Asked by alex
Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter.
x=tcos(t), y=tsin(t); t=π
x=tcos(t), y=tsin(t); t=π
Answers
Answered by
Steve
the slope is
dy/dx = (dy/dt)/(dx/dt)
= (sint + tcost)/(cost - t sint)
y'(π) = (0-π)/(-1-0) = π
at t=π, x=-π and y=0
so, using the point slope form of the line,
y = π(x+π)
dy/dx = (dy/dt)/(dx/dt)
= (sint + tcost)/(cost - t sint)
y'(π) = (0-π)/(-1-0) = π
at t=π, x=-π and y=0
so, using the point slope form of the line,
y = π(x+π)
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