Chapter 14: Q 3 (page 1153)

Find a potential function for the vector field
$G (x, y, z) = x 2 e^{5} 1 + 2 x y z e^{5} j + x e^{5} k$

Short Answer

Expert verified

The potential function is

Step by step solution

Step 1:Given information

The given expression is $G (x, y, z) = x 2 e^{5} 1 + 2 x y z e^{5} j + x e^{5} k$

Step 2:Simplification

Consider the vector field

$G (x, y, z) = x 2 e^{5} 1 + 2 x y z e^{5} j + x e^{5} k$

To find potential function

$G (x, y, z) = \nabla g (x, y, z)$

$= \frac{\partial g}{\partial x} i + \frac{\partial g}{\partial y} j + \frac{\partial g}{\partial z} k$

here

$\frac{\partial g}{\partial x} = g_{x} = z e^{y^{2}}$

$\frac{\partial g}{\partial y} = g_{y} = 2 x y z e^{y^{2}}$

$\frac{\partial g}{\partial z} = g_{z} = x e^{y^{2}}$

integrate $g_{x}$ w.r.t x

$g (x, y, z) = x z e^{y^{2}} + B (y)$ .... equation (1)

here B is constant

$Then differentiating g (x, y, z)$ w.r.t y

$g_{y} (x, y, z) = 2 x y z e^{y^{2}} + B^{'} (y)$

$Comparingthis with g_{y} = 2 x y z e^{y^{2}}$

$B^{'} (y) = 0$

$Integratingthis with respect to y,$

$Thus, B (y) = 0 + h (z)$

$Now substitute this value in (1)$

$g (x, y, z) = x z e^{y^{2}} + h (z) \dots \dots (2)$

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Find a potential function for the vector field
$G (x, y, z) = x 2 e^{5} 1 + 2 x y z e^{5} j + x e^{5} k$

Short Answer

Step by step solution

Step 1:Given information

Step 2:Simplification

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Step by step solution

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Step 2:Simplification

One App. One Place for Learning.

Most popular questions from this chapter

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Find a potential function for the vector field
$G (x, y, z) = x 2 e^{5} 1 + 2 x y z e^{5} j + x e^{5} k$