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0

There are two basic ways you can multiply a vector, the dot product, as demonstrated in the link Dot Product, which gives you a scalar, no matter if you are multiplying A.B or squaring it, A.A. Or you can have the cross product, which is A X B, which gives you another vector, perpendicular to both Cross Product. The reason there are two different ways ...


2

Why don't you work out the expansion yourself. $$ \vec{r}\cdot \vec{r} = \left( \vec{x}+\vec{y}+\vec{z} \right) \cdot \left( \vec{x}+\vec{y}+\vec{z} \right) =\\= (\vec{x} \cdot \vec{x})+(\vec{y}\cdot\vec{y}) + (\vec{z}\cdot\vec{z}) + 2(\vec{x} \cdot\vec{z}) +2(\vec{y}\cdot\vec{z}) + 2(\vec{x} \cdot\vec{y}) = \\ = \|\vec{x}\|^2 + \|\vec{y}\|^2+\|\vec{z}\|^2 ...


6

$$ {\bf r} \cdot {\bf r} = \left|{\bf r}\right|^2 = \sum_{i,j} r_i r_j {\bf e}_i \cdot {\bf e}_j = \sum_{i,j} r_i r_j \delta_{ij} = \sum_i r_i r_i $$ is often written as ${\bf r}^2$, when it's clear that you don't mean $$ {\bf r}{\bf r} = \sum_{i,j} r_i r_j {\bf e}_i {\bf e}_j $$ which is a perfectly valid quantity [1]. As for ${\bf r} = {\bf x} + {\bf y} + ...


2

First if nothing is said you can assume that $\vec{x}\vec{y}$ is a dot product. What you said ${(\vec{x}+\vec{y}+\vec{z})}^{2}=x^2+y^2+z^2$ is only true if the vectors $\vec{x},\vec{y},\vec{z}$ are perpendicular to each other but in the general case: $$ {\vec{r}}^{2}={(\vec{x}+\vec{y}+\vec{z})}^{2}\\ ...


-2

The answer and interpretation is fairly simple, Vector upon vector differentiation would look something like this, Where both of these are vectors, __> now remember a vital point, vectors have both magnitude and direction, and can be easily differentiable by time we can easily break the differentiation into parts, and solve for the 2 vectors, ...


0

This is more geometry than physics, you might be better on Mathstackexchange, but a few hints: This is simple resolving of vectors. Think about what you WOULD do if you knew the angle theta and the magnitude of Fb in order to find the magnitude of F. By my reckoning you turn out with a simultaneous equation for theta and the magnitude F. Apologies if this ...


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 ...


0

Suppose you have a 2 dimensional vector field which represents the velocity in a fluid. Let us examine two different cases and calculate the curl of the velocity vector. First, suppose the vector field $\vec{v}$ is given by $$ \vec{v}(x,y,z) = (y,-x,0). $$ If you plot this, we realize that it represents a fluid rotating in the clockwise direction. The curl ...


0

Curl is circulation per unit area just like divergence is net flux per unit volume. So any time you care about circulation per area then there you go. Wind farm, vector potentials, magnetic fields, currents.


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 ...


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 $b$.


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 ...


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 ...


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

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 ...


0

The first thing that bothers me is that for positive ${v}$ I get negative component of the velocity vector If your object was at rest then to someone moving to the right the object at rest is moving to the left in their frame. As for the second question it is same issue. Someone at rest will be moving to the left relative to someone moving faster than ...


0

Actually @Ocelo7 has already answered it; I am just showing that in a concise way: Velocity can be found by differentiating the position-vector:$$\dot{\mathbf{r}}= \mathbf{v}(t) = 10\mathtt{\hat{ i}} + (20 - 10t)\mathtt{\hat{j}}$$ . Assuming the initial time is $t_0$ & the required time is $t$, we shall use the property $$\mathbf{v}_{t_0}\cdot ...


0

One can define the (magnitude) of the cross product this way or better $$\mathbf A \times \mathbf B = AB\sin\theta\; \mathbf n $$ where $\mathbf n$ is the (right hand rule) vector normal to the plane containing $\mathbf A$ and $\mathbf B$, Another approach is to start by specifying the cross product on the Cartesian basis vectors: $$\vec e_x \times ...


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 ...


0

Yes and no. Firstly projection is an operation that takes a vector and gives another vector that lives in a subspace. Projecting doesn't give a length (though you can talk about the length of s projection). And sometimes people talk about different kinds of projections but the first (and only) projection many people learn is the orthogonal projection, so ...


0

The Minkowski metric tensor $\mathbf \eta$ takes two four-vectors as arguments and produces a real number, the inner product of the two vectors: $$\mathbf \eta(\vec u, \vec v) = \vec u \cdot \vec v = \langle\tilde u, \vec v\rangle$$ where the one-form $\tilde u$ is given by $$\tilde u = \mathbf \eta (\vec u, )$$ Geometrically, this is pictured as ...


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 ...


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. ...


0

Let the angle of the slope be $\theta$, $m$ the mass, $a$ the acceleration, $\mu$ the friction coefficient and $g=9.8 ms^{-2}$. On the mass acts a vertical gravitational force $mg$. Decompose this into a component along the line of the hill, so that is $mg\sin\theta$, and one perpendicular to that, so $mg\cos\theta$. The latter provides a friction force ...


0

The retarding force from friction is $50g \cdot 0.05$ Newtons, acting up the hill. If the particle is sliding, most brakes will not be effective as you have already accounted for the sliding friction. If the brake is effective, it will add a force of $260$ Newtons up the hill, as you are given the force. You can then resolve these into vertical and ...


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

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 ...


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 ...


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. ...


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 ...


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, ..., ...


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} ...


0

As shown in the diagram $|dr|$ represents the magnitude of the vector difference(that involves the laws of vector addition/subtraction) between $\vec{r_2}\quad \& \quad \vec{r_1}$ while $d|r|$ represents the difference between magnitudes of two vectors which is simply the difference in their lengths.


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 ...



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