Suggested languages for you:

Americas

Europe

Q 1

Expert-verifiedFound in: Page 1153

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Find a potential function for the vector field

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

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

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

Consider the vector field

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

To find potential function

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

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

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

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

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

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

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

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

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

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

$\Rightarrow B\left(y\right)=0+b$

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

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

94% of StudySmarter users get better grades.

Sign up for free