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Thread: (Help) Pure Core 3 - Maths homework

  1. #1 (Help) Pure Core 3 - Maths homework 
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    Hey everyone
    Just a little stuck on some maths work
    No, I didn't create an account just to post this, I forgot my old accounts details ):

    So yeah, I'd appreciate it if someone could give me a few hints?

    ai) Find dy/dx when y=(e^x)/sin2x

    u=e^x v=sin2x
    u'=e^x v'=2cos2x

    ((e^x)*2cos2x) + ((e^x)*sin2x) = (e^x)(sin2x+2cos2x)


    Now for what i'm stuck on...

    aii) Hence find the equation of the tangent to the curve y=(e^x)sin2x at the origin.

    So, do I let x=0, since it says at the origin?
    It doesn't imply any other x value

    If so, (e^0)(sin0+2cos0) = (1)(2) = 2, so the gradient of the curve at x=0 is 2..
    To find y, input x=0 into the original.. (e^0)(sin0) = (1)(0) = 0

    So the tangent is y = 2x ?

    Onwards to the next part

    b) Show that the equation of the normal to the curve y=(e^x)sin2x at the point where x= pi is 2(e^pi)+x=pi

    I haven't got a clue on how to do this one ):


    So yeah, some proof reading and help please?
    Thanks!


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  3. #2  
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    You used the formula for (uv)' but what you need is (u/v)' = (vu' - uv')/v^2.


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  4. #3  
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    Quote Originally Posted by mathman View Post
    You used the formula for (uv)' but what you need is (u/v)' = (vu' - uv')/v^2.
    The quotient rule is for two functions being divided, isn't it?
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  5. #4  
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    Quote Originally Posted by JNicko View Post
    ai) Find dy/dx when y=(e^x)/sin2x
    I think mathman's referring to this, unless that's a typo.

    Assuming you've got part ai right, part aii looks fine. (The origin would be x=0.)

    For part b, just do the same thing again, but use x=pi instead of x=0.
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  6. #5  
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    Quote Originally Posted by JNicko View Post
    Hey everyone
    Just a little stuck on some maths work
    No, I didn't create an account just to post this, I forgot my old accounts details ):

    So yeah, I'd appreciate it if someone could give me a few hints?

    ai) Find dy/dx when y=(e^x)/sin2x

    u=e^x v=sin2x
    u'=e^x v'=2cos2x

    ((e^x)*2cos2x) + ((e^x)*sin2x) = (e^x)(sin2x+2cos2x)
    No, this is not correct, and everything that follows is incorrect as well:



    Your mistake stems from:


    u=e^x v=sin2x
    u'=e^x v'=2cos2x

    Actually u'=e^x(v'+v) (not that it helps you in finding dy/dx)
    Last edited by Howard Roark; October 8th, 2014 at 12:33 AM.
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