To solve these questions, we first need to calculate the gradient of $T(x, y, z)$, which gives us both the direction and rate of the steepest increase in temperature.

Given:

$T(x, y, z) = 1000 e^{-x^2 - 2y^2 - z^2}$ where $T$ is measured in °C and $x, y, z$ are in meters.

Step 1: Calculate the Gradient $\nabla T$

The gradient of $T$ is: $\nabla T = \left( \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}, \frac{\partial T}{\partial z} \right)$

1. Partial derivative with respect to $x$: $\frac{\partial T}{\partial x} = 1000 \cdot e^{-x^2 - 2y^2 - z^2} \cdot (-2x) = -2000x \cdot e^{-x^2 - 2y^2 - z^2}$

2. Partial derivative with respect to $y$: $\frac{\partial T}{\partial y} = 1000 \cdot e^{-x^2 - 2y^2 - z^2} \cdot (-4y) = -4000y \cdot e^{-x^2 - 2y^2 - z^2}$

3. Partial derivative with respect to $z$: $\frac{\partial T}{\partial z} = 1000 \cdot e^{-x^2 - 2y^2 - z^2} \cdot (-2z) = -2000z \cdot e^{-x^2 - 2y^2 - z^2}$

Thus, $\nabla T = \left( -2000x \cdot e^{-x^2 - 2y^2 - z^2}, -4000y \cdot e^{-x^2 - 2y^2 - z^2}, -2000z \cdot e^{-x^2 - 2y^2 - z^2} \right)$

Question 1: Rate of Change of Temperature at $P(2, -1, 2)$ Toward $Q(3, -3, 3)$

1. Calculate $\nabla T$ at $P(2, -1, 2)$: $T_x(2, -1, 2) = -2000 \cdot 2 \cdot e^{-2^2 - 2(-1)^2 - 2^2} = -4000 \cdot e^{-12}$ $T_y(2, -1, 2) = -4000 \cdot (-1) \cdot e^{-12} = 4000 \cdot e^{-12}$ $T_z(2, -1, 2) = -2000 \cdot 2 \cdot e^{-12} = -4000 \cdot e^{-12}$

   Therefore, $\nabla T(2, -1, 2) = (-4000 e^{-12}, 4000 e^{-12}, -4000 e^{-12})$

2. Find the direction vector $\vec{PQ}$: $\vec{PQ} = (3 - 2, -3 + 1, 3 - 2) = (1, -2, 1)$

3. Unit vector in the direction of $\vec{PQ}$: $|\vec{PQ}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}$ $\hat{u}_{PQ} = \left( \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}} \right)$

4. Directional derivative $D_{\vec{PQ}} T$: $D_{\vec{PQ}} T = \nabla T \cdot \hat{u}_{PQ} = (-4000 e^{-12}, 4000 e^{-12}, -4000 e^{-12}) \cdot \left( \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}} \right)$ $= \frac{-4000 e^{-12}}{\sqrt{6}} + \frac{-8000 e^{-12}}{\sqrt{6}} - \frac{4000 e^{-12}}{\sqrt{6}} = -\frac{16000 e^{-12}}{\sqrt{6}}$

Question 2: Direction of Maximum Temperature Increase at $P$

The direction of the maximum temperature increase at $P$ is given by the direction of $\nabla T$ at $P$: $(-4000 e^{-12}, 4000 e^{-12}, -4000 e^{-12})$

Question 3: Maximum Rate of Increase at $P$

The maximum rate of increase is the magnitude of $\nabla T$ at $P$: $|\nabla T| = \sqrt{(-4000 e^{-12})^2 + (4000 e^{-12})^2 + (-4000 e^{-12})^2} = 4000 e^{-12} \sqrt{3}$

Would you like further details on each step?

Relative Questions:

1. How do you find the gradient vector of a multivariable function?
2. What is the significance of the directional derivative in physical applications?
3. How does the unit vector influence the directional derivative calculation?
4. Why is the maximum rate of increase always in the gradient direction?
5. How does one interpret the magnitude of the gradient in real-world terms?

Tip: The gradient vector points in the direction of the steepest ascent for a scalar field, providing both the direction and rate of change in that direction.