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It's because of the product rule of derivatives, that states d(fg(t))/dt=f(t)(dg/dt)+g(t)(df/dt). In this case, let's call this part: (-isinθ+jcosθ)=g(t), and solve it: d(rωg)/dt=r(dωg/dt)=r(ω*(dg/dt)+g(dω/dt)). If you substitute g for (-i*sinθ+jcosθ), you get the equation for acceleration your book presents.


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The problem is that your point view is too "mathematical". No offence, but every acoustician would jump to the ceiling hearing "one can just erase all const, they do not change anything". Oh, they do $-$ very much! Since one of them is the sound speed... But I get it, you solve that as a mathematical problem and we are undoubtedly grateful for such people. ...


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Easy way Let me first state the straight-forward way to do this computation. $$ \langle \nabla_a \nabla_b V, \partial_c\rangle = \partial_a \langle \nabla_b V, \partial_c \rangle - \langle \nabla_aV, \nabla_a \partial_c\rangle = \partial_a (\nabla_bV)_c - (\nabla_bV)_d \Gamma_{ac}^d $$ First equality follows from compatibility, second equality uses ...


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Derivative is defined at a point not for an infinitesimal interval. $$\text {If $y=f(x)$} \Longrightarrow \; y’(x)|_{x=a}=\frac {dy}{dx}|_{x=a}=f’(a)=\lim_{\Delta x\to 0}\large{\frac {f(a+\Delta x)-f(a)}{\Delta x}}$$ $\large{\frac {d\vec \Omega}{dt}}$ doesn’t represent a fraction. That represents derivative of the $\vec \Omega$ relative to $t$ at a time ...


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If you ignore the coefficients, your function will be: f(x) = x / (x2 + r2)3/2 differentiating w.r.t x will give you: simple product rule f(x) = u(x)*v(x); u(x) = x; v(x) = 1 / (x2 + r2)3/2 f'(x) = u'(x)*v(x) + u(x)*v'(x) f'(x) = [1 / (x2 + r2)3/2] + [(-3/2)2x2 / (x2 + r2)5/2] to maximize/minimize you substitute f(x) = 0: 1 / (x2 + r2)3/2 = 3x2 / (x2 ...


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What an interesting question! It depends. In modern calculus, $\frac{df}{dt}$ is just a symbol for the derivative $$\lim_{h \to 0} \frac{f(t+h)-f(t)}{h}$$ As a matter of fact, mathematicians prefer different notations for the derivative of a function $f$, as $D f$ or $f'$. But the above definition of derivative became rigorous only when the concept of ...


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Using the word fraction is a more elementary way of naming the set of rational numbers : numbers that are represented by a ratio of integers. In physics when we model physical systems with differential equations we generally work in the domain of real numbers and sometimes complex numbers since we tend to think of real physical systems as existing in a ...


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I'd say the premise is the other way around: We assume $\Delta A$ to be small, as we intend to make it infinitesimal later on, and so $\Delta \theta$ is also small. Also, when you look at $\Delta A\Delta \theta$, this is even smaller, or "quadratically small" so to say. When doing math with small deltas or infinitesimals, you usually only have to take the ...


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In physics there are no infinitesimals, so dx is always treated as a "small but finite interval" during discussions, and only when the actual calculation is being done do we switch to mathematical mode, and "take the limit." During the 17th and 18th centuries, mathematicians and physicists both did this all the time. As they say in sports "no harm, no ...


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Yes: $$ \partial_\mu \mathrm e^{ikx}=ik_\mu\mathrm e^{ikx}+\mathrm e^{ikx}\partial_\mu $$ because the differential operator acts on a test function (i.e., product rule).


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The point is that one should distinguish between a total spacetime derivative $$ \frac{d}{dx^{\mu}}~=~ \frac{\partial }{\partial x^{\mu}}+ \phi_{,\mu} \frac{\partial }{\partial \phi}+ \phi_{,\mu\nu} \frac{\partial }{\partial \phi_{,\nu}}+\ldots \tag{1}$$ (where ellipsis denotes contributions in case of higher space-time derivatives), and an explicit ...


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There are two kinds of derivatives we should differentiate: $$ \frac{\mathrm d\mathcal L}{\mathrm dx}=\lim_{h\to 0}\frac{1}{h}\big[\mathcal L(\phi(x+h),\phi'(x+h),x+h)-\mathcal L(\phi(x),\phi'(x),x)\big]\tag{1} $$ and $$ \frac{\partial\mathcal L}{\partial x}=\lim_{h\to 0}\frac{1}{h}\big[\mathcal L(\phi(x),\phi'(x),x+h)-\mathcal ...


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Indeed the Lagrangian is independent of $\phi$. However, the partial derivative w.r.t $\dot{\phi}$, \begin{equation} \frac{\partial L}{\partial\dot{\phi}}=mr^2\dot{\phi}, \end{equation} contains $r$ and $\dot{\phi}$, both depending on time $t$. Therefore, you need product rule to compute the total time derivative. That's why you have two terms. They come ...


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You can't; it isn't true as you've written it. You can trivially show this by taking some numbers (any numbers at all will do) for $T, \Delta T, v, \mu$ and plugging them in. What's going on is something else. $\Delta v/v$ can be written as a power series in $\Delta T/T$. What your professor put in the notes is the first-order term only. Thus, it is an ...


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This is a typical related rates calculation. The square root formula implies that $v^2=\frac{T}{\mu}$. Differentiating it we have $2v\,dv=\frac{dT}{\mu}$. Dividing the second formula by the first gives $2\,\frac{dv}{v}=\frac{dT}{T}$, which is equivalent to your increment formula. Of course, I replaced finite increments with differentials, so strictly ...



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