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Core 4 Help Please

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    • Thread Starter

    Hi, I've been stuck on this question for the past 30 mins:

    A curve is given by the parametric equations:


    Show that dx/dy = -3sin(t)

    I've tried (dx/dt)/(dy/dt) method and changing x into 3sint -sin3t to be able to differentiate. I've also tried substitution. These methods don't seem to work and I need a bit of help. Any help offered appreciated. Thanks!

    dy/dt= -2sin2t *what is sin2t equal to?*

    dx/dt = 12sin^2tcost

    dx/dy is indeed equal to (dx/dt)/(dy/dt).
    • Thread Starter

    Yea, I already have dy/dt= -4sintcost using sin2x rule.

    I'm struggling how you got dx/dt....

    (Original post by jammysmt)
    Yea, I already have dy/dt= -4sintcost using sin2x rule.

    I'm struggling how you got dx/dt....
    Chain rule:


    dx/dt= 4* d/dt(sin^3(t))

    To differentiate sin^3t think of it as (sint)^3 then apply the chain rule.

    (Original post by jammysmt)
    Yea, I already have dy/dt= -4sintcost using sin2x rule.

    I'm struggling how you got dx/dt....
    d/dx(un) = nun-1(du/dx)

    Using that,

    d/dx(sinn x) = n(sinn-1 x)(cos x)
    • Thread Starter

    Ah, finally got it. Thanks very much for your help

    x = 4sin^3(t)
    y = cos(2t)
    dx/dt = 12sin^2(t)cos(t)
    dy/dt = -2sin(2t)
    dx/dt / dy/dt:
    = 12sin^2(t)cos(t)
    -4sintcost sin(2t) = 2sin(t)cost(t)
    the cos(t)'s cancel out. 12/-4 = -3 and sin^2(t)/sin(t) = sin(t)
    therefore equal to -3sin(t)

    hope i helped


x = 4(sint)^3

y = 2cos2t

\dfrac {dx}{dy} = \dfrac {dx}{dt}  * \dfrac {dt}{dy}

= \dfrac {12(sint)^2cost}{-2sin2t}

= \dfrac {12(sint)^2cost}{-4sintcost}

= -3sint


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