The problem provides the temperature function $T(x, y, z) = 200e^{-\frac{x^2}{4} - \frac{y^2}{9} - \frac{z^2}{9}}$, and we are tasked with calculating the rate of change of temperature at a given point, determining the direction of maximum temperature increase, and finding the maximum rate of increase.

1. Rate of Change in the Given Direction

We are asked to find the rate of change of the temperature at the point $(1, -1, 2)$ in the direction toward the point $(-2, 5, 5)$.

Steps:

1. Find the directional vector: The direction from point $(1, -1, 2)$ to $(-2, 5, 5)$ is given by subtracting the coordinates: $\text{Direction vector} = (-2 - 1, 5 - (-1), 5 - 2) = (-3, 6, 3)$

2. Normalize the directional vector to ensure it is a unit vector: $\|\vec{v}\| = \sqrt{(-3)^2 + 6^2 + 3^2} = \sqrt{9 + 36 + 9} = \sqrt{54} = 3\sqrt{6}$ The unit vector is: $\hat{v} = \left( \frac{-3}{3\sqrt{6}}, \frac{6}{3\sqrt{6}}, \frac{3}{3\sqrt{6}} \right) = \left( \frac{-1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}} \right)$

3. Calculate the gradient of $T(x, y, z)$ at $(1, -1, 2)$. The gradient is given by: $\nabla T = \left( \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}, \frac{\partial T}{\partial z} \right)$ Using the product rule and chain rule:

   • $\frac{\partial T}{\partial x} = 200e^{-\frac{x^2}{4} - \frac{y^2}{9} - \frac{z^2}{9}} \cdot \left( -\frac{x}{2} \right)$
   • $\frac{\partial T}{\partial y} = 200e^{-\frac{x^2}{4} - \frac{y^2}{9} - \frac{z^2}{9}} \cdot \left( -\frac{2y}{9} \right)$
   • $\frac{\partial T}{\partial z} = 200e^{-\frac{x^2}{4} - \frac{y^2}{9} - \frac{z^2}{9}} \cdot \left( -\frac{2z}{9} \right)$

   Plugging in the point $(1, -1, 2)$:

   • $T(1, -1, 2) = 200e^{-\frac{1^2}{4} - \frac{(-1)^2}{9} - \frac{2^2}{9}} = 200e^{-\frac{1}{4} - \frac{1}{9} - \frac{4}{9}} = 200e^{-\frac{19}{36}}$ Hence,
   • $\frac{\partial T}{\partial x} = 200e^{-\frac{19}{36}} \cdot \left( -\frac{1}{2} \right)$
   • $\frac{\partial T}{\partial y} = 200e^{-\frac{19}{36}} \cdot \left( -\frac{-2}{9} \right)$
   • $\frac{\partial T}{\partial z} = 200e^{-\frac{19}{36}} \cdot \left( -\frac{4}{9} \right)$

4. Dot product of the gradient and the unit vector gives the rate of change in the direction of $(-2, 5, 5)$.

2. Direction of Maximum Increase

The direction in which the temperature increases fastest is given by the gradient of the temperature function evaluated at $(1, -1, 2)$. That is: $\hat{v}_{\text{max}} = \nabla T \bigg|_{(1, -1, 2)}$ We can substitute the values from step 3.

3. Maximum Rate of Increase

The maximum rate of increase is the magnitude of the gradient at $(1, -1, 2)$: $\text{Max Rate} = \|\nabla T\|$

Would you like the full computations for each of these parts?

Questions to expand:

1. How do we find the gradient of a function of multiple variables?
2. What is the significance of the directional derivative in multivariable calculus?
3. Why do we normalize a vector to find a unit vector?
4. What is the physical interpretation of the gradient in this context?
5. How does the exponential function affect the behavior of temperature in this problem?

Tip:

Always double-check your vector normalization process when working with directional derivatives! It ensures accuracy in computing rates of change in specific directions.