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    Here is the question, I don't know why but I don't understand this chapter, if anyone can solve this with working, it would be greatly appreciated.
    Thanks
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    (Original post by BroodiestPear)
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    Here is the question, I don't know why but I don't understand this chapter, if anyone can solve this with working, it would be greatly appreciated.
    Thanks
    So first you would need to create the cartesian equation, so that's y=...
    You can do this a numbers of ways but I usually do it by rearranging one equation to get t=... and substituting the value for t in to the other equation so you're dealing solely with x and ys. From there you can then rearrange to make y=... and then differentiate
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    for part a) you divide dy/dt by dx/dt
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    (Original post by BroodiestPear)
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    Here is the question, I don't know why but I don't understand this chapter, if anyone can solve this with working, it would be greatly appreciated.
    Thanks
    Find dx/dt (and hence dt/dx) and dy/dt
    Then dy/dx = dy/dt x dt/dx


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    dy/dt = 6 sec^2(t); dx/dt = 3(2cost)(-sint)=-6sintcost -> dt/dx = -1/6*sect*cosect
    Thus dy/dx = 6sec^2(t) * (-1/6)*sect*cosect = -sec^3t*cosect, as required.
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    (Original post by the bear)
    for part a) you divide dy/dt by dx/dt
    (Original post by gdunne42)
    Find dx/dt (and hence dt/dx) and dy/dt
    Then dy/dx = dy/dt x dt/dx


    Posted from TSR Mobile
    That's a much quicker and nicer way of doing it, I'd never been taught to do it like that
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    (Original post by KaylaB)
    That's a much quicker and nicer way of doing it, I'd never been taught to do it like that
    the only sensible way ...
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    It's just using the fact that d/dx(f(g(x))=df(x)/dg(x)*dg(x)/dx
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    (Original post by constellarknight)
    ...
    Full solutions are against forum guidelines, by the way.
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    (Original post by Zacken)
    Full solutions are against forum guidelines, by the way.
    Alright, whatever.
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    Im trying to dx/dt but I get -3(sin2t)
    can someone explain the steps ? Sorry:ashamed2:
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    (Original post by BroodiestPear)
    Im trying to dx/dt but I get -3(sin2t)
    can someone explain the steps ? Sorry:ashamed2:
    \displaystyle x = 3\cos^2 t = 3 (\cos t)^2, so:

    \displaystyle \frac{dx}{dt} = 3 \times 2 \times \frac{d}{dt}\left(\cos t\right) \times (\cos t)^{2-1}

    Which is basically the same thing as yours. It's just that it's easier to get the show that using the form above.
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    -3sin2t is an equivalent form of what I got, -6sintcost. It's just that it's easier to get to the required expression using -6sintcost.
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    Oh, I see it now Thanks for the help
 
 
 
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