[Solved] Unwanted minus sign in this derivative of a circular curve

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SUMMARY

The discussion addresses a common misconception regarding the derivative of a circular curve, specifically the relationship between the derivatives y'(t) and x'(t). The correct expression derived is y'(t)/x'(t) = cos t/-sin t = -cot t, which clarifies that the slope is indeed represented by -cot t rather than tan t. The user acknowledges their initial confusion about the slope from the origin to the point on the curve, leading to the resolution of the problem.

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nomadreid
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Homework Statement
Find the derivative dy/dx of the curve (circle) expressed in parametric form: (x,y)=(cos t, sin t).
Relevant Equations
dy/dx of (x(t), y(t))= y'(t)/x'(t)
y'(t)/x'(t) = cos t/-sin t = -cot t.
But as t is the angle, and the derivative is the slope, then isn't the slope supposed to be tan t?
 
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Ah, oops. Thanks, phinds. I was mistakenly thinking of the slope of the line from the origin to the point, but now I see that this was a silly mistake. Problem solved, thread closed.
 
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