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

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The discussion centers on a misunderstanding regarding the derivative of a circular curve, specifically the expression y'(t)/x'(t) equating to -cot t. The confusion arose from incorrectly associating the slope with tan t instead of recognizing the correct relationship. The contributor acknowledges the mistake, clarifying that the slope considered was from the origin to the point on the curve. This realization resolves the issue, leading to the conclusion that the problem is solved. The thread is now closed.
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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