Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 16 - Vector Calculus - 16.3 The Fundamental Theorem for Line Integrals - 16.3 Exercises - Page 1134: 5


Conservative $f(x,y)=yxe^{xy}+k$

Work Step by Step

When $F(x,y)=pi+qj$ is a conservative field, then throughout the $D$, then we have $\dfrac{\partial p}{\partial y}=\dfrac{\partial q}{\partial x}$ Here, $p$ and $q$ represents the first-order partial derivatives on a domain $D$. We are given that $F(x,y)=y^2e^{xy}i+(1+xy)e^{xy}j$ Here, $p_x=2ye^{xy}+xy^2e^{xy}; q_y=2ye^{xy}+xy^2e^{xy}$ Therefore, we can see that $\dfrac{\partial p}{\partial y} = \dfrac{\partial q}{\partial x}$ Thus, the $F$ is conservative. Need to find the function $f$ such that $F=\nabla f$ The partial derivatives of $f$ are: $f_x=y^2e^{xy}..(a) \\ f_y=(1+xy)e^{xy} ...(b)$ Integrate $f_x$ with respect to $x$. $f(x,y)=ye^{xy}+g(y)$ ...(c) Here, $g(y)$ is a constant of integration. Take derivative of the equation $c$ with respect to $y$. we have $f_y(x,y)=yxe^{xy}+e^{xy}+g'(y) ...(d)$ On comparing the equations (b) and (d) , we get $g'(y)=0$ Now, integrate it with respect to $y$. This implies that $g(y)=C$ Here, $C$ is a constant of integration. Hence, $f(x,y)=yxe^{xy}+k$
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