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Math Help - So Lost ? Derivative of Exponential and Trig Functions.

  1. #1
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    Angry So Lost ? Derivative of Exponential and Trig Functions.

    If y=Acos(kt)+Bsin(kt) , where A, B and k are constants, show that d^2y/dx^2+K^2y=0

    I just have no idea where to go with the question.

    Help would really be appreciated
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  2. #2
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    Re: So Lost ? Derivative of Exponential and Trig Functions.

    You must take the second derivative of y. Can you do that or you want help?
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    Re: So Lost ? Derivative of Exponential and Trig Functions.

    Your sine and cosine are cyclical and \frac{dy}{dx}[g(x)+f(x)]=\frac{dy}{dx}g(x)+\frac{dy}{dx}f(x), so you can break \frac{d^2y}{dx^2}+K^2y=0 up, do the first and second derivatives, and see where it goes.

    And by break up, I mean the y=Acos(kt)+Bsin(kt) part...
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    Re: So Lost ? Derivative of Exponential and Trig Functions.

    Okay ... Im going to take the derivatives and see what i get ...

    i ended up with... y''=-AK2cos(Kt)-BK2sin(Kt)

    Is this right?
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    Re: So Lost ? Derivative of Exponential and Trig Functions.

    I don't see how your supposed to deal with the other part. ...
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    Re: So Lost ? Derivative of Exponential and Trig Functions.

    Quote Originally Posted by elittlewood View Post
    If y=Acos(kt)+Bsin(kt) , where A, B and k are constants, show that d^2y/dx^2+K^2y=0

    I just have no idea where to go with the question.

    You wrote y=A\cos(k{\color{red}t})+B\sin(k{\color{red}t})

    If that is correct then \frac{dy}{dx}=0 so \frac{d^2y}{dx^2}=0.

    Did you post it correctly?
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    Re: So Lost ? Derivative of Exponential and Trig Functions.

    Yes i did post it correctly.... i don't know what you are supposed to do to prove that d^2y/Dx^2 + k^2y = 0 ... ?
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    Re: So Lost ? Derivative of Exponential and Trig Functions.

    Quote Originally Posted by elittlewood View Post
    Yes i did post it correctly.... i don't know what you are supposed to do to prove that d^2y/Dx^2 + k^2y = 0 ... ?
    Well my goodness, if there is no x in the definition then the derivative \frac{d}{dx} is the change in x and that is zero.
    This is a trick question..
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    Re: So Lost ? Derivative of Exponential and Trig Functions.

    you take the second derivative of y with respect to x or t (it depends on what is your variable) \frac{d^2 y}{d x^2} + k^2 y = -Ak^2 cos(kx) - Bk^2 sin(kx) + Ak^2 sin(kx) + Bk^2 cos(kx) = 0
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    Re: So Lost ? Derivative of Exponential and Trig Functions.

    Quote Originally Posted by ManosG View Post
    you take the second derivative of y with respect to x or t (it depends on what is your variable) \frac{d^2 y}{d x^2} + k^2 y = -Ak^2 cos(kx) - Bk^2 sin(kx) + Ak^2 sin(kx) + Bk^2 cos(kx) = 0
    My advice to the original poster: ignore the above reply.
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    Re: So Lost ? Derivative of Exponential and Trig Functions.

    I determined the second derivative... but again ... i don't know what to do with the x as brought up by Plato .
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    Re: So Lost ? Derivative of Exponential and Trig Functions.

    Here is the thing that I still see here:

    If y=A\cos(kt)+B\sin(kt) , where A, B and k are constants, show that \frac{d^2y}{dx^2}+K^2y=0
    OK: \frac{d^2}{dx^2}y+K^2y=0 Thanks to y not being a function of x, the derivative zeros that. So then you still have:

    K^2\left[A\cos(kt)+B\sin(kt)\right]=0 to deal with. It really does not make sense as a question unless there is some part missing that makes the solution something along the lines of what ManosG posted. Without a derivative to cancel out that part, it is simply not true that it = 0.
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  13. #13
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    Quote Originally Posted by elittlewood View Post
    I determined the second derivative... but again ... i don't know what to do with the x as brought up by Plato .
    If the original posting is correct, then there is no x in the definition.
    Thus, the derivative with respect to x is zero.
    As I said before, it is a trick question.

    What don't you get about that?
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    Re: So Lost ? Derivative of Exponential and Trig Functions.

    What is the source of the question?

    I ask because if something can be either true or false it generally starts with "If possible...." This one just says "show that" which usually means it is true and you are doing a proof. However, the way it has been stated seems to not allow for it to be true, so not a proof. Therefore, there must be a problem somewhere.

    The only two places I see a possible problem are either in the source or how it has been copied. If it is the source, not much can be done other than to complain or assume they meant to write it differently. If it is how it is copied, all it would take would be something like a notation for the set of problems this is on, a missing f(t) type statement, or just those words, "If possible..."
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  15. #15
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    Re: So Lost ? Derivative of Exponential and Trig Functions.

    Yes Plato, you may have a point, but the exercise does not say "find the t for which d^2y/dx^2+K^2y=0 ", it says "show that d^2y/dx^2+K^2y=0 ", so it makes no sense the "K^2y=0". I believe, he has written false the equation and the right answer is to take the second derivative
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