We are given the temperature function:

$T(x, y) = 40(3 + x^2 + y^2)$

We are asked to find the rate of change of the temperature in both the $x$-direction and the $y$-direction at the point $(3, 2)$. This means we need to compute the partial derivatives of the temperature function with respect to $x$ and $y$, then evaluate them at the given point.

(a) Rate of change in the $x$-direction

The rate of change of temperature with respect to $x$ is the partial derivative of $T(x, y)$ with respect to $x$, denoted as $\frac{\partial T}{\partial x}$.

Let's calculate it:

$T(x, y) = 40(3 + x^2 + y^2)$

Taking the partial derivative of $T(x, y)$ with respect to $x$:

$\frac{\partial T}{\partial x} = 40 \cdot \frac{\partial}{\partial x} \left( 3 + x^2 + y^2 \right)$

Since $y^2$ is independent of $x$, we only differentiate $x^2$:

$\frac{\partial T}{\partial x} = 40 \cdot 2x = 80x$

Now, evaluate this at the point $(3, 2)$:

$\frac{\partial T}{\partial x} \Bigg|_{(3, 2)} = 80 \times 3 = 240 \, \text{°C/m}$

(b) Rate of change in the $y$-direction

The rate of change of temperature with respect to $y$ is the partial derivative of $T(x, y)$ with respect to $y$, denoted as $\frac{\partial T}{\partial y}$.

Let's calculate it:

$\frac{\partial T}{\partial y} = 40 \cdot \frac{\partial}{\partial y} \left( 3 + x^2 + y^2 \right)$

Since $x^2$ is independent of $y$, we only differentiate $y^2$:

$\frac{\partial T}{\partial y} = 40 \cdot 2y = 80y$

Now, evaluate this at the point $(3, 2)$:

$\frac{\partial T}{\partial y} \Bigg|_{(3, 2)} = 80 \times 2 = 160 \, \text{°C/m}$

Final answers:

(a) The rate of change in the $x$-direction is 240°C/m.

(b) The rate of change in the $y$-direction is 160°C/m.

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Would you like more details or have any questions about this problem?

Here are five related questions to explore:

1. How would the rate of change of temperature differ at a different point, say (1, 1)?
2. How would you find the total rate of change of temperature in the direction of a vector?
3. What happens to the temperature at the origin (0, 0)?
4. How do second-order partial derivatives relate to the curvature of the temperature surface?
5. How would you calculate the rate of temperature change along a path defined by a function of $x$ and $y$?

Tip: To evaluate the rate of change of a function along a given direction, you can use the directional derivative, which requires combining the partial derivatives and the direction vector.