Thread: Derivatives
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Old January 3rd, 2009, 05:42 PM
HallsofIvy HallsofIvy is online now
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Or, since the original curve is given in terms of the parameter t, write the tangent line in the same way:
\frac{dx}{dt}= -2sin(t)
\frac{dy}{dt}= 2cos(t)
so, at t= t_0, i.e. the point (2cos(t_0), 2sin(t_0)) the tangent line is given by x= 2cos(t_0)- 2sin(t_0)(t- t_0), y= 2sin(t_0)+ 2 cos(t_0)(t- t_0).

In particular, with t= \pi/3, sin(\pi/3)= \sqrt{3}/2 and cos(\pi/3)= 1/2 so the tangent line there is x= 1- \sqrt{3}(t- \pi/3), y= \sqrt{3}+ (t- \pi/3).
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