We are given two problems. Problem 12: Given $z = \arctan(xy)$ and two points $P(-2, -0.5)$ and $Q(-2.03, -0.51)$, we are to find the value of $z$ at the given points. However, this problem is incomplete. We will assume the problem is to approximate the change in $z$ from $P$ to $Q$. Problem 13: Find all points on the surface $z = x^2 - 2xy - y^2 - 8x + 4y$ where the tangent plane is horizontal. This means finding where the partial derivatives with respect to $x$ and $y$ are both equal to zero.

1. Problem Description

We are given two problems.

Problem 12: Given $z = \arctan(xy)$ and two points $P(-2, -0.5)$ and $Q(-2.03, -0.51)$, we are to find the value of $z$ at the given points. However, this problem is incomplete. We will assume the problem is to approximate the change in $z$ from $P$ to $Q$.

Problem 13: Find all points on the surface $z = x^2 - 2xy - y^2 - 8x + 4y$ where the tangent plane is horizontal. This means finding where the partial derivatives with respect to $x$ and $y$ are both equal to zero.

2. Solution Steps

Problem 12:

First, we find the value of $z$ at $P(-2, -0.5)$.

$z_P = \arctan((-2)(-0.5)) = \arctan(1) = \frac{\pi}{4}$

Next, we find the value of $z$ at $Q(-2.03, -0.51)$.

$z_Q = \arctan((-2.03)(-0.51)) = \arctan(1.0353) \approx 0.8029$

Now, we can find the difference in $z$ values.

$\Delta z = z_Q - z_P \approx 0.8029 - \frac{\pi}{4} \approx 0.8029 - 0.7854 \approx 0.0175$

Alternatively, we can use partial derivatives to approximate the change:

$\frac{\partial z}{\partial x} = \frac{y}{1 + (xy)^2}$

$\frac{\partial z}{\partial y} = \frac{x}{1 + (xy)^2}$

At $P(-2, -0.5)$:

$\frac{\partial z}{\partial x} = \frac{-0.5}{1 + (-2 \cdot -0.5)^2} = \frac{-0.5}{1+1} = -\frac{1}{4}$

$\frac{\partial z}{\partial y} = \frac{-2}{1 + (-2 \cdot -0.5)^2} = \frac{-2}{1+1} = -1$

$\Delta x = -2.03 - (-2) = -0.03$

$\Delta y = -0.51 - (-0.5) = -0.01$

$\Delta z \approx \frac{\partial z}{\partial x} \Delta x + \frac{\partial z}{\partial y} \Delta y = -\frac{1}{4} (-0.03) + (-1) (-0.01) = 0.0075 + 0.01 = 0.0175$

Problem 13:

Given $z = x^2 - 2xy - y^2 - 8x + 4y$.

We need to find the points where the tangent plane is horizontal, i.e., where $\frac{\partial z}{\partial x} = 0$ and $\frac{\partial z}{\partial y} = 0$.

First, we find the partial derivatives:

$\frac{\partial z}{\partial x} = 2x - 2y - 8$

$\frac{\partial z}{\partial y} = -2x - 2y + 4$

Now, we set them to zero:

$2x - 2y - 8 = 0$

$-2x - 2y + 4 = 0$

Divide the first equation by 2:

$x - y - 4 = 0$ => $x = y + 4$

Divide the second equation by -2:

$x + y - 2 = 0$

Substitute $x = y + 4$ into the second equation:

$(y+4) + y - 2 = 0$

$2y + 2 = 0$

$2y = -2$

$y = -1$

Now find $x$:

$x = y + 4 = -1 + 4 = 3$

So, $x = 3$ and $y = -1$.

Now, we find $z$:

$z = (3)^2 - 2(3)(-1) - (-1)^2 - 8(3) + 4(-1) = 9 + 6 - 1 - 24 - 4 = 15 - 29 = -14$

Therefore, the point is $(3, -1, -14)$.

3. Final Answer

Problem 12: $\Delta z \approx 0.0175$

Problem 13: $(3, -1, -14)$