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## Hot answers tagged differentiation

47

It means don't be a jerk. The third derivative of position (i.e. the change in acceleration) is called "jerk", though it's a little used quantity. It's called jerk because a changing acceleration is felt as a "jerk" in that direction.

30

You are essentially asking about the material derivative when discussing a total derivative with respect to time. Let's say you are looking at the velocity of the air in your room. There is a different velocity everywhere, and it changes with time, so $$v = v(x,y,z,t)$$ When you take a derivative like $$\frac{\partial v}{\partial t}$$ you are saying "I ...

30

I usually explain it this way: $$\rho = \rho(t,x(t),p(t))$$ $$\frac{\partial\rho}{\partial t} = \lim_{\Delta t \to 0} \frac{\rho(t+\Delta t,x(t),p(t))-\rho(t,x(t),p(t))}{\Delta t}$$ $$\frac{d\rho}{d t} = \lim_{\Delta t \to 0} \frac{\rho(t+\Delta t,x(t+\Delta t),p(t+\Delta t))-\rho(t,x(t),p(t))}{\Delta t}$$

30

In physics, sometimes the third derivative of position with respect to time is called jerk.

30

Typically: $\rm d$ denotes the total derivative (sometimes called the exact differential):$$\frac{{\rm d}}{{\rm d}t}f(x,t)=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial x}\frac{{\rm d}x}{{\rm d}t}$$This is also sometimes denoted via $$\frac{Df}{Dt},\,D_tf$$ $\partial$ represents the partial derivative (derivative of $f(x,y)$ with respect to $x$ ...

24

The Laplacian measures what you could call the « curvature » or stress of the field. It tells you how much the value of the field differs from its average value taken over the surrounding points. This is because it is the divergence of the gradient..it tells you how much the rate of changes of the field differ from the kind of steady variation you expect ...

22

joseph f. johnson already gave a motivation for the one-dimensional case. I think the physical example diffusion equation $$\frac{\partial}{\partial t}n(x)=D \frac{\partial^2}{\partial x^2}n(x)$$ is the best way to illustrate it: If the function looks like $x^2$ (positive curvature), like left and right, then in the next time step the function $n(x)$ ...

21

The symbol $\Delta$ refers to a finite variation or change of a quantity – by finite, I mean one that is not infinitely small. The symbols $d,\delta$ refer to infinitesimal variations or numerators and denominators of derivatives. The difference between $d$ and $\delta$ is that $dX$ is only used if $X$ without the $d$ is an actual quantity that may be ...

12

what if distance and time both become zero as at origin in the graph is It appears that you're trying to say that the velocity is equal to the position of the particle divided by the clock time at that position: $$v= \frac{x(t)}{t}\; ?$$ But this isn't correct. Average velocity $\bar v$ is defined as displacement $\Delta x = x(t_f) - x(t_i)$ ...

12

Regarding your first question, note that $$\frac{d}{dx}\ln c = \frac{1}{c}\frac{dc}{dx}$$ Thus $$\frac{1}{a}\frac{da}{dx} + \frac{1}{b}\frac{db}{dx} = \frac{1}{c}\frac{dc}{dx}$$ Then, as physicists often do, 'cancel' the $dx$ on both sides to yield $$\frac{da}{a} + \frac{db}{b} = \frac{dc}{c}$$ Regarding your second question, it is stated that $V = ... 11 On spherical coordinates, the gradient of a general function$V$is: $$\nabla V = \frac{\partial V}{\partial r}\mathbf e_r + \frac{1}{r}\frac{\partial V}{\partial\theta}\mathbf e_\theta + \frac{1}{r\sin\theta}\frac{\partial V}{\partial\phi}\mathbf e_\phi$$ If$V(r, \theta, \phi)$only depends on$r$, that is$V = V(r)$, which is exactly the case of the ... 11 Nothing really new to add to the two great answers given already - just a particular example that helped me. When you examine the finite-difference version of Laplace's equation in 2 dimensions, you find that the discretized$\phi$satisfies Laplace's equation if, in this picture of part of the grid,$\phi_{i,j}$at the centre is the average of the ... 11 It does follow from calculus. Here's the standard way this is treated (I'm not going to be explicit about mathematical details such as smoothness assumptions here). Definition of$\delta q$. Given a parametrized path$q:t\mapsto q(t)$, we consider a deformation of the path which we call$\hat q:(t, \epsilon)\mapsto \hat q(t,\epsilon)$satisfying$\hat ...

10

It's purely notation. Given a real-valued function $f(\mathbf r) = f(x^1, \dots, x^n)$ of $n$ real variables, one defines the derivative with respect to $\mathbf r$ as follows: \begin{align} \frac{\partial f}{\partial \mathbf r}(\mathbf r) = \left(\frac{\partial f}{\partial x^1}(\mathbf r), \dots, \frac{\partial f}{\partial x^n}(\mathbf r)\right) ...

10

It seems to me that you are confonding a generic notion of total derivative and the so called Lagrangian derivative (also known as material derivative). Let us start from scratch. In Cartesian coordinates, a fluid or a generic continuous body is first of all described by a class of differentiable (smooth) maps from $\mathbb R^3$ to $\mathbb R^3$: ...

