To determine the direction in which the temperature increases the fastest at the point $(1,1,-2)$ and find the maximum rate of increase, we need to compute the gradient of the temperature function $T(x, y, z)$ at that point.

Given: $T(x, y, z) = \frac{80}{1 + x^2 + 2y^2 + 3z^2}$

Step 1: Calculate the Gradient of $T(x, y, z)$

The gradient of $T$, denoted by $\nabla T$, gives the direction of the fastest increase of the function, and its magnitude gives the maximum rate of increase.

1. Compute partial derivatives with respect to $x$, $y$, and $z$.

   Let's rewrite $T$ as: $T(x, y, z) = 80 \cdot \left(1 + x^2 + 2y^2 + 3z^2\right)^{-1}$ Using the chain rule, we get: $\frac{\partial T}{\partial x} = -80 \cdot \left(1 + x^2 + 2y^2 + 3z^2\right)^{-2} \cdot 2x$ $\frac{\partial T}{\partial y} = -80 \cdot \left(1 + x^2 + 2y^2 + 3z^2\right)^{-2} \cdot 4y$ $\frac{\partial T}{\partial z} = -80 \cdot \left(1 + x^2 + 2y^2 + 3z^2\right)^{-2} \cdot 6z$

   So, the gradient is: $\nabla T = \left( -\frac{160x}{(1 + x^2 + 2y^2 + 3z^2)^2}, -\frac{320y}{(1 + x^2 + 2y^2 + 3z^2)^2}, -\frac{480z}{(1 + x^2 + 2y^2 + 3z^2)^2} \right)$

2. Evaluate $\nabla T$ at the point $(1, 1, -2)$.

   Substitute $x = 1$, $y = 1$, and $z = -2$ into $\nabla T$: $1 + (1)^2 + 2(1)^2 + 3(-2)^2 = 1 + 1 + 2 + 12 = 16$

   Thus, $\nabla T(1, 1, -2) = \left( -\frac{160 \cdot 1}{16^2}, -\frac{320 \cdot 1}{16^2}, -\frac{480 \cdot (-2)}{16^2} \right)$ Simplifying, $16^2 = 256$, so: $\nabla T(1, 1, -2) = \left( -\frac{160}{256}, -\frac{320}{256}, \frac{960}{256} \right) = \left( -\frac{5}{8}, -\frac{5}{4}, \frac{15}{4} \right)$

Step 2: Determine the Direction and Magnitude of the Maximum Rate of Increase

1. Direction of the fastest temperature increase: This is given by the direction of the gradient vector $\nabla T(1, 1, -2) = \left( -\frac{5}{8}, -\frac{5}{4}, \frac{15}{4} \right)$.

2. Maximum rate of increase: This is the magnitude of $\nabla T(1, 1, -2)$: $\left\|\nabla T(1, 1, -2)\right\| = \sqrt{\left(-\frac{5}{8}\right)^2 + \left(-\frac{5}{4}\right)^2 + \left(\frac{15}{4}\right)^2}$ Calculating each term: $= \sqrt{\frac{25}{64} + \frac{25}{16} + \frac{225}{16}}$ Converting $\frac{25}{16}$ to $\frac{100}{64}$ and $\frac{225}{16}$ to $\frac{900}{64}$: $= \sqrt{\frac{25 + 100 + 900}{64}} = \sqrt{\frac{1025}{64}} = \frac{\sqrt{1025}}{8}$

Thus:

• Direction of fastest increase: $\left( -\frac{5}{8}, -\frac{5}{4}, \frac{15}{4} \right)$.
• Maximum rate of increase: $\frac{\sqrt{1025}}{8} \approx 4.008$ degrees Celsius per meter.

Would you like more details or explanations for specific steps?

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Further Questions

1. How would the gradient change if the function $T(x, y, z)$ had additional parameters?
2. How does the rate of change vary if we move in a direction perpendicular to the gradient?
3. What does the temperature function tell us about the nature of temperature variation in space?
4. How would the result differ if the point of evaluation were changed?
5. Can we determine the rate of temperature change along a specific direction (not the gradient)?

Tip

The gradient of a scalar field points in the direction of the maximum rate of increase of the function at any given point.