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AQA A level further maths Paper 1, 25th May

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What did people get for the loci question for the first part arg(x+2i)=tan^-1(0.5) in Cartesian form , I got y=0.5x-2 can anyone let me know if they get same or different
Original post by blitzs
For r^2 = 9sin(2theta). One loop is positive and the other is negative. Hence integrating from -pi to pi will be 0. Integrate instead from 0 to pi and double.

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yes but just because a polar curve goes below the cartesian x-axis does not mean the area is negative. the formula for the area enclosed by a polar curve squares r(t). Since A=1/2 x integral of r^2(t) dt, it shouldnt matter if the values of r be positive or negative because r(t) is squared. for example take curve r=2cos(3t). if you integrate from -pi/6 to zero, the area is not negative even though it is below the x axis.

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