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    The first part of question is to find dy/dx when given x=t-sin(t) y=1-cos(t)

    dy/dx= sin(t)/ (1-cos(t))

    Using identities sin(t)=s*sin(t/2)*cos(t/2)
    1-cos(t)=2*sin(t/2)*sin(t/2)

    therefore dy/dx= cot(t/2)

    Second part, reads Find d^2y/dx^2 at t=pi/2
    I know that dy/dx(cot(x))=-csc^2(x)

    But i think there is something else i have to do which i am missing, any help will be much appreciated, Thank you.
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    (Original post by OmegaKaos)
    The first part of question is to find dy/dx when given x=t-sin(t) y=1-cos(t)

    dy/dx= sin(t)/ (1-cos(t))

    Using identities sin(t)=s*sin(t/2)*cos(t/2)
    1-cos(t)=2*sin(t/2)*sin(t/2)

    therefore dy/dx= cot(t/2)

    Second part, reads Find d^2y/dx^2 at t=pi/2
    I know that dy/dx(cot(x))=-csc^2(x)

    But i think there is something else i have to do which i am missing, any help will be much appreciated, Thank you.
    You need to use the chain rule - write

    du/dx = (du/dt)(dt/dx)

    and use u = dy/dx in the form you found earlier.
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    (Original post by davros)
    You need to use the chain rule - write

    du/dx = (du/dt)(dt/dx)

    and use u = dy/dx in the form you found earlier.
    I am sorry, it may just be me. But i am just not getting that.
    u=cot(t/2) du/dt=-(1/2)*cosec^2(t/2)

    What function do i use as dt/dx..... does du/dx represent d^2y/dx^2?
    Again i know this should be really simple, but it just isn't happening at all for me today. Sorry for being a pain. I really do appreciate the help.
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    (Original post by OmegaKaos)
    I am sorry, it may just be me. But i am just not getting that.
    u=cot(t/2) du/dt=-(1/2)*cosec^2(t/2)

    What function do i use as dt/dx..... does du/dx represent d^2y/dx^2?
    Again i know this should be really simple, but it just isn't happening at all for me today. Sorry for being a pain. I really do appreciate the help.
    Since you want \frac{d^2y}{dx^2}
    and you have \frac{dy}{dx} as a function of t, you must use imp0licit differentiation
    i.e. if y=f(t) \mathrm{\ then\ }\frac{dy}{dx}=f'(t)\frac{dt}{dx  }
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    (Original post by brianeverit)
    Since you want \frac{d^2y}{dx^2}
    and you have \frac{dy}{dx} as a function of t, you must use imp0licit differentiation
    i.e. if y=f(t) \mathrm{\ then\ }\frac{dy}{dx}=f'(t)\frac{dt}{dx  }
    so is that d^2y/dx^2=(d/dt(dy/dx)/dx/dt)

    Final answer as -1/2

    can anyone please confirm or deny that this is indeed the correct answer?
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    (Original post by OmegaKaos)
    so is that d^2y/dx^2=(d/dt(dy/dx)/dx/dt)

    Final answer as -1/2

    can anyone please confirm or deny that this is indeed the correct answer?
    Only done this very very quickly on a piece of rough paper, but assuming your cot(t/2) was correct for dy/dx I get the final answer to be -1.

    Happy to be corrected if wrong
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    (Original post by davros)
    Only done this very very quickly on a piece of rough paper, but assuming your cot(t/2) was correct for dy/dx I get the final answer to be -1.

    Happy to be corrected if wrong
    Can you please run through your working?
    I get either -1/2 or -2 but not -1.

    So confused.
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    (Original post by OmegaKaos)
    Can you please run through your working?
    I get either -1/2 or -2 but not -1.

    So confused.
    So we agree we want [d/dt(dy/dx)] / (dx/dt)

    You claim (I haven't checked!) that dy/dx = cot(t/2), so d/dt(dy/dx) = -(1/2)cosec^2(t/2)
    and dx/dt = 1 - cos t.

    So at t = pi/2, dx/dt = 1 - 0 = 1 and
    (-1/2)cosec^2(t/2) = (-1/2)cosec^2(pi/4)
    cosec^2(pi/4) = 1/(sin^2(pi/4)) = 1/(1/2) = 2 so (-1/2)cosec^2(pi/4) = (-1/2)(2) = -1

    and the final answer = -1/1 = -1

    (if I haven't screwed up somewhere!)
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    (Original post by davros)
    So we agree we want [d/dt(dy/dx)] / (dx/dt)

    You claim (I haven't checked!) that dy/dx = cot(t/2), so d/dt(dy/dx) = -(1/2)cosec^2(t/2)
    and dx/dt = 1 - cos t.

    So at t = pi/2, dx/dt = 1 - 0 = 1 and
    (-1/2)cosec^2(t/2) = (-1/2)cosec^2(pi/4)
    cosec^2(pi/4) = 1/(sin^2(pi/4)) = 1/(1/2) = 2 so (-1/2)cosec^2(pi/4) = (-1/2)(2) = -1

    and the final answer = -1/1 = -1

    (if I haven't screwed up somewhere!)
    Oh yeah of course, i forgot about my scale factor 1/2.... thats it, you are spot on mate. Thank you very much foor your help.
 
 
 

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