Finding a scalar potential.

**Showcase_22** · May 6th 2009, 11:37 AM

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?

**NonCommAlg** · May 6th 2009, 11:49 AM

Originally Posted by Showcase_22

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 $\frac{1}{2}.$

**Showcase_22** · May 6th 2009, 11:57 AM

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.

**NonCommAlg** · May 6th 2009, 12:17 PM

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 $\mathbb{R}^3$ to $\mathbb{R}.$

**Showcase_22** · May 6th 2009, 12:25 PM

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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