Show that ∇(u v)=v ∇ u+u ∇ v, where u and v are...

Show that $\nabla(u v)=v \nabla u+u \nabla v$, where $u$ and $v$ are differentiable scalar functions of $x, y$, and $z$. (a) Show that a necessary and sufficient condition that $u(x, y, z)$ and $v(x, y, z)$ are related by some function $f(u, v)=0$ is that $(\nabla u) \times(\nabla v)=0$. (b) If $u=u(x, y)$ and $v=v(x, y)$, show that the condition $(\nabla u) \times(\nabla v)=0$ leads to the two-dimensional Jacobian $$ J\left(\frac{u, v}{x, y}\right)=\left|\begin{array}{ll} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{array}\right|=0 . $$ The functions $u$ and $v$ are assumed differentiable.

Transcript

00:01 All right, so for this problem, we are asked first to show that delta uv equals v delta u plus u delta v, where u and v are differentiable scalar functions of x, y, and z.

00:15 So we have that delta uv is going to be equal to the partial derivative with respect to x of u times v, the first component, then the partial with respect to y of uv and the second component.

00:33 And the partial with respect to z, view v in the third component.

00:39 So this is then going to be equal to partial of u with respect to x times v, plus the partial of u times vx, and similarly for y and z, where i'm using the subscript notation for partial derivative here.

01:03 Now we can see that this would be equal to ux v, uy, v, u z, v plus u vx u v y y u v z which we can recognize by the laws of scalar multiplication of vectors that this is going to be v u x u y u z plus u v x v y u z plus u v x v y v z and we can recognize now that that's going to be v times the gradient of u plus u times the gradient of v then for part a we're as to show that a necessary and sufficient condition that u and v are related by some function of u and v equal zero is that the cross product of their the cross product of their gradients must be equal to zero so if we start out with f of uv equal zero then what we need to have here is that the gradient of f uv must be equal to zero which would or yeah the if f of uv equals zero a constant then its gradient must be equal to zero so then that would have to be grade or um f u see here one woman so that would require that we have f u times ux plus u y actually let me correct myself that should be f u x plus f u x plus f v v x must be equal to zero then we have to have that f u y plus f v y plus f v y must be equal to zero and we have to have that f u u z plus fv v z must be equal to zero, which equivalent, that would be equivalent to saying that we have that f u times the gradient of u plus fv times gradient of v must be equal to zero.

03:52 Actually, let me back this up because i just realized a more effective way of writing this or a more productive way.

04:00 So we would have that, we can write that f -u must be equal to f -v -v -x, or negative f -v -x divided by u -x, and we would have that f -v -v can be written by, let's say we look at the second equation there.

04:22 We'd have that f -v can be written as negative f -u -y, divided by vy, and so we then have that fv would be equal to the negative, or we'd have double negative of fv times, or pardon me, yes, actually, it should be double negative.

04:50 Fv, we're substituting in fu here.

04:53 So we'd have fv, vx over ux times uy over vy over vy, and so we have that one.

05:06 If we divide both sides by fv, we have one must be equal to vx over ux times uy over vy.

05:15 So we have ux, vy must be equal to vx uy.

05:24 And so we have ux vy minus vx uy must be equal to zero...

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