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Hot answers tagged calculus

58

It is exactly because we have a factor of $\frac 1 2$ in the area formula of a triangle. To understand what I'm saying, consider what is the $v(t)$ graph of a particle under constant acceleration. Some say, a good plot is worth a million words! :)

48

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.

31

When I asked my undergrad analytic mechanics professor "what does it mean for a rotation to be infinitesimal?" after he hand-wavily presented this topic in class, he answered "it means it's really small." At that point, I just walked away. Later that day I emailed my TA who set me straight by pointing me to a book on Lie theory. Fortunately, I don't ...

30

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

26

There is an old tradition, going back all the way to Leibniz himself and carried on a lot in physics departments, to think of differentials intuitively as "infinitesimal numbers". Through the course of history, big minds have criticized Leibniz for this (for instance the otherwise great Bertrand Russell in Chapter XXXI of "A History of Western Philosophy" (...

17

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 ...

13

It will be the latter case, $m^2/s^3m$ which is just $m^3/s^3$. Remember that the integral is the sum of all the products $f(x)\;\text{ times } \;dx$. $dx$ is a tiny piece of the path from $0$ to $x$, so it is in units of $m$ as well. Each of the products $f(x)dx$ have units $m^3/s^3$, and the sum of all these products keeps those units.

12

We have also the same notions of derivation, curl, etc... for functions that are less regular. When you write Maxwell's equations, you are writing a system of partial differential equations. To investigate them, you have to specify the type of solution you look for (in the language of PDEs: classic, mild, weak...) and the functional space you set your ...

12

In German, this property is known as the Transformationssatz, but I do not know any appropriate translation for it. This is, however, a special case of coordinate tranformations changing the measure by the determinant of their Jacobian, since obviously $\frac{\partial y_i}{\partial x_j} = A_{ij}$. That it is the determinant that plays a role in the ...

12

The equations are entirely equivalent, as can be proven using Gauss' and Stokes' theorems. The integral forms are most useful when dealing with macroscopic problems with high degrees of symmetry (e.g. spherical or axial symmetry; or, following on from comments below, a line/surface integrals where the field is either parallel or perpendicular to the line/...

10

1. Since $x\gg p$, we see that $\sin(px)$ is highly oscillatory. In fact, the integral becomes $$\int_0 ^\infty \mathrm{d}p\ p \sin px \ e^{-it\sqrt{p^2 +m^2}}\sim \int_{-\infty} ^\infty \mathrm{d}p\ p\ e^{ipx-it\sqrt{p^2 +m^2}}$$ modulo some factor of $\pm2/i$. Observe now this integral resembles $\int f(p)\exp(g(p))\,\mathrm{d}p$. We find the point $\... 9 I think your math teacher is right. One way to see that differentials are not normal numbers is to look at their relation to so called 1-forms. I do not know if you already have had forms in calculus 2, but it is easy to look up on the internet. Since you chose a tag "integrals" in your question, let me give you an example based on an integral. Let's say ... 9 When the velocity is not constant you have: $$x(t)=\overline{v(t)} t + x_i$$ where$\overline{v(t)}$is the average velocity from$0$to$t$. When you have constant acceleration the average velocity is $$\overline{v(t)}=\frac{v(0)+v(t)}{2}=\frac{at}{2} + v_i$$ which will give the correct result. If the acceleration is non constant you will have to do the ... 9 Looking at the graph you can also see that the displacement is equal to the average velocity$\times$time. 8$a_x \Delta t = \Delta v_x = v_{xf} - v_{xi}\Delta x = v_{x,average}\Delta t = v_{xi}\Delta t + \dfrac{1}{2}a_x (\Delta t)^2\Rightarrow v_{x,average} = v_{xi} + \dfrac{1}{2}a_x \Delta t = v_{xi} + \dfrac{1}{2}(v_{xf} - v_{xi}) = \dfrac{v_{xf}+ v_{xi}}{2}$Is there a geometric interpretation or does it just work out mathematically? 8 (I'm addressing this from the point of view of standard analysis) I don't think you will have a satisfactory understanding of this until you go to multivariable calculus, because in calculus 2 it's easy to think that$\frac{d}{dx}$is all you need and that there's no need for$\frac{\partial}{\partial x}$(This is false and it has to do with why in general ... 8 The result is sometimes called Flanders' lemma. The remarkable point is that it does not need that$f$is analytic, but just that it is$C^\infty$. So it does not relies upon the Taylor series as it could seem at first glance, since that series may not converge. It works in any open star-shaped neighborhood of points in$\mathbb R^n$. A set$A\subset \...

