derive the equations of vector components in a three dimensional (spherical) co-ordinate system
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Answer:
How do you find the components of a 3d vector?
In three-dimensional space, vector →A has three vector components: the x-component →Ax=Ax^i A → x = A x i ^ , which is the part of vector →A along the x-axis; the y-component →Ay=Ay^j A → y = A y j ^ , which is the part of →A along the y-axis; and the z-component →Az=Az^k A → z = A z k ^ , which is the part of the ...
According to the assignment, we should end up with this:
v⃗ =dr⃗ dt=r˙r^+rθ˙θ^+rϕ˙sinθϕ^
In this case θ is the angle from the z axis and phi is the angle in the x-y plane.
Now, if I take it that position
r⃗ =rr^
and say
r^=x^sinθcosϕ+y^sinθsinϕ+z^cosθθ^=x^cosθcosϕ+y^cosθsinϕ−z^sinθϕ^=x^(−sinϕ)+y^cosϕ
now maybe I am making this more complex than it is. And maybe it's just a notation problem (I really hate the dot notation sometimes because I feel it obscures things, but I need to know it, I know).
If we assume that when r changes, ϕ and θ and their unit vectors stay the same, then we can safely say that dϕ^dr=0 and dθ^dr=0. (someone please tell me if i am wrong)
If we do the same thing with changing θ and ϕ though, the result is different. hen we change θ, r has to change because it changes direction, and when we change ϕ r has to change because it changes direction in that case also.
When I take the derivative of r^ with respect to θ, I get the following:
dr^dθ=x^cosθcosϕ+y^cosθsinϕ−z^sinθ
which as it happens also is equal to θ^
Now, if I look at r⃗ =rr^ and take the derivative w/r/t time, I should get dr⃗ dt=rdr^dt+drdtr^
I notice that this happens (and some of this is just seeing the notation):
dr⃗ dt=rdr^dt+drdtr^=r˙r^+rθ^