We are given the function $z = \ln(x^2y)$, and two points $P(-2, 4)$ and $Q(-1.98, 3.96)$. We need to approximate the change in $z$ as $(x, y)$ moves from $P$ to $Q$ using the total differential $dz$, and then find the exact change $\Delta z$ using a calculator.

1. Problem Description

We are given the function $z = \ln(x^2y)$, and two points $P(-2, 4)$ and $Q(-1.98, 3.96)$. We need to approximate the change in $z$ as $(x, y)$ moves from $P$ to $Q$ using the total differential $dz$, and then find the exact change $\Delta z$ using a calculator.

2. Solution Steps

First, we find the partial derivatives of $z$ with respect to $x$ and $y$:

$\frac{\partial z}{\partial x} = \frac{1}{x^2y} \cdot 2xy = \frac{2x}{x^2y}xy = \frac{2x}{x^2y}y = \frac{2x}{x^2} = \frac{2}{x}$

$\frac{\partial z}{\partial y} = \frac{1}{x^2y} \cdot x^2 = \frac{1}{y}$

Next, we evaluate these partial derivatives at point $P(-2, 4)$:

$\frac{\partial z}{\partial x}(-2, 4) = \frac{2}{-2} = -1$

$\frac{\partial z}{\partial y}(-2, 4) = \frac{1}{4}$

Now, we calculate the total differential $dz$:

$dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy$

$dz = (-1)dx + \frac{1}{4} dy$

We find $dx$ and $dy$ as the change in $x$ and $y$ from $P$ to $Q$:

$dx = -1.98 - (-2) = -1.98 + 2 = 0.02$

$dy = 3.96 - 4 = -0.04$

We plug $dx$ and $dy$ into the equation for $dz$:

$dz = (-1)(0.02) + \frac{1}{4}(-0.04) = -0.02 - 0.01 = -0.03$

Now, we calculate the exact change $\Delta z$:

$\Delta z = z(Q) - z(P)$

$z(P) = \ln((-2)^2(4)) = \ln(4 \cdot 4) = \ln(16)$

$z(Q) = \ln((-1.98)^2(3.96)) = \ln(3.9204 \cdot 3.96) = \ln(15.524784)$

$\Delta z = \ln(15.524784) - \ln(16) = \ln(\frac{15.524784}{16}) \approx \ln(0.9703)$

$\Delta z \approx -0.030146$

3. Final Answer

The approximate change in $z$ using the total differential is $dz = -0.03$.

The exact change in $z$ is $\Delta z \approx -0.030146$.