I stuck at 2nd part of the problem

[Prettygoodmilkshake]

Hi, I am trying to solve this problem. I dont know if my approach is right. I stuck second part of the question.


So I started like this to find a,b,c :


If true up to here;
for the second part of the problem I did this

and similarly for x2 and x3.
Then I stuck. Couldnt bring them together to find B. Any suggestions? Thanks.

 

[LCKurtz]

Writing what you have so far in simpler notation using x, y, and z you have
$$\vec A = \langle x+2y+4z, 2x-3y-z,4x-y+2z\rangle$$ which is correct, and you are wanting to find a scalar function $$B(x,y,z)$$ such that $\nabla B = \vec A$. So, as you have worked out, you need to start with $B_x = x+2y+4z$. Taking anti partial derivative as you have done you got$$B(x,y,z) = \frac {x^2} 2 + 2yx + 4zx +C$$Everything is correct except for the $C$. Since you took an anti partial derivative, the "constant" of integration is an unknown function of the other variables, so instead of $C$ you should have an unknown function $g(y,z)$. So at this point you have:$$B(x,y,z) = \frac {x^2} 2 + 2yx + 4zx +g(y,z)$$Everything you have done so far is correct except for that. Now using this $B(x,y,z)$, calculate $B_y$ and set it equal to the second component of $\vec A$ and take its anti partial with respect to $y$ keeping in mind what the new constant of integration must look like and continue. Post back here if you can't finish it from here.

 

[Prettygoodmilkshake]

Thank you! I did finish when I take C as a function as you said.