Suggested languages for you:

Americas

Europe

Q 1

Expert-verified
Found in: Page 1153

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# Find a potential function for the vector field$F\left(x,y\right)=\left(3{x}^{2}{y}^{2},2{x}^{3}y\right)$

The potential function is ${x}^{3}{y}^{2}$

See the step by step solution

## Step 1:Given information

The given expression is $F\left(x,y\right)=\left(3{x}^{2}{y}^{2},2{x}^{3}y\right)$

## Step 2:Simplification

Consider the vector field

$F\left(x,y\right)=\left(3{x}^{2}{y}^{2},2{x}^{3}y\right)$

To find potential function

let $\mathbf{F}\left(x,y\right)=\frac{\partial f}{\partial x}\mathbf{i}+\frac{\partial f}{\partial y}\mathbf{j}$

$={f}_{x}\left(x,y\right)\mathbf{i}+{f}_{y}\left(x,y\right)\mathbf{j}$

$=\left(3{x}^{2}{y}^{2}\right)\mathbf{i}+\left(2{x}^{3}y\right)\mathbf{j}$

now integrating ${f}_{x}\left(x,y\right)=3{x}^{2}{y}^{2}$ w.r.t x

$⇒f\left(x,y\right)=\int 3{x}^{2}{y}^{2}dx$

$={x}^{3}{y}^{2}+B\left(y\right)$

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

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

$⇒{f}_{y}\left(x,y\right)=2{x}^{3}y+B\text{'}\left(y\right)$

On comparing with

${f}_{y}\left(x,y\right)=2{x}^{3}y\phantom{\rule{0ex}{0ex}}B\text{'}\left(y\right)=0$

On integrating it

$⇒B\left(y\right)=0+b$

$⇒f\left(x,y\right)={x}^{3}{y}^{2}+b$ as constant is 0

Therefore $f\left(x,y\right)={x}^{3}{y}^{2}$