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53

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! :)

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

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

28

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 ...

24

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 ...

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 ...

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 ...

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

$a_x \Delta t = \Delta v_x = v_{xf} - v_{xi}$ $\Delta x = v_{x,average}\Delta t = v_{xi}\Delta t + \dfrac{1}{2}a_x (\Delta t)^2$ $\Rightarrow v_{x,average} = v_{xi} + \dfrac{1}{2}a_x \Delta t = v_{xi} + \dfrac{1}{2}(v_{xf} - v_{xi}) = \dfrac{v_{xf}+ v_{xi}}{2}$ Is there a geometric interpretation or does it just work out mathematically?

8

I 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 ...

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 ... 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 ... 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 ... 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 -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 ... 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 (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 ... 6 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 ... 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. 5 You may be imagining that if you push with constant force$F$, the spring will compress until the spring has such a resistive force. But since the spring was not counteracting that force, your constant force$F$was accelerating the mass. Upon reaching the point where the spring has force$F$as well, the mass does not stop but has a speed such that$KE = ...

5

$$A=\pi r^2$$ $$\frac{dA}{dr}=\pi\cdot2r$$ $$dA=2\pi rdr$$ Alternatively, you can write : $\lim_{\Delta r\to 0}\frac{\Delta A}{\Delta r}=\lim_{\Delta r\to 0}\frac{\pi\{(r+\Delta r)^2-r^2\}}{\Delta r}=\lim_{\Delta r\to 0}\frac{2\pi r\Delta r+\Delta r^2}{\Delta r}=2\pi r+0$ You have to ignore $(dr)^2$ as it is very small. Why? Because you took the limit ...

5

To add to the answer of @ACuriousMind, it may be remarked that the determinant of a square matrix is equal to the (signed) volume of the parallelepiped spanned by its columns. Since this parallelepiped is exactly the image of the unit cube, the intuitive meaning of the substitution theorem (the name under which I know it) is that you can change your ...

5

In your second derivation, the correct formulas are \begin{align} \frac{\Delta x}{v} &\approx \Delta t & \frac{\Delta v}{a} &\approx \Delta t \end{align} I'm sure you can easily find some examples to show you why $x/v = t$ and $v/a = t$ don't make any sense. Anyway, when you put these together, you get $v\Delta v \approx a\Delta x$, with ...

5

It's just conservation of energy without being called as such. $$\mbox{Energy}_{\mbox{before}}=\mbox{Energy}_{\mbox{after}}$$ $$\frac{1}{2}m v_1^2+m g h_1=\frac{1}{2}m v_2^2 +m g h_2$$ $$\frac{1}{2}m v_1^2=\frac{1}{2}m v_2^2 +m g (h_2-h_1)$$ $$v_1^2=v_2^2 +2g \Delta y$$ This seems trivial when you have calculus and a concept of energy, but without ...

5

We take: $$x=r\cos\theta\cos\phi$$ $$y=r\cos\theta\sin\phi$$ $$z=r\cos\theta$$ Now, you know the definition of the gradient in Cartesian coordinates: $\vec{\nabla}=\frac{\partial}{\partial x}\hat{x}+\frac{\partial}{\partial y}\hat{y}+\frac{\partial}{\partial z}\hat{z}$ Now, we use the chain rule or each component. For instance, \frac{\partial}{\partial ...

5

While it is true that some parts of physics are discrete; matter, energy, charge, etc. There are many others that are continuous. Distances, time, temperature, probability, and angles are just a few examples of continuous quantities. In addition, even though things such as Energy are "discrete" for one system, the smallest unit of Energy is not the same for ...

5

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 ...

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