Asked by Richard
Find the slope of the tangent line to the graph of the polar equation at the point specified by the value of π.
r = 1/π
π = π
r = 1/π
π = π
Answers
Answered by
oobleck
dy/dx = (dr/dπ) / (dx/dπ) = (r' sinπ + r cosπ) / (r' cosπ - r sinπ)
= (-sinπ/π^2 + cosπ/π) / (-cosπ/π^2 - sinπ/π)
= (-sinπ + r cosπ)/(-cosπ - r sinπ) if πβ 0
at π=π, that means
dy/dx = -π
= (-sinπ/π^2 + cosπ/π) / (-cosπ/π^2 - sinπ/π)
= (-sinπ + r cosπ)/(-cosπ - r sinπ) if πβ 0
at π=π, that means
dy/dx = -π
Answered by
oobleck
oops. did you catch my typo?
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