Problem 19 - Calculus Early Transcendentals: Pearson New International Edition

Question

The temperature at \((x, y, z)\) of a solid sphere centered at the origin is given by $$ T(x, y, z)=\frac{200}{5+x^{2}+y^{2}+z^{2}} $$ (a) By inspection, decide where the solid sphere is hottest. (b) Find a vector pointing in the direction of greatest increase of temperature at \((1,-1,1)\). (c) Does the vector of part (b) point toward the origin?

Step-by-Step Solution

Answer

(a) Hottest at (0, 0, 0); (b) Gradient: \((-\frac{25}{4}, \frac{25}{4}, -\frac{25}{4})\); (c) Yes, it points towards the origin.

Step 1: Identify the Hottest Point

The temperature function \( T(x, y, z) = \frac{200}{5+x^{2}+y^{2}+z^{2}} \) shows that the temperature decreases as the denominator increases. Therefore, the hottest point is where the denominator is minimized. Since the smallest value the sum \( x^2 + y^2 + z^2 \) can take is 0, the temperature is highest at \( (0, 0, 0) \).

Step 2: Compute the Gradient

The direction of greatest increase of a scalar field is given by its gradient. The temperature gradient is \( abla T(x, y, z) \), calculated as follows:\[abla T = \left(\frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}, \frac{\partial T}{\partial z}\right)\]Computing these partial derivatives:\[\frac{\partial T}{\partial x} = -\frac{400x}{(5+x^2+y^2+z^2)^2}\]\[\frac{\partial T}{\partial y} = -\frac{400y}{(5+x^2+y^2+z^2)^2}\]\[\frac{\partial T}{\partial z} = -\frac{400z}{(5+x^2+y^2+z^2)^2}\]

Step 3: Evaluate the Gradient at (1, -1, 1)

Substitute \((1, -1, 1)\) into the gradient calculated in Step 2:\[abla T(1, -1, 1) = \left(-\frac{400 \cdot 1}{(5+1^2+(-1)^2+1^2)^2}, -\frac{400 \cdot (-1)}{(5+1^2+(-1)^2+1^2)^2}, -\frac{400 \cdot 1}{(5+1^2+(-1)^2+1^2)^2}\right)\]Calculate the denominator:\[5 + 1^2 + (-1)^2 + 1^2 = 8\]Finally, compute:\[abla T(1, -1, 1) = \left(-\frac{400}{64}, \frac{400}{64}, -\frac{400}{64}\right) = \left(-\frac{25}{4}, \frac{25}{4}, -\frac{25}{4}\right)\]

Step 4: Check Orientation of the Gradient Vector

The gradient vector at \((1, -1, 1)\) is \(\left(-\frac{25}{4}, \frac{25}{4}, -\frac{25}{4}\right)\). This vector points in a direction opposite to each coordinate of \((1, -1, 1)\), effectively pointing towards the negative side of \(x\) and \(z\) while positive towards \(y\). Since the origin is \((0, 0, 0)\), this vector indeed points roughly towards the origin, as it moves the point in the direction towards \((0, 0, 0)\).

Key Concepts

Temperature Gradient

In the context of temperature distribution within a solid sphere, the temperature gradient is a powerful concept. It provides us insight into how temperature changes at any given point within the sphere. For the temperature equation given by \( T(x, y, z)=\frac{200}{5+x^{2}+y^{2}+z^{2}} \), the temperature gradient, denoted as \( abla T \), is a vector that indicates how temperature "slopes" in space.

The gradient vector is computed using partial derivatives of \(T\) with respect to each spatial variable \( x, y, \) and \( z \). At each point \((x, y, z)\), the components of the temperature gradient are:

• \( \frac{\partial T}{\partial x} = -\frac{400x}{(5+x^2+y^2+z^2)^2} \)
• \( \frac{\partial T}{\partial y} = -\frac{400y}{(5+x^2+y^2+z^2)^2} \)
• \( \frac{\partial T}{\partial z} = -\frac{400z}{(5+x^2+y^2+z^2)^2} \)

These components tell us how the temperature decreases in relation to each coordinate direction as you move away from the origin.
By understanding the gradient, one gets a clear picture of where the temperature increases or decreases the most, akin to finding the "steepest incline" on a hill.

Direction of Greatest Increase

The direction in which the temperature increases most rapidly at any given point is aligned with the temperature gradient vector. For the given function \( T(x, y, z) \), the gradient vector points in the direction where temperature rises fastest from a specific location.

To find the direction of greatest increase of temperature at \((1, -1, 1)\), we calculate the gradient at this point:

• Substitute \((1, -1, 1)\) into the gradient formulas, resulting in \( abla T(1, -1, 1) = \left(-\frac{25}{4}, \frac{25}{4}, -\frac{25}{4}\right) \).

This vector indicates the direction of steepest ascent in temperature. It shows that small movements in that direction will increase the temperature at the fastest rate. Understanding this vector's orientation gives insights into how temperature behaves locally.

Partial Derivatives

Partial derivatives are a key mathematical tool used to understand how a function changes when we vary one variable while keeping the others constant. They are especially useful in multivariable calculus, such as when examining the temperature function \( T(x, y, z) \) of a three-dimensional sphere.

The partial derivatives \( \frac{\partial T}{\partial x} \), \( \frac{\partial T}{\partial y} \), and \( \frac{\partial T}{\partial z} \) are components of the gradient vector and demonstrate how the temperature at a point changes along each coordinate axis individually. These derivatives help us dissect the influence of each variable on the temperature:

• \( \frac{\partial T}{\partial x} \) shows how the temperature changes as we slightly alter \( x \), with \( y \) and \( z \) held fixed.
• \( \frac{\partial T}{\partial y} \) reveals the change in temperature due to tiny shifts in \( y \), with \( x \) and \( z \) held constant.
• \( \frac{\partial T}{\partial z} \) tells us about the temperature change with small variations in \( z \), while \( x \) and \( y \) remain unchanged.

Each partial derivative captures a slice of how the overall temperature pattern shifts, crucial for understanding the temperature dynamics of the sphere.