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    Line c has equation x=5cos a y=4 sin a (a=theta) 0<=a<2pi

    The curve look like below.

    Gradient of tangent = -4/5 at P
    equation of tangent = y-2(,/2) = -4/5 (x-5/,/2))
    co-ordinates or R = (5,/2 , 0)

    question s find the shaded area, leawing your answer in terms of pi.

    well the answer is 10-25pi but i cant get it dont even know what im doing wrong.
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    (Original post by lexazver203)
    Line c has equation x=5cos a y=4 sin a (a=theta) 0<=a<2pi

    The curve look like below.

    Gradient of tangent = -4/5 at P
    equation of tangent = y-2(,/2) = -4/5 (x-5/,/2))
    co-ordinates or R = (5,/2 , 0)

    question s find the shaded area, leawing your answer in terms of pi.

    well the answer is 10-25pi but i cant get it dont even know what im doing wrong.
    What's the value of theta for P?
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    (Original post by meepmeep)
    What's the value of theta for P?
    dy/da = 4cosa
    dx/da = -5sina
    dy/dx = -4cosa/5sina = -4/5

    therefore cosa = sina
    sina/cosa = 1
    tana = 1
    a = pi/4 or 5pi/4

    now you an finish it.

    MB
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    (Original post by musicboy)
    dy/da = 4cosa
    dx/da = -5sina
    dy/dx = -4cosa/5sina = -4/5

    therefore cosa = sina
    sina/cosa = 1
    tana = 1
    a = pi/4 or 5pi/4

    now you an finish it.

    MB
    didnt quite follow that how can u make sina=cosa?
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    (Original post by lexazver203)
    didnt quite follow that how can u make sina=cosa?
    he didnt.. look again, they're not the same thing

    remember when you're parametrically differentiating:

    dy/dx = dy/da * da/dx
    due to the chain rule
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    (Original post by kimoni)
    he didnt.. look again, they're not the same thing

    remember when you're parametrically differentiating:

    dy/dx = dy/da * da/dx
    due to the chain rule

    yeah i know the chain rule but it still doesnt make much sense
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    No he was right...

    dy/da = 4cosa
    dx/da = -5sina

    .'. da/dx = -1/5sina

    dy/dx = -4cosa/5sina = -4/5 <--here you just compare coefficients

    therefore cosa = sina
    sina/cosa = 1
    tana = 1
    a = pi/4 or 5pi/4
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    (Original post by jonas123)
    No he was right...

    dy/da = 4cosa
    dx/da = -5sina

    .'. da/dx = -1/5sina

    dy/dx = -4cosa/5sina = -4/5 <--here you just compare coefficients

    therefore cosa = sina
    sina/cosa = 1
    tana = 1
    a = pi/4 or 5pi/4
    i actually understood it but i still cant get the right answer forthe area of the thing
 
 
 

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