11 Calculus in Polar Coordinates - Calculus of Polar Curves...

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Calculus of Polar Curves Mathematics 54–Elementary Analysis 2 Institute of Mathematics University of the Philippines-Diliman 1 / 1
Tangent Lines to Polar Curves Goal : obtain slopes of tangent lines to polar curves of form r = f ( θ ) 2 / 1
Tangent Lines to Polar Curves Parametrization of a Polar Curve A polar curve r = f ( θ ) can be parametrized as x = f ( θ )cos θ y = f ( θ )sin θ Recall. In parametric form, dy dx = dy d θ dx d θ . Meanwhile, dx d θ = f 0 ( θ ) cos θ - f ( θ )sin θ and dy d θ = f 0 ( θ ) sin θ + f ( θ )cos θ Slope of a Tangent Line to a Polar Curve The slope of the tangent line to a polar curve r = f ( θ ) is given by m = (sin θ ) dr d θ · + r cos θ (cos θ ) dr d θ · - r sin θ 3 / 1
Example Find the slope of the tangent line to the cardioid r = 2 + 2sin θ at the point where θ = π 3 .
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Example Determine all points on the cardioid r = 2 + 2sin θ where the tangent lines are horizontal or vertical.
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horizontal tangent line: (cos θ )(2sin θ + 1) = 0 vertical tangent line: (1 + sin θ )(1 - 2sin θ ) = 0 provided that they are not simultaneously zero . (cos θ )(2sin θ + 1) = 0 cos θ = 0 or sin θ = - 1 2 θ = π 2 , 3 π 2 , 7 π 6 , 11 π 6 (1 + sin θ )(1 - 2sin θ ) = 0 sin θ = - 1 or sin θ = 1 2 θ = 3 π 2 , π 6 , 5 π 6 What happens at θ = 3 π 2 ? lim θ 3 π 2 m = lim θ 3 π 2 (cos θ )(2sin θ + 1) (1 + sin θ )(1 - 2sin θ ) form: 0 0 = lim θ 3 π 2 (cos θ )(2cos θ ) + (2sin θ + 1)( - sin θ ) (1 + sin θ )( - 2cos θ ) + (1 - 2sin θ )(cos θ ) form: - 1 0 ± = ±∞ 6 / 1
horizontal tangent lines at ( 4, π 2 ) , ( 1, 7 π 6 ) , ( 1, 11 π 6 ) vertical tangent lines at ( 0, 3 π 2 ) , ( 3, π 6 ) ( 3, 5 π 6 )

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