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?

Short answer

(a) Hottest at (0,0,0). (b) Gradient: (-6.25, 6.25, -6.25). (c) Yes, it points toward the origin.

Step-by-step solution

Find the Hottest Point

The temperature function is \( T(x, y, z)=\frac{200}{5+x^{2}+y^{2}+z^{2}} \). Since the denominator \( 5 + x^2 + y^2 + z^2 \) is smallest when \( x=y=z=0 \), the temperature is highest at the origin. Thus, the sphere is hottest at the origin, \( (0,0,0) \).

Compute the Gradient of T

The gradient \( abla T(x, y, z) \) points in the direction of greatest increase of \( T \). Calculate the partial derivatives: \[ \frac{\partial T}{\partial x} = -\frac{400x}{(5 + x^2 + y^2 + z^2)^2}, \\frac{\partial T}{\partial y} = -\frac{400y}{(5 + x^2 + y^2 + z^2)^2}, \\frac{\partial T}{\partial z} = -\frac{400z}{(5 + x^2 + y^2 + z^2)^2}. \] Thus, the gradient is \( abla T(x, y, z) = \left(-\frac{400x}{(5 + x^2 + y^2 + z^2)^2}, -\frac{400y}{(5 + x^2 + y^2 + z^2)^2}, -\frac{400z}{(5 + x^2 + y^2 + z^2)^2}\right). \)

Evaluate the Gradient at (1, -1, 1)

Substitute \( x = 1 \), \( y = -1 \), and \( z = 1 \) into the gradient: \[ abla T(1, -1, 1) = \left(-\frac{400 \cdot 1}{(5 + 1^2 + (-1)^2 + 1^2)^2}, -\frac{400 \cdot (-1)}{(5 + 1^2 + (-1)^2 + 1^2)^2}, -\frac{400 \cdot 1}{(5 + 1^2 + (-1)^2 + 1^2)^2}\right). \]The common factor in the denominator is \((5+3)^2=64\), so the evaluated gradient is \( \left(-\frac{400}{64}, \frac{400}{64}, -\frac{400}{64}\right) = \left(-6.25, 6.25, -6.25\right). \)

Determine if the Gradient Vector Points Toward the Origin

The evaluated vector \( (-6.25, 6.25, -6.25) \) can be expressed as a scalar multiple of the vector \( (1, -1, 1) \), indicating it points in the direction towards the origin. This vector is in the opposite direction of \( (1, -1, 1) \), which is pointing towards \( (0,0,0) \). Therefore, the gradient vector does point towards the origin.

Key concepts

Gradient Vector

In multivariable calculus, the gradient vector is a crucial tool for understanding how a function changes in space. It is represented by \( abla T \), where \( T \) is the function.

The gradient vector in vector calculus is a collection of all the partial derivatives of a function. It tells us the direction in which the function increases most steeply.

For a function \( T(x, y, z) \), the gradient \( abla T(x, y, z) \) is a vector of partial derivatives:

• \( \frac{\partial T}{\partial x} \)
• \( \frac{\partial T}{\partial y} \)
• \( \frac{\partial T}{\partial z} \)

Let's look at an example with the temperature function, where the gradient vector shows us how temperature changes in space.

Vector Calculus

Vector calculus is the branch of mathematics that deals with vector fields and differential operators like the gradient. In the context of three-dimensional space, it is used to analyze fields like velocity, electromagnetic fields, and temperature distributions.

It allows us to determine not just how much a quantity like temperature changes, but in what direction it changes the most. We can use gradients, div, and curl to analyze these fields.

• The gradient finds the direction of greatest change.
• Other tools like divergence and curl help understand different aspects of vector fields.

Utilizing these ideas, we can solve practical problems like determining how heat moves through an object.

Temperature Function

A temperature function models how temperature varies in space. In our example, it is represented as:\[ T(x, y, z) = \frac{200}{5 + x^2 + y^2 + z^2} \]

This function shows that the temperature depends on the coordinates \((x, y, z)\) within a solid sphere.The function value will be highest when the denominator \(5 + x^2 + y^2 + z^2\) is minimum, that is, at the origin \((0, 0, 0)\). Hence, it indicates the sphere is hottest at the center.
This kind of function is common in physics for modeling temperature distribution. Its simplicity allows for easy derivative calculation and insight into the hottest and coolest points of an object.

Partial Derivatives

Partial derivatives are fundamental to understanding how a function changes concerning one variable while keeping others constant. In a multivariable function \( T(x, y, z) \), there are three partial derivatives:

• \( \frac{\partial T}{\partial x} \) - how \( T \) changes with a change in \( x \)
• \( \frac{\partial T}{\partial y} \) - how \( T \) changes with a change in \( y \)
• \( \frac{\partial T}{\partial z} \) - how \( T \) changes with a change in \( z \)

Each partial derivative reveals how the function partially increases or decreases.
In the temperature example, these partial derivatives formed the components of the gradient vector. They help find the direction in which the temperature increases most rapidly.