# Tag Info

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The kinetic energy of a particle whose motion is described by $\textbf{r}(t) = \left(x(t),\,y(t),\, z(t)\right)$ is, at the point $(x,y,z)\in\mathbb{R}^3$, defined as $$T(x,y,z) = \frac{1}{2}m\left(\dot{x}^2 + \dot{y}^2 + \dot{z}^2\right) = \frac{1}{2}m\,\textbf{v}\cdot\textbf{v}$$ and yes, the square of a vector means (by abuse of notation) its scalar ...

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I am going to assume that you have not yet studied linear algebra, sorry if it seems as if I am talking down to you at any point. You are correct in that we can split a vector into two components in the plane. This is because any two linearly independent(not parallel or anti-parallel) vectors form a basis(a set of vectors from which you can "build" other ...

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Here are a couple of links on some important vector algebra, such as vector addition, dot product, etc. https://www.mathsisfun.com/algebra/vectors.html http://emweb.unl.edu/Math/mathweb/vectors/vectors.html

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Hint: Calculate the vertical velocity as a function of time ($v_y=-g\cdot t$). Two vectors are perpendicular when their dot product is zero, so $$v_{x1}\cdot v_{x2} + v_{y1}\cdot v_{y2}=0$$ Now $v_{y1} = v_{y2} = gt$ since both are free falling. So you find that $gt = \sqrt{12}$ when they are perpendicular. From this, you calculate the time; and from the ...

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Your equation is not valid, see Figure below With equations \Vert\mathbf{r}\Vert^{2}=\mathbf{r}\circ \mathbf{r} \Longrightarrow 2\cdot \Vert\mathbf{r}\Vert\cdot d\Vert\mathbf{r}\Vert =2\cdot\left(\mathbf{r}\circ d\mathbf{r}\right)\Longrightarrow d\Vert \mathbf{r}\Vert =\dfrac{\mathbf{r}}{\Vert\mathbf{r}\Vert}\circ d \mathbf{r} \tag{01} ...

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is $\left|\frac{d\vec{r}}{dt}\right| = \frac{d|\vec{r}|}{dt}$ [?] There's a subtlety here ... It depends on how (you want that) the right-hand side of your suggested equation is interpreted. We know, of course, by to the rigorous definition of what the notation you've used is supposed to mean, that the value of a derivative is evaluated "at" a ...

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Actually you're asking two different questions. Is the magnitude of instantaneous velocity the same as instantaneous speed? Well, yes, that's the definition of instantaneous speed. Is this equation true? $$\biggl\lvert\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\biggr\rvert = \frac{\mathrm{d}\lvert\vec{r}\rvert}{\mathrm{d}t}$$ No, it's not - but instantaneous ...

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Obviously not: think to a very simple (2d) example, $r(t)=(t,t)$. Componentwise, the derivative yields $1,1$, and hence $\lvert dr/dt\rvert=\sqrt{2}$. On the other hand, $\lvert r(t)\rvert = \sqrt{2}\lvert t \rvert$. And the absolute value function is not differentiable in zero. Hence the two derivative functions coincide almost everywhere, but not in ...

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I am guessing that you are encountering vectors for the first time. Vectors have been presented as things with magnitude and direction, like little arrows. You are asking for clarification on what vectors are all about. Little arrows are a good example of vectors. But yes there are vectors that do not have magnitude and direction. To a mathematician, ...

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Tensors have angles and magnitudes but are more complicated than vectors. One may have a quantity with a direction and magnitude which will not follow vector rules because a tensor is defined with 9 components, not three. The most recent physical tensor I have seen are the E and B fields in the BICEP2 experiment, an attempt to get the imprint of the ...

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Yes. Electric current is an example. It has a direction and magnitude but it doesn't follow vector summation rule, so it's not a vector. However current density is a vector.

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The answers are exactly equivalent, the dot product between two vectors produces the product between magnitude of the vectors and the cosine of the angle between the vectors... In this case the angle between two c vectors will be 0 and the angle between two (bt) vectors will also be 0, hence the dot product will be equal to the square of the magnitude of ...

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The answers are actually equivalent. $|\vec{c}|^2 = \vec{c} \cdot \vec{c} = \Sigma_i x_i^2$ Where the $i$'s run over whatever number of dimensions you have. So you're both right.

