Question

Find the divergence of the vector field F(x,y) = (x^2 − y^2) i + 8xy j .

div F =

Answer 1

\[\nabla\cdot \mathbf F(x,y) =\frac{\partial F_x}{\partial x}+\frac{\partial F_y}{\partial y}\]

Answer 2

\[F_x = x^2-y^2\implies\frac{\partial F_x}{\partial x} = ?\]
\[F_y = 8xy\implies\frac{\partial F_y}{\partial y} = ??\]

Answer 3

\[Div F=(F_x \hat{i}+F_y \hat{j}+F_z \hat{k})(\frac{ \partial }{ \partial x }\hat{i}+\frac{ \partial }{ \partial y }\hat{j}+\frac{ \partial }{ \partial z}\hat{k})\\\]
that how the expression @UnkleRhaukus wrote is coming :)

Answer 4

@Ldaniel, can you take the partial derivatives of the components?

Answer 5

\[\frac{ \partial }{ \partial x}=take~derivative~of~"x"~terms~keeping~"y"~as ~constant\]

Answer 6

F_x= 2x
F_y=8y

Answer 7

yes correct :)

Answer 8

nope
\[F_x = x^2-y^2\implies\frac{\partial F_x}{\partial x} = 2x\]\[F_y = 8xy\implies\frac{\partial F_y}{\partial y} =8x\]

Answer 9

okey pit fall

Answer 10

oh yeah sorry

Answer 11

8xy(8x) +2x(x^2-y^2)?

Answer 12

@sidsiddhartha

Answer 13

so its 10x?

Answer 14

\[\nabla\cdot \mathbf F(x,y) \\
=\left(\frac{ \partial }{ \partial x }\hat{\boldsymbol\imath}+\frac{ \partial }{ \partial y }\hat{\boldsymbol\jmath}\right)\cdot\left(F_x \hat{\mathbf {\boldsymbol\imath} }+F_y \hat{\mathbf {\boldsymbol\jmath} }\right)\\
=\frac{\partial F_x}{\partial x}+\frac{\partial F_y}{\partial y}\]

\[{\mathbf F}=(x^2 − y^2) \hat{\boldsymbol\imath} + 8xy \hat{\boldsymbol\jmath} \]\[F_x = x^2-y^2\implies\frac{\partial F_x}{\partial x} = 2x\]\[F_y = 8xy\implies\frac{\partial F_y}{\partial y} =8x\]

\[\nabla\cdot \mathbf F(x,y) =2x+8x =10x\]