Finding the Radial Velocity Component of a Point P with Given Velocity Vector v

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To find the radial velocity component of a point P with a given velocity vector v, the correct approach involves using the dot product between the unit vector of P and the velocity vector. The radial component can be calculated as v_r = v · (P/|P|), where P is the position vector and |P| is its magnitude. The initial suggestion of using angles to determine the radial component is not necessary. The discussion highlights some confusion with the mathematical notation but concludes that the method is straightforward. Understanding the dot product is key to solving this problem effectively.
matteo86bo
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Hi there,
I'm really ashamed of doing this stupid question but I really need help with this thing. And, even if it's simple, it's not homework.

I have a point P=(x,y,z) with velocity v=(vx,vy,vz)
how can I determine the radial component of the velocity?

My answer is that, v_r=vcos(\theta-\alpha) where \theta,\alpha are the angle between the x direction and the radial direction and atan(vx/vy).

Is this right?
 
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Make the dot product between the unit vector \frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}(x,y,z) and the velocity vector, and you have the radial component of the velocity.

Hmm..doesn't LateX work todaY?
 
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thanks arildno,
It was really easy! It's just that sometimes I get confused with this staff.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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