The question asks for the greatest rate of increase of the function $f = xy^{2}z^{3}$ at the point $(0, -1, -2)$. This is determined by the magnitude of the gradient vector of the function at that specific point.

Calculate the Gradient Vector

First, find the partial derivatives of $f$ with respect to $x$, $y$, and $z$. The gradient vector $\nabla f$ is given by:

$\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)$

• $\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (xy^{2}z^{3}) = y^{2}z^{3}$
• $\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (xy^{2}z^{3}) = x(2y)z^{3} = 2xyz^{3}$
• $\frac{\partial f}{\partial z} = \frac{\partial}{\partial z} (xy^{2}z^{3}) = xy^{2}(3z^{2}) = 3xy^{2}z^{2}$

So, the gradient vector is $\nabla f = (y^{2}z^{3}, 2xyz^{3}, 3xy^{2}z^{2})$.

Evaluate Gradient at the Point

Now, substitute the coordinates of the point $(0, -1, -2)$ into the gradient vector:

• $\frac{\partial f}{\partial x}\Big|_{(0,-1,-2)} = (-1)^{2}(-2)^{3} = (1)(-8) = -8$
• $\frac{\partial f}{\partial y}\Big|_{(0,-1,-2)} = 2(0)(-1)(-2)^{3} = 0$
• $\frac{\partial f}{\partial z}\Big|_{(0,-1,-2)} = 3(0)(-1)^{2}(-2)^{2} = 0$

The gradient vector at the point $(0, -1, -2)$ is $\nabla f(0, -1, -2) = (-8, 0, 0)$.

Find the Magnitude of the Gradient

The greatest rate of increase is the magnitude (or length) of this gradient vector:

$||\nabla f(0, -1, -2)|| = \sqrt{(-8)^2 + 0^2 + 0^2}$

$||\nabla f(0, -1, -2)|| = \sqrt{64}$

$||\nabla f(0, -1, -2)|| = 8$

The greatest rate of increase of the function at the given point is 8.