Phy 121
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Introductory Problem Set 5 #4
Problem
What are the magnitude and angle of a vector whose x and y components are respectively -3.1 and 4.9?
Solution
The magnitude of the vector is found by the Pythagorean Theorem to be
magnitude = `sqrt( ( -3.1) ^ 2 + ( 4.9) ^ 2) = 5.79.
The angle made by the vector with the x axis is one of two angles:
angle = tan-1 ( 4.9 m/s / (-3.1 m/s) ) = -57.69 degrees or (-57.69 + 180) degrees.
Since the x component of the vector is negative, the vector is in the second or third quadrant.
Its angle with the positive x axis cannot therefore be equal to the -57.69 degrees found from the tan-1 function, which is always a first- or fourth- fourth-quadrant angle.
The angle is therefore (-57.69 + 180) degrees = 122.31 degrees, which is as required in the second or third quadrant.
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You say since the x component is negative, it has to be in the 2nd or 3rd quadrant but don't we know it's in the 2nd quadrant since the y component is positive?
Please bear with me because it's been 12 years since I've had trig.
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The angle is in the second quadrant.
However what we know from the fact that the x component is negative is that it's in either the second or third quadrant. The point is that since the arcTangent is in the fourth or first quadrant, we have to add 180 degrees. It doesn't matter whether the angle is in the second or third quadrant. Adding 180 degrees will put us in the correct quadrant, without our ever having to think about it.
Not that we don't want to think about it, but even if we don't, the rule will work. If the x component is negative, we add 180 degrees to the arcTangent.