Problem 1.42 from Classical Mechanics by John R. Taylor. The
last sentence of the problem was not answered in the answer
provided here by answers. "Explain why the equation for phi is
not quite complete and give a complete version."
Could you please provide this solution?
I came up with my own version of phi that tries to account for
the different quadrants.
This equation I came up with looks really complicated but I can
explain. For the first time I chose arccos rather than arctan(y\x)
because when x = 0 arctan is undefined. Using arccos avoids this
issue. Notice that if x > 0 then the second and third terms are
gone. When x > 0 The first term is all we need.
The issue is when x < 0. When x<0 then the arccos function
will only give us a value in the 1st and 4th quadrant. So if x <
0 then we need to add pi to phi. This is exactly what the second
term does. If x < 0 then the second term adds pi.
The equation also needs to account for special cases. Like when x=0
or y=0. When y = 0 and x>0 the equation provides the correct
value of phi = 0. When y=0 and x<0 the equation gives pi, which
is correct.
Then, we have the case where x = 0. This is where the last term
finally becomes important. Notice that the last term only appears
when x = 0 and y<0.
When x = 0 and y > 0 the last term cancels out and phi equals
pi/2 using just the first two terms.
Finally, when x = 0 and y <0, Then the first two terms give pi/2
and the last term ads pi making phi = 3pi/2. Which is
correct.
And when x = 0 and y = 0 then phi should be undefined, and my
equation provides that as well.
I don't know if there was an easier way to go about this and there
is a simpler equation. If so let me know.
Problem 1.42 from Classical Mechanics by John R. Taylor. The last sentence of the problem was not answered in the answer
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