Problem 80 - Calculus Early Transcendentals

Question

The temperature at a point \((x, y)\) on a flat metal plate is given by \(T(x, y)=60 /\left(1+x^{2}+y^{2}\right),\) where \(T\) is measured in \(^{\circ} \mathrm{C}\) and \(x, y\) in meters. Find the rate of change of temperature with respect to distance at the point \((2,1)\) in (a) the \(x\) -direction and (b) the \(y\) -direction.

Step-by-Step Solution

Answer

At (2,1), the rate of change in the x-direction is \(-\frac{20}{3}\), and in the y-direction is \(-\frac{10}{3}\).

Step 1: Understand the Temperature Function

The temperature function is given by \( T(x, y) = \frac{60}{1 + x^2 + y^2} \). This means the temperature depends on the distance from the origin, with temperature decreasing as \( x \) and \( y \) increase.

Step 2: Determine Partial Derivatives

To find the rate of change in specific directions, we use partial derivatives. The partial derivative with respect to \( x \) (\( \frac{\partial T}{\partial x} \)) and the partial derivative with respect to \( y \) (\( \frac{\partial T}{\partial y} \)) represent the rate of change of temperature in the \( x \) and \( y \) directions, respectively.

Step 3: Calculate the Partial Derivative with Respect to x

Using the quotient rule for differentiation, we find \( \frac{\partial T}{\partial x} = \frac{-120x}{(1 + x^2 + y^2)^2} \). This derivative tells us how temperature changes as we move in the \( x \)-direction.

Step 4: Evaluate \( \frac{\partial T}{\partial x} \) at (2, 1)

Substitute \( x = 2 \) and \( y = 1 \) into \( \frac{\partial T}{\partial x} \). This yields \( \frac{-120 \times 2}{(1 + 2^2 + 1^2)^2} = \frac{-240}{36} = -\frac{20}{3} \).

Step 5: Calculate the Partial Derivative with Respect to y

Similarly, differentiate \( T(x, y) \) with respect to \( y \) to get \( \frac{\partial T}{\partial y} = \frac{-120y}{(1 + x^2 + y^2)^2} \). This gives us the temperature change as we move in the \( y \)-direction.

Step 6: Evaluate \( \frac{\partial T}{\partial y} \) at (2, 1)

Substitute \( x = 2 \) and \( y = 1 \) into \( \frac{\partial T}{\partial y} \). This yields \( \frac{-120 \times 1}{(1 + 2^2 + 1^2)^2} = \frac{-120}{36} = -\frac{10}{3} \).

Key Concepts

Rate of Change

The rate of change refers to how a quantity varies with respect to another. In the context of our problem, it's about how the temperature at a specific point on the plate changes as we move in certain directions. This is particularly important in fields like thermodynamics and physics where understanding how variables change relative to others provides significant insights.

When we talk about rate of change in this exercise, we're analyzing how the temperature varies as we move along the x- or y- axis. To measure this change, we use partial derivatives. This is because they allow us to see how the temperature varies with respect to one variable while keeping the other constant. It’s a localized measurement, giving us a snapshot of the temperature's behavior in a defined direction at a point.

Quotient Rule

The Quotient Rule is an essential tool for differentiating functions of the form \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of a variable. In this exercise, we use it to determine the partial derivatives of the temperature function with respect to x and y.

The general formula for the Quotient Rule is given by:

• \[ \left( \frac{u}{v} \right)' = \frac{vu' - uv'}{v^2} \]

Applying this rule ensures we correctly account for both the numerator's and the denominator's rates of change. This rule allows us to systematically and accurately compute partial derivatives for the temperature function, critical for understanding changes along the x and y directions on the metal plate.

Temperature Function

The temperature function provided in the exercise is \( T(x, y) = \frac{60}{1 + x^2 + y^2} \). It’s the mathematical representation of how temperature varies depending on the point's location on the plate. The form of this function indicates that the temperature decreases as the distance from the origin (0,0) increases, which is characteristic of many physical phenomena.

This setup is often seen in problems dealing with heat diffusion and thermal dynamics, where spatial position dramatically influences the observed temperature. Functions like this help in modeling the behavior of temperatures in response to position, which is useful for solving real-world engineering problems where control of heat is necessary.

Directional Derivatives

Directional derivatives extend the concept of partial derivatives by considering the rate of change of a function not just along the axes but in any given direction. In simple terms, they give us the slope of the function in the direction of our choosing. This lays the groundwork for understanding gradients and complex changes in multidimensional spaces.

In our exercise, even though we primarily focused on changes in the x and y directions separately, directional derivatives would allow us to analyze changes in a combination of these directions. It's essentially taking the partial derivative to a more general case, allowing exploration in any direction in the xy-plane.

Understanding directional derivatives is key to applying calculus in physics and engineering, where motion or change isn’t confined to simple directional components but instead follows more arbitrary paths.