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Math Help - Finding a scalar potential.

  1. #1
    Super Member Showcase_22's Avatar
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    Finding a scalar potential.

    Am I doing this correctly?

    Find a scalar potential for the vector field \underline{u}(x,y,z):= \left( \frac{xy^2}{1+x^2 y^2}, \frac{x^2y}{1+x^2 y^2}, \frac{1}{4+z^2} \right).
    f(x,y,z)=\int_0^xf_x(t,0,0)~dt+\int_0^yf_y(x,t,0)~  dt+\int_0^z f_z(x,y,t)~dt

    f(x,y,z)=\int_0^x0~dt+\int_0^y \frac{x^2 t}{1+x^2 t^2}~dt+\int_0^z \frac{1}{4+t^2}~dt

    f(x,y,z)=C+\int_0^y \frac{x^2t}{1+x^2 t^2}~dt+\frac{1}{4}\int_0^z \frac{1}{1+ \left( \frac{t}{2} \right)^2}~dt

    Let u=1+x^2 t^2 \Rightarrow \ \frac{du}{dt}=2t x^2

    f(x,y,z)=C+\int_1^{1+x^2 y^2} \frac{x^2 t}{u} \frac{du}{2tx^2}+\frac{1}{4}\int_0^z\frac{1}{1+ \left( \frac{t}{2} \right)^2}~dt

    f(x,y,z)=C+\int_1^{1+x^2 y^2} \frac{1}{2u}~du+\int_0^z\frac{1}{1+ \left( \frac{t}{2} \right)^2}~dt

    f(x,y,z)=C+\frac{1}{2} \left[ \ln u \right]_1^{1+x^2 y^2}+\frac{1}{4}\int_0^z\frac{1}{1+ \left( \frac{t}{2} \right)^2}~dt

    Let \frac{t}{2}=\tan \theta \Rightarrow \ \frac{dt}{d \theta}= 2 \sec^2 \theta

    f(x,y,z)=C+\frac{1}{2} \ln (1+x^2y^2)+ \frac{1}{4}\int_0^{\arctan \left( \frac{z}{2} \right)} 2~d \theta

    f(x,y,z)=C+\frac{1}{2} \ln (1+x^2 y^2)+\frac{1}{2} \arctan \left( \frac{z}{2} \right)

    Is this the correct way of finding a scalar potential?
    Last edited by Showcase_22; May 6th 2009 at 12:00 PM.
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  2. #2
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    Quote Originally Posted by Showcase_22 View Post
    Am I doing this correctly?



    f(x,y,z)=\int_0^xf_x(t,0,0)~dt+\int_0^yf_y(x,t,0)~  dt+\int_0^z f_z(x,y,t)~dt

    f(x,y,z)=\int_0^x0~dt+\int_0^y \frac{x^2 t}{1+x^2 t^2}~dt+\int_0^z \frac{1}{4+t^2}~dt

    f(x,y,z)=C+\int_0^y \frac{x^2t}{1+x^2 t^2}~dt+\int_0^z \frac{1}{1+ \left( \frac{t}{2} \right)^2}~dt

    Let u=1+x^2 t^2 \Rightarrow \ \frac{du}{dt}=2t x^2

    f(x,y,z)=C+\int_1^{1+x^2 y^2} \frac{x^2 t}{u} \frac{du}{2tx^2}+\int_0^z\frac{1}{1+ \left( \frac{t}{2} \right)^2}~dt

    f(x,y,z)=C+\int_1^{1+x^2 y^2} \frac{1}{2u}~du+\int_0^z\frac{1}{1+ \left( \frac{t}{2} \right)^2}~dt

    f(x,y,z)=C+\frac{1}{2} \left[ \ln u \right]_1^{1+x^2 y^2}+\int_0^z\frac{1}{1+ \left( \frac{t}{2} \right)^2}~dt

    Let \frac{t}{2}=\tan \theta \Rightarrow \ \frac{dt}{d \theta}= 2 \sec^2 \theta

    f(x,y,z)=C+\frac{1}{2} \ln (1+x^2y^2)+ \int_0^{\arctan \left( \frac{z}{2} \right)} 2~d \theta

    f(x,y,z)=C+\frac{1}{2} \ln (1+x^2 y^2)+ \color{red}2 \arctan \left( \frac{z}{2} \right)

    Is this the correct way of finding a scalar potential?
    that 2 should be \frac{1}{2}.
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  3. #3
    Super Member Showcase_22's Avatar
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    Oh no, I see why it should be 0.5. Give me a second to change my initial thread.

    On another note, this is the correct method for finding a scalar potential isn't it? I keep thinking that "scalar" implies that I should be finding a number of some sort.
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  4. #4
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    Quote Originally Posted by Showcase_22 View Post
    Oh no, I see why it should be 0.5. Give me a second to change my initial thread.

    On another note, this is the correct method for finding a scalar potential isn't it? I keep thinking that "scalar" implies that I should be finding a number of some sort.
    scalar here means a scalar function, meaning a real-valued and not vector-valued function. in other words function from \mathbb{R}^3 to \mathbb{R}.
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  5. #5
    Super Member Showcase_22's Avatar
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    There's your 800th thanks!!

    Thanks very much, I just needed someone to tell me I was doing them correctly (even if my integration was a bit sloppy).
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