Problem 49 - Thomas Calculus

Question

Temperature change It took 14 sec for a mercury thermometer to rise from \(-19^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) when it was taken from a freezer and placed in boiling water. Show that somewhere along the way the mercury was rising at the rate of \(8.5^{\circ} \mathrm{C} / \mathrm{sec}.\)

Step-by-Step Solution

Answer

Somewhere, the rate is exactly \(8.5^{\circ} \mathrm{C}/\mathrm{sec}\) by the Mean Value Theorem.

Step 1: Understand the Average Rate of Change

The thermometer rises from \(-19^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) in \(14\) seconds. First, calculate the average rate of change. This is done by dividing the total change in temperature by the total change in time:\[ \text{Average Rate} = \frac{100 - (-19)}{14} = \frac{119}{14} \approx 8.5^{\circ} \mathrm{C}/\mathrm{sec} \] This average rate suggests that there is a time when the instantaneous rate of change is exactly this value.

Step 2: Explain the Mean Value Theorem

The Mean Value Theorem for derivatives states that if a function \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then there exists some \(c\) in \((a,b)\) such that:\[ f'(c) = \frac{f(b) - f(a)}{b-a} \]This means there is at least one point \(c\) where the instantaneous rate of change (the derivative) equals the average rate of change over the interval \([a, b]\).

Step 3: Apply the Mean Value Theorem

Apply the Mean Value Theorem to the temperature \(T\) of the thermometer over time. Let \( T(t) \) be the temperature at time \( t \). We know that \(T(0) = -19\) and \(T(14) = 100\). Since \(T(t)\) is differentiable and continuous, there must be a time \(c\) in \((0,14)\) where:\[ T'(c) = \frac{T(14) - T(0)}{14 - 0} = \frac{119}{14} = 8.5^{\circ} \mathrm{C}/\mathrm{sec} \] Thus, there is a moment when the mercury rate rises at exactly \(8.5^{\circ} \mathrm{C}/\mathrm{sec}\).

Step 4: Conclude with the Existence of the Rate

Having applied the Mean Value Theorem correctly, we've shown that the rate of change in temperature reaches exactly \(8.5^{\circ} \mathrm{C}/\mathrm{sec}\) at some point during the interval. This proves the statement of the exercise.

Key Concepts

Understanding the Average Rate of Change

The average rate of change measures how a quantity, like temperature, changes over a specified period. It provides an overview of how fast or slow the change occurs between two points.

To calculate the average rate of change, you take the difference between the final value and the initial value, and divide it by the total time taken for this change:

• In our example, the thermometer's temperature rises from \(-19^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) over \(14\) seconds.
• Calculate by using the formula:\[ \text{Average Rate} = \frac{100 - (-19)}{14} = \frac{119}{14} \approx 8.5^{\circ} \mathrm{C}/\mathrm{sec} \]

This value indicates that on average, the temperature changes at approximately \(8.5^{\circ} \mathrm{C}/\mathrm{sec}\). It doesn't mean the rate is constantly \(8.5^{\circ} \mathrm{C}/\mathrm{sec}\), but it gives a sense of the overall change speed.

Understanding the Instantaneous Rate of Change

The instantaneous rate of change is about finding the rate at a specific instant, rather than an overall average over time.

Think of it like the speedometer showing the exact speed of a car at a moment, instead of its average speed across a trip.

• In mathematical terms, this exact change at any instant is found using derivatives, denoted as \(f'(c)\).
• The Mean Value Theorem tells us that if a function is continuous and differentiable, there must be a point where the instantaneous rate equals the average rate.

In our temperature scenario, despite fluctuations, there must be some instant where the rate of \(8.5^{\circ} \mathrm{C}/\mathrm{sec}\) is exactly achieved. That's the magic of the Mean Value Theorem. It assures us even with average data, exact points can be identified.

Calculating Temperature Change

Temperature change calculations are straightforward if you follow a structured approach:

The task here involved finding evidence that the exact rate of temperature change would be equal to the average rate at some point.

• We start by knowing initial and final temperatures and time: initial \(-19^{\circ} \mathrm{C}\), final \(100^{\circ} \mathrm{C}\), and total time \(14\) seconds.
• The average rate tells us \(8.5^{\circ} \mathrm{C}/\mathrm{sec}\).
• Using the Mean Value Theorem, we assert there must be a moment where this change is precisely occurring.

This calculated rate is crucial as it simplifies understanding how temperature varies over time, with both average and specific rates of change considered. Knowing these concepts ensures that students can not only solve this kind of problems but also appreciate the broader application of rates in various physical processes.