Problem 29 - Calculus Early Transcendentals: Pearson New International Edition

Question

The temperature \(T\) in degrees Celsius at \((x, y, z)\) is given by \(T=10 /\left(x^{2}+y^{2}+z^{2}\right)\), where distances are in meters. A bee is flying away from the hot spot at the origin on a spiral path so that its position vector at time \(t\) seconds is \(\mathbf{r}(t)=\) \(t \cos \pi t \mathbf{i}+t \sin \pi t \mathbf{j}+t \mathbf{k}\). Determine the rate of change of \(T\) in each case. (a) With respect to distance traveled at \(t=1\). (b) With respect to time at \(t=1\). (Think of two ways to do this.)

Step-by-Step Solution

Answer

(a) The rate of change w.r.t. distance at t=1 is \(-k\). (b) The rate of change w.r.t. time using both methods gives \(-m\).

Step 1: Understand the problem

The problem requires us to find the rate of change of temperature with respect to two different variables: distance and time, given a temperature function and the bee's path described as a function of time.

Step 2: Parameterize distance traveled

The distance traveled can be represented by the speed. The bee's speed is the magnitude of the derivative of its position vector: \[ \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = \left(\cos(\pi t) - \pi t\sin(\pi t)\right)\mathbf{i} + \left(\sin(\pi t) + \pi t\cos(\pi t)\right)\mathbf{j} + \mathbf{k}. \] Next, compute the magnitude of \( \mathbf{v}(t) \).

Step 3: Compute speed at t=1

Calculate the magnitude of the velocity vector at \(t=1\): \[ \|\mathbf{v}(1)\| = \sqrt{(\cos(\pi) - \pi \sin(\pi))^2 + (\sin(\pi) + \pi \cos(\pi))^2 + (1)^2}. \] Knowing \(\cos(\pi) = -1\) and \(\sin(\pi) = 0\), this simplifies and the magnitude is \(\sqrt{2 + \pi^2}\).

Step 4: Rate of change with respect to distance at t=1

The rate of change of \(T\) with respect to distance is given by \(-abla T \cdot \frac{\mathbf{v}}{\|\mathbf{v}\|}\). First, calculate \(abla T\), the gradient of \(T\): \[ abla T = \left( \frac{-20x}{(x^2 + y^2 + z^2)^2}, \frac{-20y}{(x^2 + y^2 + z^2)^2}, \frac{-20z}{(x^2 + y^2 + z^2)^2} \right). \]Then, evaluate at \((1,0,1)\) from \(\mathbf{r}(1)\), and compute the rate of change for \(t=1\) using the dot product.

Step 5: Rate of change with respect to time (method 1)

To find the rate of change with respect to time directly, differentiate \(T(t)\) using the chain rule to get \(\frac{dT}{dt}\): \[ \frac{dT}{dt} = \frac{\partial T}{\partial x} \frac{dx}{dt} + \frac{\partial T}{\partial y} \frac{dy}{dt} + \frac{\partial T}{\partial z} \frac{dz}{dt}. \] Substitute \(t=1\) into the equation with \(\mathbf{r}(1)\) and its derivatives.

Step 6: Rate of change with respect to time (method 2)

Another approach is to use \( \frac{dT}{dt} = -abla T \cdot \mathbf{v} \). We already computed \(abla T\) and \(\mathbf{v}\); simply plug these into the formula. Evaluate at \(t=1\) with previously obtained gradient and velocity values.

Key Concepts

Gradient vector

The gradient vector is a fundamental concept in vector calculus that relates to how a function of several variables behaves in space. For a temperature function such as \( T(x, y, z) = \frac{10}{x^2 + y^2 + z^2} \), the gradient \( abla T \) tells us the direction and rate of the greatest increase of the temperature.

• The gradient is calculated by taking the partial derivatives with respect to each variable.
• For our temperature function, this results in: \[abla T = \left( \frac{-20x}{(x^2 + y^2 + z^2)^2}, \frac{-20y}{(x^2 + y^2 + z^2)^2}, \frac{-20z}{(x^2 + y^2 + z^2)^2} \right).\]
• The vector points in the direction where the temperature increases the fastest.

Understanding the gradient vector is crucial for determining how variables like temperature change as you move in space.

Directional derivative

The directional derivative measures how a function changes as you move in a specific direction. In the context of temperature, it helps understand the rate at which the temperature changes in that direction.

• The directional derivative of a function \( T \) in the direction of a vector \( \mathbf{v} \) is given by the dot product \( abla T \cdot \mathbf{v} \).
• For our example, this shows how temperature changes as the bee moves.
• Calculated as: \[\frac{dT}{ds} = abla T \cdot \frac{\mathbf{v}}{\|\mathbf{v}\|},\]during which \( \|\mathbf{v}\| \) is the magnitude of the velocity vector.

This concept allows you to predict changes in a physical property such as temperature as an object moves along a path.

Velocity vector

A velocity vector describes the rate of change of position with respect to time. In our scenario, the bee’s spiral path is defined by its position vector \( \mathbf{r}(t) \).

• The velocity vector \( \mathbf{v}(t) \) is obtained by differentiating the position vector with respect to time:
\[\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = \left( \cos(\pi t) - \pi t \sin(\pi t) \right)\mathbf{i} + \left( \sin(\pi t) + \pi t \cos(\pi t) \right)\mathbf{j} + \mathbf{k}.\]

This vector shows the direction and speed of the bee at each point in time.

The magnitude of this vector at a specific time \( t \) gives the speed.

Understanding the velocity vector is crucial for computing how changes in time relate to the bee's movement along its path.

Chain rule differentiation

Chain rule differentiation is a method used to find the derivative of a composite function. It is particularly useful when one variable depends on others, which depend on another variable.

• In the given problem, the temperature function \( T(x, y, z) \) changes as the bee moves based on time \( t \).
• The chain rule helps relate these changes via:\[\frac{dT}{dt} = \frac{\partial T}{\partial x} \frac{dx}{dt} + \frac{\partial T}{\partial y} \frac{dy}{dt} + \frac{\partial T}{\partial z} \frac{dz}{dt}.\]
• This equation helps find the rate of change of temperature with respect to time directly.

The chain rule is fundamental for problems involving interdependent variables and provides a systematic way to compute derivatives.