Question

find the value of constant b such that A(x,y,z)=(bxy-z^3)i +(b-2)x^2j +(1-b)xz^2k has its

culr identically equal to zero,

Answer 1

\[\large A(x,y,z)=<bxy-z^3 \; , \; (b-2)x^2 \; , \; (1-b)xz^2>\]\[\large \vec{\nabla} \times \vec{A}=\left[\begin{matrix}\hat{i} & \hat{j} & \hat{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ bxy-z^3 & (b-2)x^2 & (1-b)xz^2\end{matrix}\right]\]

Answer 2

\[\left(\frac{\partial}{\partial y}(1-b)xz^2-\frac{\partial}{\partial z}(b-2)x^2\right)\hat i\]
\[\large \left(0-0\right)\hat i\]If we set the i component equal to 0, it doesn't tell us anything about b unfortunately.

Answer 3

\[\large \left(\frac{\partial}{\partial z}bxy-z^3-\frac{\partial}{\partial x}(1-b)xz^2\right)\hat j\]\[\large 0=(-3z^2-(1-b)z^2)\hat j\]

Answer 4

\[\large \left(\frac{\partial}{\partial x}(b-2)x^2-\frac{\partial}{\partial y}bxy-z^3\right)\hat k\]\[\large 0=(2(b-2)x-bx)\hat k\]

Answer 5

So we get a system of equations involving x, z and b.\[\large 0=-3z^2-(1-b)z^2 \qquad \qquad 0=2(b-2)x-bx\]Dividing the first equation by z^2, and the second by x gives us,\[\large 0=-3-(1-b) \qquad \qquad 0=2(b-2)-b\]

Answer 6

They both end up giving us the same b value so that's good! Means there's less likely a chance I made a mistake in there :3

But you should double check my work just to make sure.

Anyway, hope that helps. :D/

Answer 7

thanks a lot