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) At \(t=1\), the rate of change with respect to distance is zero because \(|\mathbf{v}(1)| = 0\). (b) The rate with respect to time is \(-10\). Alternate method yields the same result.

Step 1: Compute the Temperature Function at Path

First, substitute the position vector components into the temperature function. This gives the temperature along the bee's path as a function of time: \[T(t) = \frac{10}{{(t \cos \pi t)^2 + (t \sin \pi t)^2 + t^2}}\]This simplifies to \[T(t) = \frac{10}{2t^2} = \frac{5}{t^2}\] when the path is substituted into the formula and simplified.

Step 2: Calculate the Total Distance Traveled

For the distance traveled, first find the velocity vector, which is the derivative of the position vector with respect to time:\[\mathbf{v}(t) = \frac{d}{dt}[t \cos \pi t\, \mathbf{i} + t \sin \pi t\, \mathbf{j} + t \mathbf{k}]\]Computing this, we get:\[\mathbf{v}(t) = (\cos \pi t - \pi t \sin \pi t)\, \mathbf{i} + (\sin \pi t + \pi t \cos \pi t)\, \mathbf{j} + \mathbf{k}\]The magnitude of this velocity vector \(|\mathbf{v}(t)|\) is needed to determine the distance:\[|\mathbf{v}(t)| = \sqrt{(\cos \pi t - \pi t \sin \pi t)^2 + (\sin \pi t + \pi t \cos \pi t)^2 + 1}\]For calculation at \(t=1\), solve the magnitude expression.

Step 3: Differentiate Temperature with Respect to Distance

To find the rate of change of temperature with respect to distance, calculate the derivative of the temperature with respect to \(t\) and divide it by the magnitude of the velocity vector determined in Step 2. Using the chain rule:\[\frac{dT}{ds} = \frac{\frac{dT}{dt}}{|\mathbf{v}(t)|}\]At \(t=1\), substitute the simplified expression and complete the differentiation.

Step 4: Differentiate Temperature with Respect to Time Directly

Differentiate the expression \(T(t) = \frac{5}{t^2}\) with respect to \(t\):\[\frac{dT}{dt} = \frac{d}{dt}\left(\frac{5}{t^2}\right) = -\frac{10}{t^3}\]Evaluate this expression at \(t=1\) to find the rate of change of temperature with respect to time.

Step 5: Alternative Method for Temperature Derivative with Respect to Time

The second way to find \(\frac{dT}{dt}\) involves using the chain rule on the path vector and temperature function. First, calculate the gradient of \(T\):\[abla T = -20\, \frac{x \mathbf{i} + y \mathbf{j} + z \mathbf{k}}{(x^2 + y^2 + z^2)^2}\]Then, compute \(abla T \cdot \mathbf{v}(1)\) using the values of the position vector at \(t=1\) to find another expression for \(\frac{dT}{dt}\) when the evaluations are made.

Key Concepts

Vector Calculus

Vector calculus is an essential branch of multivariable calculus that involves differentiation and integration of vector fields. It deals with vector functions and scalar functions that depend on multiple variables. In this context, the path of a bee is described by a vector function, \( \mathbf{r}(t) \), indicating its position in 3D space as time changes. This function is a combination of unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \), each multiplied by corresponding coefficients.

• \( \mathbf{r}(t) = t \cos \pi t \mathbf{i} + t \sin \pi t \mathbf{j} + t \mathbf{k} \)

When working with vector calculus, it's crucial to understand how to calculate derivatives of vector functions. These derivatives, like the velocity in this case, provide insights into how quantities change with respect to each parameter.The key part of vector calculus here is dealing with the derivative of \( \mathbf{r}(t) \). This derivative, \( \mathbf{v}(t) \), tells us the velocity vector:

• \( \mathbf{v}(t) = \frac{d}{dt}[t \cos \pi t \mathbf{i} + t \sin \pi t \mathbf{j} + t \mathbf{k}] \)

By breaking down the expression, we can compute how fast and in which direction the bee moves away from the origin.

Temperature Gradient

The temperature gradient is a vector that represents the direction and rate of fastest increase of temperature in a field. It's calculated using the gradient operator, \( abla \), which combines partial derivatives with respect to each spatial variable: \( x \), \( y \), and \( z \).For the given temperature equation:\[ T(x, y, z) = \frac{10}{x^2 + y^2 + z^2} \]Its gradient, \( abla T \), is determined to be:\[ abla T = -20 \frac{x \mathbf{i} + y \mathbf{j} + z \mathbf{k}}{(x^2 + y^2 + z^2)^2} \]Here, \( abla T \) gives us a vector pointing towards where the temperature decreases the fastest. This is fundamentally important when calculating how a temperature changes over a path, as the bee flies.In practical calculations, knowing \( abla T \) allows us to apply the chain rule to derive other rate expressions, such as the rate of change of temperature with time while accounting for the complexity of the path.

Rate of Change

Rate of change is a fundamental concept in calculus which describes how a quantity changes with respect to another quantity. In this exercise, we consider the rate of temperature change with respect to both time and distance. These rates provide insights into how quickly the temperature around the moving bee is dropping.To determine the rate of change of temperature with respect to time directly, we differentiate the simplified temperature function \( T(t) = \frac{5}{t^2} \):

• \( \frac{dT}{dt} = - \frac{10}{t^3} \)

Evaluating this at \( t=1 \), you can ascertain how rapidly the temperature decreases as time progresses.Furthermore, the temperature's rate of change concerning the distance traveled by the bee involves differentiating \( T \) with reference to \( t \), then normalizing it by the speed:

• \( \frac{dT}{ds} = \frac{\frac{dT}{dt}}{|\mathbf{v}(t)|} \)

Utilizing all these formulas enables us to tackle the problem both fundamentally and in respect to the distinct paths taken by the bee. This approach underscores the versatility and depth of multivariable calculus in real-world scenarios.