M. next
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I googled it, and it says:
[itex]\dot{x}[/itex]=[itex]\dot{r}[/itex]sinθcos∅ + (rcosθcos∅)[itex]\dot{θ}[/itex] - (rsinθsin∅)[itex]\dot{∅}[/itex]
.
.
and so on for [itex]\dot{y}[/itex] & [itex]\dot{z}[/itex]
And then they wrote "We will also need the inverse transformation obtained by solving the equations above w.r.t [itex]\dot{r}[/itex], [itex]\dot{θ}[/itex], and [itex]\dot{∅}[/itex]
for example they got:
[itex]\dot{r}[/itex]=sinθcos∅[itex]\dot{x}[/itex]+sinθsin∅[itex]\dot{y}[/itex] + cosθ[itex]\dot{z}[/itex]
.
.
and so on, my question how did they deduced the latter from the former equations?
Here's the link if you want to see them clearer:
http://www.physics.sc.edu/~yar/Phys701_2009/homework/hw9_solutions.pdf
Just what's the procedure?
Thanks
[itex]\dot{x}[/itex]=[itex]\dot{r}[/itex]sinθcos∅ + (rcosθcos∅)[itex]\dot{θ}[/itex] - (rsinθsin∅)[itex]\dot{∅}[/itex]
.
.
and so on for [itex]\dot{y}[/itex] & [itex]\dot{z}[/itex]
And then they wrote "We will also need the inverse transformation obtained by solving the equations above w.r.t [itex]\dot{r}[/itex], [itex]\dot{θ}[/itex], and [itex]\dot{∅}[/itex]
for example they got:
[itex]\dot{r}[/itex]=sinθcos∅[itex]\dot{x}[/itex]+sinθsin∅[itex]\dot{y}[/itex] + cosθ[itex]\dot{z}[/itex]
.
.
and so on, my question how did they deduced the latter from the former equations?
Here's the link if you want to see them clearer:
http://www.physics.sc.edu/~yar/Phys701_2009/homework/hw9_solutions.pdf
Just what's the procedure?
Thanks
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