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5

The general requirement you are looking for is that the particular function be of class $C^1$, where ...if all order $p$ partial derivatives evaluated at a point $\mathbf a$: $$\frac{\partial^p}{\partial x_1^{p1}\partial x_1^{p2}\cdots\partial x_n^{pn}}f\left(\mathbf x\right)\vert_{\mathbf x=\mathbf a}$$ exist and are continuous, where $p1,\,p2, ..., ...


5

The places in physics where commutation of partial derivatives tends to be important are in the identities of vector calculus. The situations where these identities might seem to break down is when there is some kind of topological winding. Then the partial derivatives commute at almost all points except some small set where they are undefined but still can ...


4

The only orthonormal coordinate basis is the Cartesian coordinate basis. The basis vectors for the, e.g., polar coordinate basis are orthogonal but not normalized. That doesn't mean that one can't normalize the polar basic vectors to get the polar unit basis but such a basis isn't a coordinate basis. For the Cartesian coordinate basis, the basis vectors ...


2

Consider the angle between the two vectors as $\theta$ and the following rules $$\begin{align} \vec{a}\cdot\vec{b} &= \|\vec{a}\| \|\vec{b}\| \cos\theta & \|\vec{a}\times\vec{b}\| & = \|\vec{a}\| \|\vec{b}\| \sin\theta \end{align} $$ Now to construct the parallel vector use the direction of $\vec{b}$ and the adjacent side of the triangle ...


2

If you want to be physical, you'd have to have a physical interpretation of the derivatives. If you've already taken two derivatives you can ask yourself whether it is possible to take the gradient of those second derivatives. If so, then the second derivatives commuted, if not then the second derivatives are weird (if something wasn't weird you could take ...


2

A discontinuity in the flow of water could be a wall, or a clog that water is still getting around, but not flowing directly through. In finite electric current it could be a substance with a different conductivity, notably zero or ∞. In theory, the mixed partial second derivatives would not be generally equal, just on the cusp of a boundary such as these. ...


1

When introducing the stress tensor to undergraduate engineers, I note that, if a body is deformed by a stress state, and we imagine it sliced by a plane, then there is a (force) vector acting across (but not normal to) the plane. To fully specify the stress state, you need to cut the body with three planes, each of which can be specified by its normal ...


1

Avoiding integration is not physics. And that isn't what is going on. What is happening is you are doing the integration, and you are integrating vectors. Sometimes all the vectors point in the same direction so you can add then up like numbers to get the overall vector. Other times you get a bunch of vectors to add up and they point in different ...


1

Timaeus's answer could be correct. The $\Lambda$ matrix from your book may have been intended as a passive transformation (one that acts on the coordinate system) and you mistakenly used it as an active transformation (one that acts on the object). Alternatively, the $\Lambda$ matrix from your book may have been intended as the active transformation of a ...


1

Singularities in functions often lead to non commuting second derivatives. As for a Physical interpretation I think the following exercise may help: The partial derivative can be from First Principles can be written as df(x,y)/dx = (f(x+h,y)-f(x,y))/h i.e the function is incremented by h and then the derivative is found. (x,y+h). .(x+h,y+h) (x,y). ...


1

If you have a coordinate system you could move along a coordinate, which indicates some vectors you could use for a basis. These vectors might be orthogonal, that depends on your coordinates (think, does the metric look diagonal in those coordinates)? But even if your coordinates are orthogonal then you still have to pick a magnitude for these vectors. ...


1

Using polar coordinates it holds $ |d{\bf r}| = \sqrt{(d|{\bf r}|)^2 + |{\bf r}|^2(d{\bf \phi})^2}$. From this equation you can see, the two expressions you are asking about are actually only equal (in absolute value) for a straight line through the origin, thus otherwise different. For them to be exactly equal, the $d|{\bf r}|$ should be moreover pointing ...


1

If $$ \overrightarrow{r}=r_{x}\widehat{i}+r_{y}\widehat{j} $$ then $$ \left | \overrightarrow{r} \right |=\sqrt{r_{x}^{2}+r_{y}^{2}} $$ and $$ d\left | \overrightarrow{r} \right |=\frac{r_{x}dr_{x}+r_{y}dr_{y}}{\sqrt{r_{x}^{2}+r_{y}^{2}}} $$ on the other hand $$ d\overrightarrow{r}=dr_{x}\widehat{i}+dr_{y}\widehat{j} $$ and $$ \left | d\overrightarrow{r} ...


1

Curl is a measure of the rate at which a(n infinitesimally small) region of fluid rotates about its own centre. You might measure it by inserting a (very) small paddlewheel in the fluid - the speed at which it rotates is the curl. For example, on a fairground Ferris wheel, the big wheel rotates (non-zero curl) the gondolas gyrate (zero curl). Swirl ...


1

Since this is a specific question, I can't answer it directly for you but I may give you a hint: you already have the displacement formula $\stackrel{\to }{r}\left(t\right)$. From there you may find the relation y(x) using t as a parameter and use the formula for the arc length from calculus: $L=\underset{a}{\overset{b}{\int }}\sqrt{1+{\left({y}^{\prime ...


1

Length: $\int_{t_0}^{t_1}| \vec{r} '(t)|dt$. In your case: $ \int_{0}^{t}\sqrt{(20+4t')^2+(15)^2}dt'$ Use Wolframalpha if you are now sure how to deal with it: https://www.wolframalpha.com/input/?i=int+sqrt(%2820%2B4x%29^2%2B%2815%29^2)dx


1

There's a small error here: you say, "...these two vectors $v$ and $\mathbf{v}$" (emphasis mine). The problem there is that $v$ is not a vector. Rather, it's the magnitude of a vector; specifically, it is the magnitude of the velocity vector $\mathbf{v}$. This is actually implicit in your derivation: you created a unit vector in the direction of the motion ...


1

When you calculate the "parallel" vector, you should not use the dot product of $a\cdot b$ but instead the normalized dot product $$\frac{a\cdot b}{|b|}$$ times the unit vector $b$. The projection of $a$ onto $b$ should always be independent of the length of $b$.



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