Finding the potential function of $F$
$F= langle ye^{xy}+x^2,xe^{xy}+2y rangle$. Find the potential function of $F$.
My Try:
$varphi_x=f(x,y)=ye^{xy}+x^2 $ and $varphi_y=g(x,y)=xe^{xy}+2y$
Now integrated the first equation with respect to $x$
$$intvarphi_xdx=int ye^{xy}+x^2dx=frac{x^3}{3}+e^{xy}+c(y)$$
To find $c(y)$, I differentiated with respect to $y$
$$varphi_y=xe^{xy}+c^1(y)=xe^{xy}+2y$$
So, from above $c^1(y)=2y$
$$int2ydy=y^2+k$$
where k is constant and let $k=0$
So, finally I got the potential function as $frac{x^3}{3}+e^{xy}+y^2$
Is my above attempt correct?
calculus integration multivariable-calculus line-integrals
edited Dec 4 '18 at 2:10
caverac
14k21130
asked Dec 4 '18 at 2:07
user982787user982787
1117
add a comment |
$F= langle ye^{xy}+x^2,xe^{xy}+2y rangle$. Find the potential function of $F$.
My Try:
$varphi_x=f(x,y)=ye^{xy}+x^2 $ and $varphi_y=g(x,y)=xe^{xy}+2y$
Now integrated the first equation with respect to $x$
$$intvarphi_xdx=int ye^{xy}+x^2dx=frac{x^3}{3}+e^{xy}+c(y)$$
To find $c(y)$, I differentiated with respect to $y$
$$varphi_y=xe^{xy}+c^1(y)=xe^{xy}+2y$$
So, from above $c^1(y)=2y$
$$int2ydy=y^2+k$$
where k is constant and let $k=0$
So, finally I got the potential function as $frac{x^3}{3}+e^{xy}+y^2$
Is my above attempt correct?
calculus integration multivariable-calculus line-integrals
edited Dec 4 '18 at 2:10
caverac
14k21130
asked Dec 4 '18 at 2:07
user982787user982787
1117
2
Try computing the gradient of $frac{x^3}{3}+e^{xy}+y^2$. Do you recover $F$? Also, you could leave $k$ as a constant to have the general potential function.
– Dave
Dec 4 '18 at 2:15
@Dave Thanks, for suggesting.
– user982787
Dec 4 '18 at 2:30
add a comment |
$F= langle ye^{xy}+x^2,xe^{xy}+2y rangle$. Find the potential function of $F$.
My Try:
$varphi_x=f(x,y)=ye^{xy}+x^2 $ and $varphi_y=g(x,y)=xe^{xy}+2y$
Now integrated the first equation with respect to $x$
$$intvarphi_xdx=int ye^{xy}+x^2dx=frac{x^3}{3}+e^{xy}+c(y)$$
To find $c(y)$, I differentiated with respect to $y$
$$varphi_y=xe^{xy}+c^1(y)=xe^{xy}+2y$$
So, from above $c^1(y)=2y$
$$int2ydy=y^2+k$$
where k is constant and let $k=0$
So, finally I got the potential function as $frac{x^3}{3}+e^{xy}+y^2$
Is my above attempt correct?
calculus integration multivariable-calculus line-integrals
edited Dec 4 '18 at 2:10
caverac
14k21130
asked Dec 4 '18 at 2:07
user982787user982787
1117
$F= langle ye^{xy}+x^2,xe^{xy}+2y rangle$. Find the potential function of $F$.
My Try:
$varphi_x=f(x,y)=ye^{xy}+x^2 $ and $varphi_y=g(x,y)=xe^{xy}+2y$
Now integrated the first equation with respect to $x$
$$intvarphi_xdx=int ye^{xy}+x^2dx=frac{x^3}{3}+e^{xy}+c(y)$$
To find $c(y)$, I differentiated with respect to $y$
$$varphi_y=xe^{xy}+c^1(y)=xe^{xy}+2y$$
So, from above $c^1(y)=2y$
$$int2ydy=y^2+k$$
where k is constant and let $k=0$
So, finally I got the potential function as $frac{x^3}{3}+e^{xy}+y^2$
Is my above attempt correct?
