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Math Help - Differentiation

  1. #1
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    Differentiation

    In my assignment, I got a few questions similar to the one shown below. I know I have to use partial differentiation but I have no idea where to start given these particular kind of questions.If you show me how to do these two, I think I'll do the rest at ease. Your help will be highly appreciated.


    1.Determine the values of a and b in the functions
    u(x,y) = 16x^3 - 48xy^2+16x^2 - 16y^2, v(x,y) = ax^2y -16y^3 +b xy such that u and v satisfy the Cauchy-Riemann equations
    u/ x
    = v/ y
    u/ y
    = - v/ x
    .
    2.Determine the POSITIVE values of a and b in the function
    u(x,t) = e^(x+ at)(6x+bt) such that u satisfies the wave equation
    ^2 u/ t^2
    = 17^2 (^2 u/ x^2)
    .
    Last edited by McCoy; August 3rd 2008 at 01:56 AM.
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  2. #2
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    Quote Originally Posted by McCoy View Post
    In my assignment, I got a few questions similar to the one shown below. I know I have to use partial differentiation but I have no idea where to start given these particular kind of questions.If you show me how to do these two, I think I'll do the rest at ease. Your help will be highly appreciated.


    1.Determine the values of a and b in the functions
    u(x,y) = 16x^3 - 48xy^2+16x^2 - 16y^2, v(x,y) = ax^2y -16y^3 +b xy such that u and v satisfy the Cauchy-Riemann equations
    /u x
    = v y
    u y
    = - v x
    .
    2.Determine the POSITIVE values of a and b in the function
    u(x,t) = ex+ at(6x+bt) such that u satisfies the wave equation
    ^2 u t^2
    = 172 2 u x^2
    .
    Do the partial derivatives, plug into the equations and then the coefficients of like powers of x and y (x and t in the second question) must be equal. That will give you a set of simultaneous equations for the unknowns independent of x and y which you should solve.

    In fact for question 1, you will obtain a solution for a and b from the first of the C-R equations, and you will just need to show that this is also a solution for the second

    RonL
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by McCoy View Post
    In my assignment, I got a few questions similar to the one shown below. I know I have to use partial differentiation but I have no idea where to start given these particular kind of questions.If you show me how to do these two, I think I'll do the rest at ease. Your help will be highly appreciated.

    2.Determine the POSITIVE values of a and b in the function
    u(x,t) = ex+ at(6x+bt) such that u satisfies the wave equation
    2 u t2
    = 172 2 u x2
    .
    I take it that you mean to find a and b such that u(x,t)=ex+at(6x+bt) is a solution to \frac{\partial^2u}{\partial t^2}=172\frac{\partial^2u}{\partial x^2}?

    well, find \frac{\partial^2u}{\partial t^2} and \frac{\partial^2u}{\partial t^2}.

    u(x,t)=ex+6axt+abt^2

    \frac{\partial u}{\partial t}=6ax+2abt

    \frac{\partial^2 u}{\partial t^2}=2ab

    \frac{\partial u}{\partial x}=e+6at

    \frac{\partial^2 u}{\partial x^2}=0

    Thus, 2ab=0...

    ...are you sure you wrote out the problem right?

    I'm getting zero as an answer for both!

    --Chris
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Chris L T521 View Post
    I take it that you mean to find a and b such that u(x,t)=ex+at(6x+bt) is a solution to \frac{\partial^2u}{\partial t^2}=172\frac{\partial^2u}{\partial x^2}?

    well, find \frac{\partial^2u}{\partial t^2} and \frac{\partial^2u}{\partial t^2}.

    u(x,t)=ex+6axt+abt^2

    \frac{\partial u}{\partial t}=6ax+2abt

    \frac{\partial^2 u}{\partial t^2}=2ab

    \frac{\partial u}{\partial x}=e+6at

    \frac{\partial^2 u}{\partial x^2}=0

    Thus, 2ab=0...

    ...are you sure you wrote out the problem right?

    I'm getting zero as an answer for both!

    --Chris

    May-be ex means e^x?

    Actually I see that the original poster has modified the question.

    RonL
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  5. #5
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    Quote Originally Posted by Chris L T521 View Post
    I take it that you mean to find a and b such that u(x,t)=ex+at(6x+bt) is a solution to \frac{\partial^2u}{\partial t^2}=172\frac{\partial^2u}{\partial x^2}?

    well, find \frac{\partial^2u}{\partial t^2} and \frac{\partial^2u}{\partial t^2}.

    u(x,t)=ex+6axt+abt^2

    \frac{\partial u}{\partial t}=6ax+2abt

    \frac{\partial^2 u}{\partial t^2}=2ab

    \frac{\partial u}{\partial x}=e+6at

    \frac{\partial^2 u}{\partial x^2}=0

    Thus, 2ab=0...

