# 11 Calculus in Polar Coordinates - Tangent Lines Arclength...

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Tangent LinesArclengthAreaExercisesCalculus in Polar CoordinatesMathematics 54–Elementary Analysis 2Institute of MathematicsUniversity of the Philippines-Diliman1 / 23
Tangent LinesArclengthAreaExercisesTangent Lines to Polar CurvesObtain a formula for the slope of the tangent line to a given polarcurve whose equation is of the formr=f(θ).2 / 23
Tangent LinesArclengthAreaExercisesTangent Lines to Polar CurvesParametrization of a Polar CurveA polar curver=f(θ) can be parametrized asx=rcosθ=f(θ)cosθy=rsinθ=f(θ)sinθRecall.slope of a parametric curvedydx=dy/dθdx/dθdxdθ=f0(θ) cosθ-f(θ)sinθanddydθ=f0(θ) sinθ+f(θ)cosθSlope of a Tangent Line to a Polar CurveGiven thatdy/dθanddx/dθare continuous anddx/dθ6=0, then theslope of the tangent line to the polar curver=f(θ) at a point (r,θ) ism=(sinθ)drdθ·+rcosθ(cosθ)drdθ·-rsinθ.3 / 23
Tangent LinesArclengthAreaExercisesTangent Lines to Polar CurvesExample 1.Find the (Cartesian) equation of the tangent line to the limaçonr=3-2cosθat the point whereθ=π3.
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Tangent LinesArclengthAreaExercisesExample 1. (continued) point of tangency: (1,p3)The slope is given bym=(sinθ)drdθ·+rcosθ(cosθ)drdθ·-rsinθ=(sinθ)(2sinθ)+rcosθ(cosθ)(2sinθ)-rsinθmflflfl(2,π3)=p32·(p3)+2(12)(12)(p3)-2p32·=-5p3.Hence, the tangent line has equationy-p3= -5p3(x-1).5 / 23
Tangent LinesArclengthAreaExercisesTangent Lines to Polar CurvesExample 2.Determine all points on the cardioidr=1-cosθwhere the tangentlines are horizontal; vertical.
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Tangent LinesArclengthAreaExercisesExample 2. (continued)m=sin2θ+cosθ-cos2θ2cosθsin

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Term
Winter
Professor
NoProfessor
Tags
Polar coordinate system, Tangent lines to circles, polar curves
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