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anonymous

  • 4 years ago

Find the maximum value of the y-coordinate of the points on the limacon r=1+2cos(theta)

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  1. anonymous
    • 4 years ago
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    R u there?

  2. dumbcow
    • 4 years ago
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    Let theta =A y= rsin(A) y = sinA(1+2cosA) = sinA +2sinAcosA set dy/dA = 0 dy/dA = cosA+2cos^2A-2sin^2A = 0 **substitute sin^2 = 1-cos^2 cosA +2cos^2A-2(1-cos^2A) = 0 4cos^2A +cosA -2 = 0 using quadratic formula cosA = -.843 or .593 max at .593 since it would lead to greater r A = 53.625 degrees r = 1+2(.593) = 2.186 max_y = 2.186*sin(53.625) = 1.76

  3. Zarkon
    • 4 years ago
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    if you want an exact answer it is \[\frac{\sqrt{6(11\sqrt{33}+69)}}{16}\]

  4. anonymous
    • 4 years ago
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    ok thanks :D

  5. anonymous
    • 4 years ago
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    Thanks dumbcow :D

  6. anonymous
    • 4 years ago
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    See only once someone helps me out it becomes so clear. I dont know y i cldnt figure this one out on my own

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