The problem asks us to approximate the change in $z$ using the total differential $dz$ as $(x, y)$ moves from point $P$ to point $Q$. Then, we need to calculate the exact change $\Delta z$ using the given function and points. We will solve problem 10, where $z = x^2 - 5xy + y$, $P(2, 3)$, and $Q(2.03, 2.98)$.

1. Problem Description

The problem asks us to approximate the change in $z$ using the total differential $dz$ as $(x, y)$ moves from point $P$ to point $Q$. Then, we need to calculate the exact change $\Delta z$ using the given function and points. We will solve problem 10, where $z = x^2 - 5xy + y$, $P(2, 3)$, and $Q(2.03, 2.98)$.

2. Solution Steps

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

$\frac{\partial z}{\partial x} = 2x - 5y$

$\frac{\partial z}{\partial y} = -5x + 1$

The total differential $dz$ is given by:

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

$dx$ and $dy$ are the changes in $x$ and $y$, respectively:

$dx = x_Q - x_P = 2.03 - 2 = 0.03$

$dy = y_Q - y_P = 2.98 - 3 = -0.02$

We evaluate the partial derivatives at point $P(2, 3)$:

$\frac{\partial z}{\partial x}(2, 3) = 2(2) - 5(3) = 4 - 15 = -11$

$\frac{\partial z}{\partial y}(2, 3) = -5(2) + 1 = -10 + 1 = -9$

Now, we can approximate the change in $z$ using the total differential:

$dz = (-11)(0.03) + (-9)(-0.02) = -0.33 + 0.18 = -0.15$

Next, we calculate the exact change in $z$, $\Delta z = z(Q) - z(P)$.

First, we find $z(P)$:

$z(2, 3) = (2)^2 - 5(2)(3) + 3 = 4 - 30 + 3 = -23$

Then, we find $z(Q)$:

$z(2.03, 2.98) = (2.03)^2 - 5(2.03)(2.98) + 2.98 = 4.1209 - 30.247 + 2.98 = -23.1461$

Finally, we calculate $\Delta z$:

$\Delta z = z(2.03, 2.98) - z(2, 3) = -23.1461 - (-23) = -23.1461 + 23 = -0.1461$

3. Final Answer

The approximate change in z is $dz = -0.15$.

The exact change in z is $\Delta z = -0.1461$.