10

Maybe this graph helps to illustrate the point. Consider a constant force that is applied for a finite interval $\Delta t$. The slope in the $p$ graph at any given time is $\frac{dp}{dt}$. Now, in order to have a step in $p$ (an instantaneous impulse), it is necessary to get an infinite slope, hence an infinite force is required which is physically not ...

9

The reason you see the $da$, $db$, or $dc$ is precisely because it isn't specified what the derivative is taken with respect to. What happens is the following: First, they write the differential element of the logarithmic function: $$d\,\ln f=\frac{df}{f}$$ This is always true no matter what function $f$ is. When we take a derivative with respect to some ...

9

This is an axiom, not a result. When defining the covariant derivative, we choose for it to obey a number of properties. Carroll's Spacetime and Geometry summarizes these well -- look at the preprint for Chapter 3 here. We of course want $\nabla$ to act linearly on its argument and to obey the product rule. We also demand that it commute with contractions ...

9

Consider an $n$-dimensional space (two dimension in the picture), and let $f(\vec x)$ be a non-constant scalar function, like a temperature distribution in your case. Let $\vec y(t)$ be any curve in the space such that the function $f(\vec y(t))=c$ is constant along that trajectory (the colored lines). Now compute the scalar product $\left\langle ., ... 9 This is true - in fact you could define$\nabla^\sigma = g^{\sigma\rho} \nabla_\rho$. I assume this meant to say $$g^{\sigma\rho} \nabla_\nu \nabla_\sigma = \nabla_\nu \nabla^\rho.$$ Again, this is true, but for a slightly less trivial reason than (1). To employ (1) to prove this, you need to be able to switch$g^{\sigma\rho}$with$\nabla_\nu$, which you ... 9 Yes, it's just the second derivative of some function, it doesn't matter that this function is organized as a component of a tensor,$h_{\mu\nu}$. The identity above – assuming the function is differentiable and smooth etc. (add some "niceness" conditions on the function) – follows from the rules of calculus and is formally proven by the ... 9 I) Here we discuss the problem of defining a connection on a conformal manifold$M$. We start with a conformal class$[g_{\mu\nu}]$of globally$^{1}$defined metrics $$\tag{1} g^{\prime}_{\mu\nu}~=~\Omega^2 g_{\mu\nu}$$ given by Weyl transformations/rescalings. Under mild assumption about the manifold$M$(para-compactness), we may assume that there ... 9 Curl can be equated with the closed line integral in the limit that the encircled area$\Delta S$goes to zero. However, we would have to do this in three components because curl is a vector. $$(\nabla \times \vec{v})_x = \lim_{\Delta S \rightarrow 0} \frac{1}{\Delta S} \oint \vec{v}\cdot d\vec{l}$$ in the$yz$plane and so on. But what does it mean? ... 8 Despite what some of the other answers are mentioning, the following equation you have is correct $$\vec{r} \cdot d \vec{r} = r dr$$ You can check this by noting $$\vec{r} = x {\hat i} + y {\hat j} + z {\hat k} \implies d \vec{r} = d x {\hat i} + d y {\hat j} + d z {\hat k}$$ Then $$\vec{r} \cdot d \vec{r} = x dx + y dy + z dz$$ Further note $$r = ... 8 If you have just given the voltage signal with$$ \def\l{\left}\def\r{\right} v(t) = \l(2-\l|\frac t{\rm s}-2\r|\r)\rm V $$then the current at t=2\rm s is undefined. Right. But, in most cases really nobody cares. What we learn theoretically about the current from the above voltage signal definition is that$$ i(t) = \begin{cases} C\cdot 1\frac{\rm V}{\rm ... 8 If you want a simple intuitive explanation, you can get a lot from vehicles. In a car traveling at a constant speed, suppose there is a white dot painted on the top of the steering wheel. If that dot is in the center, you are traveling in a straight line. If you turn it some angle to the left, say 90 degrees, then the car is traveling in a circular arc at a ... 8 Maybe an intuitive answer is best given in terms of classical physics. Suppose you are looking at the movement of a classical particle. The relevant variables here are position and momentum. If you solve the motion of your system, you are presented with functions$x(t)$and$p(t)$. Now, there are a lot of derived quantities you can build from these ... 8 Your second and third equations are the same equation. They just use a different notation for the time derivative. Since in this "abstract" form$|\Psi \rangle$only depends on time perhaps it is more correct to use the last one, but it is matter of taste. In order to get your first equation (a wave equation), you must project on$\langle x|\$: $$H(P=-i\hbar ... 7 Here we will only consider the added last subquestion (v4):$$\tag{1} \frac{d}{d\lambda}e^{\hat{A}} ~=~ \int_0^1\!ds~e^{(1-s)\hat{A}}\frac{d\hat{A}}{d\lambda}e^{s\hat{A}} .$$The identity (1) follows by setting t=1 in the following identity$$\tag{2} e^{-t\hat{A}} \frac{d}{d\lambda}e^{t\hat{A}} ~=~ ...

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