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2

Let's do this explicitly for both cases. For these examples, the classical formula for the geodesic curvature $k_g$ suffices. Let $\gamma(t)$ be a curve in a surface $S \subset \mathbb{R}^3$, and let $n(t)$ be the unit normal to $S$ at the point $\gamma(t)$. Then $$ k_g = \frac{\ddot{\gamma}(t) .(n(t) \times \dot{\gamma}(t))}{|\dot{\gamma}(t)|^3} $$ First ...


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You don't need to use the metric of the hemisphere. This is because the pullback of arbitrary forms onto the submanifold is the trivial pullback operator. All you need to do is apply the projection operator. Therefore, the extrinsic curvature tensor is just $K_{ab} = - \gamma_{a}{}^{c}\gamma_{b}{}^{d}\nabla_{c}n_{d}$, where $\gamma_{ab}$ is the metric of ...


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We can call the first object P, and the second object Q. Then we can say that the location of the particles can be notated as $ <P_x, P_y>, <Q_x, Q_y> $ Therefore, the distance, $d$, between A and B is $d=\sqrt{(A_x-B_x)^2+(A_y-B_y)^2}$. So, because $x(t)=v_x*t+x(0)$, and $v=<U_1*\frac{B_x-A_x}{d},U_1*\frac{B_y-A_y}{d}>$ Combining all ...


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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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The situation you are describing is an example of Fresnel diffraction (or near-field diffraction). In general, when a wave propagates every point of the wave front can be thought of as its own source of waves traveling in all directions (called Huygens construction). It turns out that neighboring point sources along an infinite straight wave front reinforce ...


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You will benefit by finding some tutorials on wave theory. In brief, assuming a spherical wavefront from the emitter, you are correct there's no direct path to the receiver. However, the edge of yourabsorber there causes diffraction (Huygen's principle), so thatsome of the sound wave (energy) will make its way to the receiver. You can see a demo of this, ...


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Sourisse's comment answers your question, but just for the record I'll expand on it here as a Wiki answer. Note that this is a physicist's answer - any mathemticians present would be wise to avert their gaze now. Remember that when we say that the volume element is: $$ dV = 4\pi r^2 dr \tag{1} $$ We are talking about the limit in which $dr \rightarrow 0$. ...


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Ellipses, parabolas and hyperbolas are both: Defined as the conic sections you speak of: work out the intersection between the cone $\vec{R}.(\cos\phi\,\hat{X} + \sin\phi\,\hat{Z}) = \sin\theta\,|\vec{R}|$ and the plane $z=const$ where $\theta$ is your polar angle and $\phi$ the angle between the cone's axis of symmetry and the slicing plane and you'll ...


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The polar angle of the plane (the angle of the plane with respect to the symmetry axis) should relate to the eccentricity of the orbit. For $90^o$, the section is a circle and the eccentricity is zero. For an angle between the $90^o$ and the angle of the cone, you will get an ellipse with an eccentricity $0<\epsilon<1$. In both of these cases, the ...



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