calculus integration multivariable-calculus line-integrals
calculus integration multivariable-calculus line-integrals
edited Dec 4 '18 at 2:10
caverac
14k21130
asked Dec 4 '18 at 2:07
user982787user982787
1117
edited Dec 4 '18 at 2:10
caverac
14k21130
asked Dec 4 '18 at 2:07
user982787user982787
1117
edited Dec 4 '18 at 2:10
caverac
14k21130
edited Dec 4 '18 at 2:10
caverac
14k21130
edited Dec 4 '18 at 2:10
caverac
14k21130
14k21130
asked Dec 4 '18 at 2:07
user982787user982787
1117
asked Dec 4 '18 at 2:07
user982787user982787
1117
asked Dec 4 '18 at 2:07
user982787user982787
1117
1117
2
Try computing the gradient of $frac{x^3}{3}+e^{xy}+y^2$. Do you recover $F$? Also, you could leave $k$ as a constant to have the general potential function.
– Dave
Dec 4 '18 at 2:15
@Dave Thanks, for suggesting.
– user982787
Dec 4 '18 at 2:30
add a comment |
2
Try computing the gradient of $frac{x^3}{3}+e^{xy}+y^2$. Do you recover $F$? Also, you could leave $k$ as a constant to have the general potential function.
– Dave
Dec 4 '18 at 2:15
@Dave Thanks, for suggesting.
– user982787
Dec 4 '18 at 2:30
2
2
Try computing the gradient of $frac{x^3}{3}+e^{xy}+y^2$. Do you recover $F$? Also, you could leave $k$ as a constant to have the general potential function.
– Dave
Dec 4 '18 at 2:15
Try computing the gradient of $frac{x^3}{3}+e^{xy}+y^2$. Do you recover $F$? Also, you could leave $k$ as a constant to have the general potential function.
– Dave
Dec 4 '18 at 2:15
@Dave Thanks, for suggesting.
– user982787
Dec 4 '18 at 2:30
@Dave Thanks, for suggesting.
– user982787
Dec 4 '18 at 2:30
add a comment |
1 Answer
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Yes, it is correct, except that you do not need to force $k=0$, the solution is just
$$
phi(x, y) = frac{x^3}{3} + e^{xy} + y^2 + color{blue}{k}
$$
answered Dec 4 '18 at 2:13
caveraccaverac
14k21130
add a comment |
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1 Answer
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active
oldest
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votes
Yes, it is correct, except that you do not need to force $k=0$, the solution is just
$$
phi(x, y) = frac{x^3}{3} + e^{xy} + y^2 + color{blue}{k}
$$
answered Dec 4 '18 at 2:13
caveraccaverac
14k21130
add a comment |
Yes, it is correct, except that you do not need to force $k=0$, the solution is just
$$
phi(x, y) = frac{x^3}{3} + e^{xy} + y^2 + color{blue}{k}
$$
answered Dec 4 '18 at 2:13
caveraccaverac
14k21130
add a comment |
Yes, it is correct, except that you do not need to force $k=0$, the solution is just
$$
phi(x, y) = frac{x^3}{3} + e^{xy} + y^2 + color{blue}{k}
$$
answered Dec 4 '18 at 2:13
caveraccaverac
14k21130
Yes, it is correct, except that you do not need to force $k=0$, the solution is just
$$
phi(x, y) = frac{x^3}{3} + e^{xy} + y^2 + color{blue}{k}
$$
answered Dec 4 '18 at 2:13
caveraccaverac
14k21130
answered Dec 4 '18 at 2:13
caveraccaverac
14k21130
answered Dec 4 '18 at 2:13
caveraccaverac
14k21130
answered Dec 4 '18 at 2:13
caveraccaverac
14k21130
14k21130
add a comment |
add a comment |
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2
Try computing the gradient of $frac{x^3}{3}+e^{xy}+y^2$. Do you recover $F$? Also, you could leave $k$ as a constant to have the general potential function.
– Dave
Dec 4 '18 at 2:15
@Dave Thanks, for suggesting.
– user982787
Dec 4 '18 at 2:30