# Thread: Finding a scalar potential.

1. ## Finding a scalar potential.

Am I doing this correctly?

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

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

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

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

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

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

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

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

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

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

Is this the correct way of finding a scalar potential?

2. Originally Posted by Showcase_22
Am I doing this correctly?

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

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

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

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

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

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

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

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

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

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

Is this the correct way of finding a scalar potential?
that 2 should be $\displaystyle \frac{1}{2}.$

3. 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.

4. Originally Posted by Showcase_22
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 $\displaystyle \mathbb{R}^3$ to $\displaystyle \mathbb{R}.$

5. 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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