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
[math]\vec A = \langle x+2y+4z, 2x-3y-z,4x-y+2z\rangle[/math] which is correct, and you are wanting to find a scalar function [math]B(x,y,z)[/math] such that [imath]\nabla B = \vec A[/imath]. So, as you have worked out, you need to start with [imath]B_x = x+2y+4z[/imath]. Taking anti partial derivative as you have done you got[math]B(x,y,z) = \frac {x^2} 2 + 2yx + 4zx +C[/math]Everything is correct except for the [imath]C[/imath]. Since you took an anti partial derivative, the "constant" of integration is an unknown function of the other variables, so instead of [imath]C[/imath] you should have an unknown function [imath]g(y,z)[/imath]. So at this point you have:[math]B(x,y,z) = \frac {x^2} 2 + 2yx + 4zx +g(y,z)[/math]Everything you have done so far is correct except for that. Now using this [imath]B(x,y,z)[/imath], calculate [imath]B_y[/imath] and set it equal to the second component of [imath]\vec A[/imath] and take its anti partial with respect to [imath]y[/imath] 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.