FInd a function f such that F is the gradient of f

Question

perl · Answer

that is called antiderivative , or primitive in some countries

caozeyuan · Answer

$$F(x,y)=(xycos(xy)+\sin(xy))i+(x ^{2}\cos(xy))j$$

caozeyuan · Answer

@perl

caozeyuan · Answer

My calculation goes awful and I don' know why

caozeyuan · Answer

@perl , any idea?

caozeyuan · Answer

@Loser66

anonymous · Answer

Well, we would say: $$
\nabla f = \frac{\partial f}{\partial x}\mathbf i + \frac{\partial f}{\partial y}\mathbf j  = F 
$$

caozeyuan · Answer

yea, I know that

anonymous · Answer

That means: $$
\frac{\partial f }{\partial x} = xy\cos(xy)+\sin(xy)
$$If we integrate by $x$, we get: $$
f = \int \frac{\partial f }{\partial x} \;dx + g(y)
$$where $g(y)$ is our constant of integration.

anonymous · Answer

Then do the partial derivative with respect to $y$, on that: $$
\frac {\partial f}{\partial y} = \frac{\partial }{\partial y}\left[\int xy\cos(xy) + \sin(xy)\;dx\right] + g'(y) = x^2\cos(xy)
$$

anonymous · Answer

You can use this equation to find $g'(y)$.

caozeyuan · Answer

I see！， thx

anonymous · Answer

We can use $g(y) = \int g'(y)\;dy+C$.  There is no means of solving for $C$.

anonymous · Answer

However, this is fine, since the anti-gradient, much like the anti-derivative, has a family of solutions.