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This is simply the chain rule for derivatives, that is, for instance : $$\partial_t \,g(y_1(x, t), y_2(x, t)) = \partial_{y_1} \,g(y_1(x, t), y_2(x, t)) \,\partial_t y_1(x, t) + \partial_{y_2} \,g(y_1(x, t), y_2(x, t)) \,\partial_t y_2(x, t) $$ Now, you may apply it with $y_1(x,t)= x+bt$ and $y_2(x,t)= t$, and you will recover your formula.


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How about a radar gun? It works by measuring the Doppler shift of a radio signal incurred when it bounces off a moving object. The frequency shift of each photon encodes the instantaneous speed of the object when the photon scattered off of it. Difficult to get much more instantaneous than that (especially when you compare to a measurement of position, which ...


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There are many devices that can measure rate of a quantity without using approximate derivative and integral. I will give some example here: Pitot tube Pitot tube is a device used to measure velocity of a body with respect to flow. This device uses Bernoulli's equation. For computing the velocity using Pitot tube, the total pressure and static pressure ...


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Years ago, the speedometer in a car moved the needle by spinning a magnet. The physical rotation of the driveshaft turned the cable inside the assembly. The spinning cable is attached to a magnet. The needle is mounted on a disk attached to a spring which provides rotational counter-force. The spinning magnet attempts to spin the disk, but the spring ...


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The fundamental equation of thermodynamics says that \begin{equation}\mathrm{d}U = T\mathrm{d}S-P\mathrm{d}V \end{equation} It is a mathematical identity that \begin{equation}\mathrm{d}U = \left(\frac{\partial U}{\partial S}\right)_V\mathrm{d}S+\left(\frac{\partial U}{\partial V}\right)_S\mathrm{d}V \end{equation} Comparing the two we find that ...


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They are not the same $-$ those are the Maxwell relations, which are a consequence of $S$, $U$ and other thermodynamical potentials being functions of state. The thermodynamical notation of parital derivatives can be defined as follows: $$ \left( \frac{\partial U}{\partial S} \right)_V := \frac{\partial U(S,V)}{\partial S} $$ This means it is only a first ...


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The function $U = U(x_1, x_2, \dots x_k, t)$ would be an example of a potential energy function explicitly dependent on time. In your case, you have the function $U = U(x_1, x_2,\dots x_k)$, where it is understood that for each $x_i,\, i \in \left\{1,\dots k\right\}$, we have an implicit dependence $x_i = x_i(t)$. The total derivative of $U$ is $$dU = ...


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Lets make up an example. $$U(x,y,z,t) = E_0\left(x^2+y^2+\alpha\,z^4 -\beta\,t\,z^2\right)$$ In that case parital derivatives are: $$\frac{\partial U}{\partial x} = 2xE_0,\; \frac{\partial U}{\partial x} = 2yE_0,\; \frac{\partial U}{\partial z} = \left(4\alpha z^3 -2\beta z\right)E_0,\; \frac{\partial U}{\partial t} = -\beta\,z^2E_0 $$ Now you have a ...


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Partial derivative signifies its (functions) explicit time dependence. And since potential energy does not depend on time implicitly its partial time derivative will be zero. Hence we add partial time derivative. Hope it is helpful.



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