https://file.aiquickdraw.com/mathbot/file/17274031175245d84590e.jpg

The problem describes the temperature \( T(x, y) \) on a flat metal plate, given by the formula:

\[
T(x, y) = \frac{35}{2 + x^2 + y^2}
\]

You are asked to find the rate of change of temperature at the point \( (2, 1) \) in both the x-direction and the y-direction. These rates correspond to the partial derivatives of \( T \) with respect to \( x \) and \( y \).

### 1. Partial derivative with respect to \( x \) (for the x-direction):

To compute \( \frac{\partial T}{\partial x} \), we'll apply the chain rule. Let's differentiate the expression:

\[
T(x, y) = \frac{35}{2 + x^2 + y^2}
\]

The denominator is \( D = 2 + x^2 + y^2 \). Using the quotient rule, we get:

\[
\frac{\partial T}{\partial x} = \frac{-35 \cdot 2x}{(2 + x^2 + y^2)^2}
\]

### 2. Partial derivative with respect to \( y \) (for the y-direction):

Similarly, for the derivative with respect to \( y \):

\[
\frac{\partial T}{\partial y} = \frac{-35 \cdot 2y}{(2 + x^2 + y^2)^2}
\]

### Evaluate at the point \( (x, y) = (2, 1) \):

Now, substitute \( x = 2 \) and \( y = 1 \) into both partial derivatives.

#### For \( \frac{\partial T}{\partial x} \):

\[
\frac{\partial T}{\partial x} = \frac{-35 \cdot 2(2)}{(2 + 2^2 + 1^2)^2} = \frac{-140}{(2 + 4 + 1)^2} = \frac{-140}{7^2} = \frac{-140}{49} = -\frac{20}{7} \, \text{°C/m}
\]

#### For \( \frac{\partial T}{\partial y} \):

\[
\frac{\partial T}{\partial y} = \frac{-35 \cdot 2(1)}{(2 + 2^2 + 1^2)^2} = \frac{-70}{(2 + 4 + 1)^2} = \frac{-70}{7^2} = \frac{-70}{49} = -\frac{10}{7} \, \text{°C/m}
\]

### Final answers:

- (a) The rate of change in the x-direction: \( -\frac{20}{7} \, \text{°C/m} \)
- (b) The rate of change in the y-direction: \( -\frac{10}{7} \, \text{°C/m} \)

Would you like further details or have any other questions?

---

Here are 5 related questions to deepen understanding:

1. What is the physical interpretation of a negative rate of change in this context?
2. How would you find the directional derivative at \( (2, 1) \) in a specific direction?
3. What happens to the temperature rate of change as you move further from the origin?
4. Can you verify the solution using a graph of the temperature function?
5. How would you compute the second-order partial derivatives for this function?

**Tip:** When dealing with multivariable functions, always double-check that you're applying the chain and quotient rules correctly when differentiating!

The temperature at a point (x, y) on a flat metal plate is given by T(x, y) = 35 / (2 + x^2 + y^2). Find the rate of change of temperature at the point (2, 1) in the x-direction and the y-direction.

Learn how to compute the rate of change of temperature at point (2, 1) on a flat metal plate using partial derivatives. This problem involves multivariable calculus concepts, including the quotient rule and chain rule. Suitable for undergraduate calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation