Calc10_6

# Calc10_6 - 10.6 The Calculus of Polar Curves Try graphing...

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Unformatted text preview: 10.6: The Calculus of Polar Curves Try graphing this on the TI-89. ( 29 2sin 2.15 16 r θ θ π = Greg Kelly, Hanford High School, Richland, Washington To find the slope of a polar curve: dy dy d dx dx d θ θ = sin cos d r d d r d θ θ θ θ = sin cos cos sin r r r r θ θ θ θ ′ + = ′- We use the product rule here. To find the slope of a polar curve: dy dy d dx dx d θ θ = sin cos d r d d r d θ θ θ θ = sin cos cos sin r r r r θ θ θ θ ′ + = ′- sin cos cos sin dy r r dx r r θ θ θ θ ′ + = ′- → Example: 1 cos r θ = - sin r θ ′ = ( 29 ( 29 sin sin 1 cos cos Slope sin cos 1 cos sin θ θ θ θ θ θ θ θ +- =-- 2 2 sin cos cos sin cos sin sin cos θ θ θ θ θ θ θ θ +- =- + 2 2 sin cos cos 2sin cos sin θ θ θ θ θ θ- + =- cos 2 cos sin 2 sin θ θ θ θ- + =- → The length of an arc (in a circle) is given by r . θ when θ is given in radians. Area Inside a Polar Graph: For a very small θ , the curve could be approximated by a straight line and the area could be found using the triangle formula: 1 2 A bh = r d θ ⋅ r ( 29 2 1 1 2 2 dA rd r r d θ θ = = → We can use this to find the area inside a polar graph....
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## This note was uploaded on 03/10/2008 for the course MATH 215 taught by Professor Riggs during the Fall '05 term at Cal Poly Pomona.

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Calc10_6 - 10.6 The Calculus of Polar Curves Try graphing...

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