Essay Example on We can use the idea of polar curves So instead of x y we will have r r

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We can use the idea of polar curves So instead of x y we will have r r will represent the length and the angle Consider r f cos θ x r 1 sin θ y r when r 2 x2 y 2 where x y and r are real numbers 2 θ cos 1 x r And sin 1 y r Also tan 1 y x On the other hand z x iy rxr iryr From 1 and 2 This becomes z r cos i sin It is presupposed knowledge that z r cos i sin rei we can find the intersections of different curves more easily and identify the nature of the curve in a more efficient way rather than directly equate them Continuing with the previous result z rei if this equation represents the points then we can find the point of intersection by equating two such different z Here we have to remember that this is for the function r f If r f and t k k is a function then to find the common intersection points we can equate the two points represented by z reior z teiwhere and represent angles Then rei tei In such a scenario both can be equal only if r t and 2n n ℤ Or if r t and when 2n n ℤ If put differently f ei k 2n ei 2n or f ei k 2n ei 2n f k 2n or f k 2n because ei2n 1and ein 1 Polar curves are usually used for complex numbers because of its ease of use in case of complex algebra but here complex numbers are used to represent a point even though i is not directly involved in this problem The number of points of intersection of the two rose curves given by sin p rand cos q r can be found out by equating them as shown previously This is only for intersection if both p and q are odd positive integers The results can be extrapolated to other cases too by changing the bounds below cos p sin q sin q cos 2 q cos p cos 2 q 



The general solution will then be p 2 q 2l l ℤ here l represents integers which is chosen specifically to satisfy 0 this is because of the periodic nature of this curve as seen in the earlier example This leads to different sets of inequalities which if summed together will gives the number of intersections The solutions are p 2 q 2l together with p 2 q 2l p q 12 2l and p q 12 2l This becomes 12 2l p q and 12 2l p q Since the solutions are limited with 0 we have 0 12 2l p q and 0 12 2l p q This can be algebraically manipulated to give 0 12 2l p q and 0 12 2l p q This gives 12 2l p q 12 and 12 2l p q 12 Here 2lcan take the value of only integers and not fraction So the inequality should be reconciled to the nearest integer This will give 0 2l p q and 1 2l p q If we notice 12 2l p q 12 becomes 1 2l p q here the inequality becomes this can be attributed to the fact that 2lis an integer and that removing 12 will make p q 12 it s previous integer and therein lies the reason So all the integers within the range of the two inequalities add up to give the number of solutions Therefore p q2 p q2 1 1 p Therefore there are p number of intersections if the two rose curves have positive odd p and positive odd q 



But the problem here is we don t know if the set of inequalities we are adding have a set of numbers that are common in both the inequalities Also the points of intersection does not include intersections at the point of origin This discussion on intersection can then be continued for intersection with some lines of axes And it is surprisingly easy because of the use of angles to describe lines INTERSECTION WITH THE AXES If we observe the graphs of different rose curves cos n it can be noticed that curves with a odd positive integer n intersect the x axis alone and not the y axis On the other hand if n is a even integer then it intersects with both the axes The standard equation is r p cos n q where p qz Case 1 Let n be a odd integer n ℤ Then 2k 1 n r p cos 2k 1 q To find the intersection with the x axis to the left r p cos 2k 1 q r q p So it is at this point where the curve intersects with the axis And for 2 r p q Similarly for 1 12 and 2 r q because in r p cos 2k 1 q cos 2k 1 becomes 0 Case 2 Let n be a even integer n2 4 6 Then 2k n k ℤ r p cos 2k q To find the intersection with the line from angle r p cos 2k q r q p So it is at this point where the curve intersects with the axis And for 2 r p q For 1 12 rcan either be p q or q p depending on the n On further investigation it was found that if n is divisible by 4 then r p cos n q would give out r p qfor all the axes This can be shown by keeping n 4k where k ℤ Now r p cos 4k 1 12 q p q


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