Question

Consider f and c below. f(x, y, z) = yzexzi + exzj + xyexzk,

Answer 1

$\mathbf f(x,y,z)=yze^{xz}\,\mathbf i+e^{xz}\,\mathbf j+xye^{xz}\,\mathbf k$

We're looking for a scalar function $f(x,y,z)$ such that its gradient is equal to the given vector-valued function $\mathbf f(x,y,z)$:

$\nabla f(x,y,z)=\mathbf f(x,y,z)$

That is, we require

$\dfrac{\partial f}{\partial x}=yze^{xz}$
$\dfrac{\partial f}{\partial y}=e^{xz}$
$\dfrac{\partial f}{\partial z}=xye^{xz}$

Integrating the first equation with respect to $x$ gives

$f(x,y,z)=\dfrac{yz}xe^{xz}+g(y,z)$

Differentiating with respect to $y$ gives

$\dfrac{\partial f}{\partial y}=e^{xz}=\dfrac zxe^{xz}+\dfrac{\partial g}{\partial y}$

but from this step we can see that it's not possible for $g$ to be a function of $y$ and $z$, independent of $x$. So there is no such function $f(x,y,z)$.

Answer 2

Answer:

a) ye^(xz)

b) 16

Step-by-step explanation:

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