Value of $a,b,c$ that makes the vector field a gradient of a function

Question

I am given vector field $F = (ay^2 + 2cxz)i + (ybx+ycz)j + (ay^2+cx^2)k$ and I am supposed to find values $a,b,c$ that would make $f(x,y,z)$ a function.

I first took partial derivatives and I get the following

$f_{x} = ay^2 + 2cxz$ meaning $F_{x} = ay^2x+czx^2$

$f_{y} = ybx+ycz$ meaning $F_{y} = \frac{1}{2} bxy^2 + \frac{1}{2}czy^2$

$f_{z} = ay^2 + cx^2$ meaning $F_{z} = ay^2z+cx^2z$

After this, I guess I can add the three to make $f(x,y,z)$, but I am not sure how to progress from here. Any help as to what I should do next? Thanks.

Answer 1

As suggested by amd, we can make use of the fact that the curl of a gradient identically vanishes, so, being $F$ a gradient, we have, $\nabla\times F=\vec0$. In components.

$\partial_y(ay^2+cx^2)-\partial_z(ybx+ycz)=0$

$\partial_z(ay^2+2cxz)-\partial_x(ay^2+cx^2)=0$

$\partial_x(ybx+ycz)-\partial_y(ay^2+2cxz)=0$

So is,

$2ay-cy=0$

$2cx-2cx=0$

$by-2ay=0$

Or

$c=2a$ and $b=2a$

Answer 2

As @RafaBudria and @amd mention, for a vector field defined on a simply connected domain, being a gradient is the same as having the curl vanish. Even though it is not needed for this problem, if you really want to find a function $f(x,y,z)$ whose gradient is $F(x,y,z)$, you can follow the following procedure.

• Starting with the first coordinate of $F$, we find that $f_x=ay^2+2cxz$. Integrating this with respect to $x$ gives $$ f=\int f_xdx=\int (ay^2+2cxz)dx=axy^2+cx^2z+g(y,z). $$ Here, the constant of integration is a function in $y$ and $z$ because both $y$ and $z$ are constants with respect to $x$.

• We can take the derivative of this with respect to $y$ to get $f_y=2axy+g_y(y,z)$. Setting this equal to the second coordinate of $F$ gives $$ 2axy+g_y(y,z)=ybx+cyz. $$ Since $g_y(y,z)$ doesn't depend on $x$, the parts that depend on $x$ must be equal. Therefore, $2axy=bxy$ or $2a=b$. After making this substitution, we have $g_y(y,z)=cyz$. Integrating this with respect to $y$ gives $$ g=\int g_ydy=\int cyz dy=\frac{1}{2}cy^2z+h(z). $$ Here, the constant of integration is a function of $z$ since $g$ is only a function of $y$ and $z$ and $z$ is constant with respect to $y$. Therefore, we get by substitution: $$ f=axy^2+cx^2z+\frac{1}{2}cy^2z+h(z). $$

• Finally, we can take the derivative of $f$ with respect to $z$ to get $f_z=cx^2+\frac{1}{2}cy^2+h'(z)$. Setting this equal to the third coordinate of $F$ gives $$ cx^2+\frac{1}{2}cy^2+h'(z)=ay^2+cx^2. $$ Since $h'(z)$ only depends on $z$, the parts depending on $x$ and $y$ must be equal. In other words, $\frac{1}{2}cy^2=ay^2$ so $c=2a$. Therefore, we have $h'(z)=0$, so $h$ is a constant. Putting this all together gives $$ f=axy^2+2ax^2z+K, $$ where $K$ is a constant.

Putting this all all together, we get that the relationships that others have found, i.e., $b=2a$ and $c=2a$. We also have a family of functions whose gradient is $F$ (for any $a$).