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If you have the position vector along a path $\vec{r}(q)$ parametrized by $q$, where $q$ can be time, angle, distance, or whatever then the derivatives are: $$\vec{v}(q,\dot{q}) = \frac{\partial \vec{r}(q)}{\partial q} \dot{q}$$ $$v = \| \vec{v} \|$$ $$\vec{e} = \frac{ \vec{v}}{v}$$ $$\vec{a}(q,\dot{q},\ddot{q}) = \frac{\partial ... 2 \newcommand{\dv}[2]{\frac{\mathrm{d} #1}{\mathrm{d}#2}}Elaborating on Mikael's answer, note that equations like \vec v = \dv{\vec r}{t} are sort of shorthand notations to make the life of a physicist easier. Note that there are two (three in 3d) equations in this condensed notation. What we mean by such equations is simply the following:$$v_x = ...

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You first write your position vector as $\vec r = (x_0 -t\cdot10~\mathrm{m/s},y_0)$ and then take the derivative of that. This produces $$\vec v = \frac{d\vec r}{dt}=(-10,0)~\mathrm{ms^{-1}}=-10\hat i\,~\mathrm{ms^{-1}}$$ which is what you want.

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Mathematicians don't talk about rotations or isometries or whatnot because they already know if they're talking about a scalar or something else; a physicist has to determine whether a physical quantity has the properties of a scalar or a vector or something else. The easiest way to do that is to look at transformation laws--to devise some experiment (real ...

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Here taking $|0\rangle$ and $|1\rangle$ as orthonormal basis for 2 dimensional hilbert space. Now $|00\rangle ,|01\rangle,|10\rangle,|11\rangle$ are orthogonal to each other ( take the inner product of any two it will be zero, eg. $\langle 00|01 \rangle= \langle0|0\rangle \langle 0|1\rangle =0$ ). Thus any vector of a 4 dimensional dimensional hilbert space ...

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$\newcommand{\ket}[1]{\left| #1 \right>}$Note that you can write a tensor product as a matrix in the following way: $$A\otimes B = \begin{pmatrix} A_{11}B & \ldots & A_{1m}B\\ \vdots & \ddots & \vdots\\ A_{m1}B & \ldots & A_{mm} B \end{pmatrix}$$ where $A$ is a $m\times m$ matrix and $B$ is a $n\times n$ matrix. Notice that ...

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When talking about scalars, mathematicians usually use your definition, that is, something which doesn't vary with coordinate changes. (Basically, that there's some mapping to the actual points in space, in which the scalar is well defined) When physicists talk about scalars, we usually refer to Lorentz scalars, which requires two things: Invariance under ...

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There seem to be two different questions here: (1) How are the components of a vector actually defined, and (2) Why does the component of the force perpendicular to the velocity lead only to changes in the direction of motion and not the speed. The first question has essentially been answered in one of the other answers, but let me make a few comments ...

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But I just don't get it. I mean as long as you can make a triangle with three vectors, any two can be called the components of the remaining one. Actually, you do get it: what you have said is perfectly correct, as long as you exclude degenerate triangles (ones that collapse to a line, which happens when the sum of two side lengths equals the third ...

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The vectors ${\bf F}$ and $\bf v$ cannot be added. That is to say: you can't draw a geometric line connecting the two because they cannot be added. They have different units and represent very different things. You actually have to use Newton's laws: ${\bf F}=m \dot{\bf v}$ to figure out how $\bf v$ changes.

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Just going to guess what you are getting at, sorry if I misunderstand you. Perpendicular vectors are orthogonal, you cannot express one in terms of the other and you can extend this idea into 3,4 or more dimensions. They are called basis vectors. If you Google basis vectors that might help. As an example, I can walk in the left or right directions in a ...

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Draw a picture... From here the answer should be obvious. Don't focus just on the math - look at what is actually going on.

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Gauss' Law says $\iint_S \mathbf{E} \cdot \mathbf{dA} = Q/\epsilon_0$ whereas the total vector area is $\iint_S \mathbf{dA}=\mathbf{0}$ for some closed surface $S$. The total vector area is taking vector sum of all the differential area vectors that are normal to the surface, whereas $\iint_S \mathbf{E} \cdot \mathbf{dA} = Q/\epsilon_0$ is taking the dot ...

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When you find the vector area of a closed surface you are dealing with vectors and the total sum is zero. However, when you use Gauss' law with a constant E what you are integrating is a scalar, and that gives you the surface (not zero).

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Let's assume that the rain is pretty consistent in the interim you have to go between two points. Your front side absorbs most of the water as you move along the path. Every moment you're moving, your shirt gets some water in it. The water density of the environment is pretty consistent now, so if you move faster, you trace out a larger volume differential ...

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