Parametric Curves and Tangent line equations

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SUMMARY

The discussion focuses on finding the equation of the tangent line to the parametric curve defined by the equations x = t sin(t) and y = t cos(t) at the point (0, -π). The correct parameter value t is determined to be π, which satisfies the conditions for both x and y at the specified point. The derivative dy/dt is calculated as (cos(t) - t sin(t)) and dx/dt as (sin(t) + t cos(t)). The tangent line can then be derived using these derivatives at t = π.

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ms. confused
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Find an equation of the tangent line to the curve with parametric equations x=tsint, y=tcost at the point (0,-π).

went dy/dt / dx/dt --> cost - tsint/sint + tcost

t not given so figured it could be:

x=t(sin(1)) --> t= x/sin(1)

or

y=t(cos(1)) --> t= y/cos(1)

wondering if this is the right way to do it...looks really messy...help would be nice. Thanks.
 
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ms. confused said:
t not given so figured it could be
You can find t by checking which t is necessary to be get at (0,-pi). If x has to be 0, either t or sint has to be zero. But with t = 0, y is 0 as well. So sint has to be zero, this is for t = 0 (but that couldn't be) or t = pi. Use t = pi and x will be 0 while y will be pi*cos(pi) = pi*(-1) = -pi, as was given. Then you have your t :smile:
 

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