Calc10_3day2

# Calc10_3day2 - 10.3 day 2 Calculus of Polar Curves Greg...

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Unformatted text preview: 10.3 day 2 Calculus of Polar Curves Greg Kelly, Hanford High School, Richland, Washington Photo by Vickie Kelly, 2007 Lady Bird Johnson Grove, Redwood National Park, California Try graphing this on the TI-89. ( 29 2sin 2.15 16 r θ π = ≤ ≤ To find the slope of a polar curve: dy dy d dx dx d θ = sin cos d r d d r d = sin cos cos sin r r r r ′ + = ′-We use the product rule here. To find the slope of a polar curve: dy dy d dx dx d θ = sin cos d r d d r d = sin cos cos sin r r r r ′ + = ′-sin cos cos sin dy r r dx r r ′ + = ′-→ Example: 1 cos r θ = -sin r ′ = ( 29 ( 29 sin sin 1 cos cos Slope sin cos 1 cos sin +-=--2 2 sin cos cos sin cos sin sin cos +-=-+ 2 2 sin cos cos 2sin cos sin-+ =-cos 2 cos sin 2 sin-+ =-→ 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 = × = → We can use this to find the area inside a polar graph....
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Calc10_3day2 - 10.3 day 2 Calculus of Polar Curves Greg...

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