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18

The result you've got would be better known as this: $$\int_0^t\biggl(\int_0^{t'} a\mathrm{d}t''\biggr)\mathrm{d}t' = \frac{1}{2}at^2$$ In other words, it's a derivation of the formula for uniformly accelerated motion. This derivation, or something like it, is one of the first things students in a good calculus-based introductory physics class learn. The ...

8

Here is a brief historical ideosyncratic intro to calculus. Calculus of finite differences Consider this problem from a typical IQ test: 2 5 10 17 26 ? What's the next number you expect in the sequence (this is not hard, you should do it). The n-th term in the sequence is given by: $$n^2 + 1$$ as you can see by substituting n=1,2,3,4,5, so the next ...

5

Consider Tsiolkovsky's rocket equation $\Delta v = v_e \ln \left( m_0/m_f \right)$ with $\Delta V$ the total change in velocity, $v_e$ the exhaust speed of the reaction products, $m_0$ the initial mass (structure+payload+propellant) and $m_f$ the final mass (structure+payload). If you ignore the atmosphere and other such "nuisances", it should be ...

4

This is more like a maths question to me. This is just an identity, which is true and facilitates the calculation and it is valid for any vector field. The proof, using Einstein summation convention would be something like: $$(\nabla \times \vec u )\times \vec u = \epsilon_{ijk}(\nabla \times u)_j u_k = \\ \epsilon_{ijk}\epsilon_{jlm}\partial_l (u_m) u_k ... 4 1.) The differentiation operator acting will give rise to Kronecker-Deltas since \frac{\partial x_a}{\partial x_b}=\delta_{ab} This will kill one summation. More specifially: \frac{\partial U}{\partial x_a}=-1/2 \sum_{ij}b_{ij}(\delta_{ai}x_j+\delta_{aj}x_i)=-1/2( \sum_{j}b_{aj}x_j+\sum_{i}b_{ia}x_i)=-\sum_{j}b_{aj}x_j. Rename j to be i and you're ... 4 You must first rewrite the old partial derivatives in terms of the new ones. A priori, they're some linear combinations with coefficients that could depend on the spacetime coordinates in general but here they don't depend because the transformation is linear. The rules$$ t'=t, \quad x'=x-Vt,\quad y'=y $$get translated to$$ \frac{\partial}{\partial t} = ...

3

One way to see that considering the dependence of $\dot{x}$ on $x$ is problematic is as follows: $x(t)$ maps a real number $t$ to another real number $x$. So $\dot{x}=dx/dt$ is the derivative of that map, meaning we take $$\lim_{\Delta t \to 0} \frac{x(t+\Delta t) - x(t)}{\Delta t}$$ So we can see that $dx/dt$ is itself another map from a real number $t$ ...

2

There isn't a simple answer to your question. The sort of basic calculus you need to understand the equations of motion isn't especially hard maths, but it takes a while to get comfortable with it. Your profile says you're 16, so I'd guess that if you do Physics and Maths at school you'll soon be learning calculus. In the UK you learn it as part of your ...

2

$$\text d f(x)=f'(x)\text d x\equiv\frac{\partial f}{\partial x}\text d x.$$ The second equation you posted is of course only a component representation, the vectors in the first equation are most definately higher dimensional (3 dimensional). And so the derivative in the second equation is chosen such that it goes in the direction of the path, denoted y s. ...

2

Using $v^2 = \vec v \cdot \vec v$ and $\dfrac{d}{dt} (f g) = f \dfrac{dg}{dt} + \dfrac{df}{dt} g$ write: $\dfrac{d}{dt}(v^2) = \dfrac{d}{dt}(\vec v \cdot \vec v) = \vec v \cdot \dfrac{d\vec v}{dt} + \dfrac{d\vec v}{dt} \cdot \vec v = 2 (\dfrac{d\vec v}{dt} \cdot \vec v)$ The fact that the $dt$ and $1/dt$ "cancel" in the RHS integral means that you're ...

2

Let $v(t)$ be velocity vector and define the scalar function $$f(t)~:=~\langle v(t),v(t)\rangle,$$ which is the absolute value of velocity (speed) squared. The time derivative is $${d \over dt}f(t)~=~ \left\langle {d \over dt}v(t),v(t)\right\rangle + \left\langle v(t),{d \over dt} v(t) \right\rangle ~=~ 2\ \left\langle{d \over ... 2 No need to make it that complicated--calculate the potential at the surface of a sphere with a uniform mass distribution with mass M. Then calculate the energy required to add a bit of mass with mass dm and radius dr to the top of this sphere. Express M and dm in terms of r and dr and integrate. 2 I don't have time for a detailed derivation, so the following can contain errors, so take it for what it's worth...In the following I assume that \mathbf{B} is constant in time. If not, the difference will just give (in the first approximation) the first term in the integral in the right-hand side. Let us consider the volume formed by \Sigma(t_0) and ... 1 The question seems to be ill-posed. From perspective I must say that mathematics knowledge requires constant improvement. I am doing a PhD now. As a student I did MSc in physics and separate MSc in mathematics. I think I have a good background to study new things, but I have to do that often. For example in quantum mechanics there is a notion of boundedness ... 1 See How Apollo flew to the moon the first lander and LCM had a large fuel reserve because they weren't sure and this was reduced to the final mission which returned to earth orbit with some very small amount left. Sorry don't have the book here for the actual figures. 1 The identity you're looking for is$$ \nabla \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times \mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B})  so indeed, if the curl of both factors in the cross product vanish, the divergence of the cross product vanishes, too.

1

To build upon @akhmeti's excellent answer, let us relax the assumption that $\partial \Sigma$ is constant in time. We let the boundary of the surface move along with the fluid. Then we must correct the divergence formula for the flux coming out of this edge strip: $\int_{\partial \Sigma} \mathbf{B} \cdot [d\mathbf{l} \times \mathbf{v} dt] = ... 1 In the left hand side, both B and the surface A are time dependant. In the right hand side, the first term is due to B changing (even if A is fixed), the second and third term come from the fact that A(t) is not constant. To take into account the variation of A with time, you need to use the convective derivative: D/Dt=d/dt+(V.grad) The integrand on the ... 1 What tolerances are acceptable? If I were the man that calculates this, I would calculate the computer error range. Then human-error range. Then think of the worst-case (maximum error). Then multiply that error-fraction with the total amount of fuel. Then add that much fuel onto the existing fuel. Errors would come from: Measurement (both human and ... 1 Very briefly... I'm following the advice of others here by presenting a short list of the resources I hunted for a couple of questions back! Fairly light and significantly incomplete, constantly update, though: Wikibook on Calculus A bit disconnected between lessons but fairly thorough and the videos are marvelous! MIT OpenCourseWare Basically, a ... 1 Perhaps the simplest and most grandiose way of describing calculus is: 'the mathematical study of change'. In practice, and simply, this general means integrals and derivates of functions. If a given function corresponds to some value of interest$f$which is a function of a parameter$x$, then the derivative of that function describes the change of$f\$ ...

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