8

In order to express the position as a function of the velocity you have to integrate with respect to time. When the velocity is constant this integral is simple, namely $vt+C$. However once the velocity becomes a function of time this integral will change and will in general not be equal to $v(t)t+C$. You actually have to integrate $v(t)$ with respect to $t$ ...

8

The dimensions of the integral are simply those of $f(x)dx$, so in this case they would be $m^2/s^3 \times m = m^3/s^3$.

7

Technically, the equation $$d = \frac{\mathrm{d}x}{\mathrm{d}t}t + \frac{\mathrm{d}^2x}{\mathrm{d}t^2}\frac{t^2}{2}$$ is not right. Instead, for constant acceleration, you need $$d = \left(\left.\frac{\mathrm{d}x}{\mathrm{d}t}\right|_0\right) t + \left(\left.\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\right|_0\right) \frac{t^2}{2}$$ In other words, a quantity ...

7

-What is an infinitesimal quantity like $\delta$ to the physicist? To most physicists, it means the same thing it meant to Newton, Leibniz, and Euler. It means something that's small enough that we can apply a certain informally defined body of techniques to it and get correct answers. To physicists who know more about math after 1960, it means the same ...

7

I can give you an intuitive view from a physicist. Charges are the sources and sinks for the electrical field. Consider the extreme case where the volume enclosed by the surface is empty space, so no charges. Then any field line that enters the volume must exit the volume somewhere else. Thus, the integral of the field over the entire surface is 0. If ...

6

Pretty sure the question is about $\frac{\hat{r}}{r^2}$, i.e. the electric field around a point charge. Naively the divergence is zero, but properly taking into account the singularity at the origin gives a delta-distribution.

6

There are three cases here: The acceleration is a function of time $a(t)$. Then the velocity is $$v(t)=\int a(t)\,{\rm d}t$$ and the position as a function of time $$x(t)=\int v(t)\,{\rm d}t$$ The distance is calculated from $x(t)$. The acceleration is function of position $a(x)$. Then the velocity as a function of position is $$\frac{1}{2}v(x)^2 = \int a(... 6 You must first rewrite the old partial derivatives in terms of the new ones. A priori, they're some linear combinations with coefficients that could depend on the spacetime coordinates in general but here they don't depend because the transformation is linear. The rules$$ t'=t, \quad x'=x-Vt,\quad y'=y $$get translated to$$ \frac{\partial}{\partial t} = \...

6

Nice question! The answer to this depends on the version of Newton's first law you use. In the Principia, the statement of the first law, as translated by Machin, is: Every body perseveres in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed thereon. This is immediately followed by ...

6

This is not an equality, strictly speaking. Looks like your lecturer used spherical coordinates. If the integrand is spherically symmetric, i.e. it only depends on the magnitude of $\mathbf{p}$, then the integration over the angular coordinates is trivial and just gives you the solid angle subtended by a sphere, $4\pi$.

6

Since the force is a function of distance, you need to integrate: $$F = kx\\ W = \int F\ dx\\ W = \int k\ x\ dx\\ W = \frac12kx^2$$ Add signs as needed... Your work considered the force to be constant - and that's not how springs work.

6

Whenever you see a function that looks like: $$y = \tfrac{1}{2}kx^2$$ there's a good chance it came from integrating the function: $$\frac{dy}{dx} = kx$$ For example your distance function comes from integrating the velocity $v = at$: $$y(t) = \int v\,dt = \int at\,dt = \tfrac{1}{2}at^2$$ The spring energy function comes from integrating the ...

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