Explained: "Find a potential function f for the field..."

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nuais6lfp

Answered question

2021-11-11

Find a potential function f for the field F.

F = 2 ξ + 3 y j + 4 z k

Answer & Explanation

Abel Maynard

Beginner2021-11-12Added 19 answers

If $F = \nabla f$ for some scalar function f, then
$M = 2 x = \frac{d f}{d x}$
$N = 3 y = \frac{d f}{d y}$
$P = 4 z = \frac{d f}{d z}$
From equation we have that
$\frac{d f}{d x} = 2 x \Rightarrow f = \int 2 x d x = x^{2} + C (y, z)$
where C(y,z) does not depended on x. Therefore,
$f (x, y, z) = x^{2} + C (y, z)$
Substituting this result into equation we have that
$\frac{d f}{d y} = 3 y \Rightarrow \frac{d}{d y} [x^{2} + C (y, z)] = 3 y$
$\Rightarrow \frac{d C}{d y} = 3 y \Rightarrow C = 3 \frac{y^{2}}{2} + C_{0} (z)$
Where $C_{0}$ is a function of z. Therefore,
$f (x, y, z) = x^{2} + \frac{3}{2} y^{2} + C_{0} (z)$
Finally, by substituting this result into equation we have that
$\frac{d f}{d z} = 4 z \Rightarrow \frac{d}{d z} (x^{2} + \frac{3}{2} y^{2} + C_{0} (z)) = 4 z$
$\Rightarrow \frac{d C_{0}}{d z} = 4 z \Rightarrow C_{0} \int 4 z d z$
$\Rightarrow C_{0} = 2 z^{2} + c$
where c is a real constant. Letting c=0 we have that
$f (x, y, z) = x^{2} + \frac{3}{2} y^{2} + 2 z^{2}$
is a potential function for F.

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