectors and 1-forms, the usual way$$\eta_\mu = g_{\mu\nu} \eta^\mu.$$We also have an isomorphism between 1-forms and 2-forms, given by \star : dz\mapsto dx\wedge dy and cyclically. This isomorphism has a fancy name, the Hodge dual, if you want to know about it in general. Then if ... 1 Without specifying any particular scenario, and ignoring any proportionality constants, simply consider some general differential form \omega, and let this represent the electric flux through a closed surface which bounds some volume V. In classical electromagnetism, the Gauss law tells us that the flux through a closed surface is proportional to the ... 0 I will give you the general strategy for a problem of this kind. You know that acceleration is the instantaneous rate of change of position:$$a=\frac{dv}{dt}$$Through change of variables, this can also be equal to:$$a=\frac{dv}{dx}\times\frac{dx}{dt}.$$By chain rule:$$a=v\frac{dv}{dx}$$Knowing the acceleration as a function of position means you have: ... 3 The integral of a scalar function / vector field along a curve is defined with reference to a parametrization, e.g for a scalar$$ \int_C f \mathrm{d}s := \int_a^bf(r(t))\lvert r'(t)\rvert \mathrm{d}t$$where$ r: (a,b)\rightarrow \mathbb{R}^n $is a parametrization of the curve$C$. It is then shown, that the integral is invariant under reparametrization. ... 0 Note that all your integrals are Gaussians in differences of positions at successive instants$(x_k - x_{k-1})$so implement a change of integration variable from$x_k \longrightarrow (x_k - x_{k-1})$. You will have$N-1$(straightforward) integrations to perform with$x_0$and$x_N\$ held fixed.

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