    ...are you sure you wrote out the problem right?

    I'm getting zero as an answer for both!

    --Chris
    Thanks guys but I'm sorry the way I've typed the questions was wrong. I have done some corrections just now.e.g for the second question, it's 17^2, not 172 .e^(x+at), not ex. I'm sorry guys for the inconvenience.
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by McCoy View Post
    In my assignment, I got a few questions similar to the one shown below. I know I have to use partial differentiation but I have no idea where to start given these particular kind of questions.If you show me how to do these two, I think I'll do the rest at ease. Your help will be highly appreciated.

    2.Determine the POSITIVE values of a and b in the function
    u(x,t) = e^(x+ at)(6x+bt) such that u satisfies the wave equation
    ^2 u/ t^2
    = 17^2 (2 u/ x^2)
    .
    I figured you were missing some sort of exponential function here.

    Well, you still need to find \frac{\partial^2u}{\partial t^2} and \frac{\partial^2u}{\partial x^2}

    \because u(x,t)=e^{x+at}(6x+bt)

    \frac{\partial u}{\partial t}=e^{x+at}(b)+ae^{x+at}(6x+bt)=(b+6ax+abt)e^{x+at  }

    \frac{\partial^2u}{\partial t^2}=e^{x+at}(ab)+ae^{x+at}(b+6ax+abt)=(2ab+6a^2x+  a^2bt)e^{x+at}

    \frac{\partial u}{\partial x}=e^{x+at}(6)+e^{x+at}(6x+bt)=(6+6x+bt)e^{x+at}

    \frac{\partial^2 u}{\partial x^2}=e^{x+at}(6)+e^{x+at}(6+6x+bt)=(12+6x+bt)e^{x+  at}

    Thus,

    (2ab+6a^2x+a^2bt)e^{x+at}=17^2(12+6x+bt)e^{x+at}

    \implies 2ab+6a^2x+a^2bt=3468+1734x+17^2bt)

    We can come up with 3 simultaneous equations equations:

    \left\{\begin{array}{lcl}<br />
2ab & = & 3468\\<br />
6a^2 & = & 1734 \\<br />
a^2b & = & 17^2b \\<br />
\end{array}<br />
\right.<br />

    It turns out that the last 2 equations give the same value: a^2=17^2\implies a=.......

    Once you find a, then b=\frac{1734}{a}\implies b=.......

    Can you finish the rest of this?

    Does this make sense?

    --Chris
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  7. #7
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    Quote Originally Posted by CaptainBlack View Post
    Do the partial derivatives, plug into the equations and then the coefficients of like powers of x and y (x and t in the second question) must be equal. That will give you a set of simultaneous equations for the unknowns independent of x and y which you should solve.

    In fact for question 1, you will obtain a solution for a and b from the first of the C-R equations, and you will just need to show that this is also a solution for the second

    RonL
    I'm sorry but i've done some corrections now. My original post was wrong. I'm sorry for that.Thanks for your time.
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  8. #8
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    Quote Originally Posted by Chris L T521 View Post
    I figured you were missing some sort of exponential function here.

    Well, you still need to find \frac{\partial^2u}{\partial t^2} and \frac{\partial^2u}{\partial x^2}

    \because u(x,t)=e^{x+at}(6x+bt)

    \frac{\partial u}{\partial t}=e^{x+at}(b)+ae^{x+at}(6x+bt)=(b+6ax+abt)e^{x+at  }

    \frac{\partial^2u}{\partial t^2}=e^{x+at}(ab)+ae^{x+at}(b+6ax+abt)=(2ab+6a^2x+  a^2bt)e^{x+at}

    \frac{\partial u}{\partial x}=e^{x+at}(6)+e^{x+at}(6x+bt)=(6+6x+bt)e^{x+at}

    \frac{\partial^2 u}{\partial x^2}=e^{x+at}(6)+e^{x+at}(6+6x+bt)=(12+6x+bt)e^{x+  at}

    Thus,

    (2ab+6a^2x+a^2bt)e^{x+at}=17^2(12+6x+bt)e^{x+at}

    \implies 2ab+6a^2x+a^2bt=3468+1734x+17^2bt)

    We can come up with 3 simultaneous equations equations:

    \left\{\begin{array}{lcl}<br />
2ab & = & 3468\\<br />
6a^2 & = & 1734 \\<br />
a^2b & = & 17^2b \\<br />
\end{array}<br />
\right.<br />

    It turns out that the last 2 equations give the same value: a^2=17^2\implies a=.......

    Once you find a, then b=\frac{1734}{a}\implies b=.......

    Can you finish the rest of this?

    Does this make sense?

    --Chris
    Thank you very much Chris.I think I can do the rest.
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