Chapter 14: Q 2 (page 1153)

Find a potential function for the vector field
$F (x, y) = \frac{y^{2}}{x} i + 2 y \ln x j$

Short Answer

Expert verified

The potential function is $y^{2} \ln x$

Step by step solution

Step 1:Given information

The given expression is $F (x, y) = \frac{y^{2}}{x} i + 2 y \ln x j$

Step 2:Simplification

Consider the vector field

$F (x, y) = \frac{y^{2}}{x} i + 2 y \ln x j$

To find potential function

let $F (x, y) = \frac{\partial J}{\partial x} i + \frac{\partial J}{\partial y}$ $j$

width="139" height="25" role="math">=fx(x,y)i+fy(x,y)j

$= (\frac{y^{2}}{x}) i + (2 y \ln x) j$

now integrating $f_{x} (x, y) = \frac{y^{2}}{x}$ w.r.t x

$f (x, y) = \int (\frac{y^{2}}{x}) d x$

$= y^{2} \ln x + B (y)$

Here B(y) is integral w.r.t y

Differentiating $f (x, y)$ partially w.r t y

$f_{y} (x, y) = 2 y \ln x + B^{'} (y)$

On comparing with $f_{y} (x, y) = 2 y \ln x$

$B' (y) = 0$

On integrating it

$B (y) = 0 + β$

Setting constants equal to zero, then

$f (x, y) = y^{2} \ln x$

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Find a potential function for the vector field
$F (x, y) = \frac{y^{2}}{x} i + 2 y \ln x j$

Short Answer

Step by step solution

Step 1:Given information

Step 2:Simplification

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Find a potential function for the vector field
$F (x, y) = \frac{y^{2}}{x} i + 2 y \ln x j$