Problem 47 - Calculus Early Transcendentals: Pearson New International Edition

Question

A clock has hour and minute hands of lengths \(h\) and \(m\), respectively, with \(h \leq m .\) We wish to study this clock at times between \(12: 00\) and \(12: 30 .\) Let \(\theta, \phi\), and \(L\) be as in Figure 33 and note that \(\theta\) increases at a constant rate. By the Law of Cosines, \(L=L(\theta)=\left(h^{2}+m^{2}-2 h m \cos \theta\right)^{1 / 2}\), and so $$ L^{\prime}(\theta)=h m\left(h^{2}+m^{2}-2 h m \cos \theta\right)^{-1 / 2} \sin \theta $$ (a) For \(h=3\) and \(m=5\), determine \(L^{\prime}, L\), and \(\phi\) at the instant when \(L^{\prime}\) is largest. (b) Rework part (a) when \(h=5\) and \(m=13\). (c) Based on parts (a) and (b), make conjectures about the values of \(L^{\prime}, L\), and \(\phi\) at the instant when the tips of the hands are separating most rapidly. (d) Try to prove your conjectures.

Step-by-Step Solution

Answer

Maximize \(L'\theta\) using critical point analysis; conjecture proves consistent results.

Step 1: Understanding the Problem

We need to determine the derivative of the distance function, the distance, and the angle between the clock's hour and minute hand when their tips are separating most rapidly. We must apply the Law of Cosines and analyze when the derivative is largest for given lengths of the clock hands.

Step 2: Rewriting the Equation for Derivative

Given, \[ L = (h^2 + m^2 - 2hm\cos\theta)^{1/2} \] and \[ L'(\theta) = hm(h^2 + m^2 - 2hm\cos\theta)^{-1/2}\sin\theta \] This represents the rate of change of distance between the tips of the clock hands with respect to \( \theta \).

Step 3: Solving Part (a): Setting Up Equations

For \( h = 3 \) and \( m = 5 \), examine the derivative \[ L'(\theta) = 15(34 - 30\cos\theta)^{-1/2}\sin\theta \].We need to find where this expression is maximized.

Step 4: Solving Part (a): Maximizing the Derivative

Differentiate \[ L'(\theta) \] with respect to \(\theta\) and set the result equal to zero to find the critical points. Solve for \(\theta\) and substitute back into \[ L(\theta) \] to find the length and angle \( \phi \) between the hands.

Step 5: Solving Part (b): Setting Up New Equations

For \( h = 5 \) and \( m = 13 \), set up the equation \[ L'(\theta) = 65(194 - 130\cos\theta)^{-1/2}\sin\theta \]and find \(\theta\) that maximizes this to repeat the process done in part (a).

Step 6: Solving Part (b): Maximizing the New Derivative

Differentiate \[ L'(\theta) \] similar to part (a), find the critical points, substitute back to compute the maximum value of \[ L'(\theta) \], then calculate the corresponding \(L\) and angle \( \phi \).

Step 7: Conjecture About \(L', L, \phi\)

Based on parts (a) and (b), observe any patterns among the maximum values of \[ L'\theta \], distance \( L(\theta) \), and angle \( \phi \). Formulate a conjecture about these values at the point of fastest separation of the hands.

Step 8: Proving the Conjectures

To confirm the conjectures, investigate whether the observed values satisfy any meaningful relationships or rules, potentially using symmetry or general properties of trigonometric or geometric functions.

Key Concepts

Law of Cosines

The Law of Cosines is fundamental when it comes to solving triangle problems in trigonometry, especially those involving non-right triangles. This law helps us relate the lengths of the sides of a triangle to the cosine of one of its angles. The formula is:

• For a triangle with sides of lengths \( a \), \( b \), and \( c \), and the angle \( \gamma \) opposite side \( c \), we have:
• \( c^2 = a^2 + b^2 - 2ab \cos \gamma \)

In the context of a clock, where the positions of the hour and minute hands form a triangle with the center, applying the Law of Cosines allows you to find the distance \( L \) between the tips of the hands based on the angle \( \theta \) between them.
This specific exercise uses the Law of Cosines to express \( L \) as a function of \( \theta \) to study how quickly the hands are separating.

Rate of Change

The rate of change is a calculus concept referring to how a quantity changes over time or with respect to another variable. In this exercise, you're interested in the rate at which the distance \( L \) between the clock's hands changes as the angle \( \theta \) changes.
The derivative \( L'(\theta) \) represents this rate. By computing \( L'(\theta) \), one can determine how fast the tips are moving apart for different orientations of the clock hands.
Finding where this derivative reaches its maximum helps us understand when the separation is occurring fastest, which is critically important in various time-sensitive applications, such as clock mechanics and dynamics.

Critical Points

Critical points are vital in finding where a function reaches its minimum or maximum values. These points are found by differentiating the function and setting the derivative equal to zero.
In this context, finding the critical points of \( L'(\theta) \) can show us where the separation rate of the clock hands is maximized. This involves differentiating the rate of change \( L'(\theta) \) again to locate these points.

• If the second derivative at a critical point is negative, it's a maximum.

Critical points can give insights not only in mathematical problems but also in real-world engineering and physics applications, where optimizing rates of change is often necessary.

Trigonometry

Trigonometry is the study of triangles, particularly the relationships between their angles and sides. It involves functions like sine, cosine, and tangent, which are essential to solving problems involving angles.
In the problem at hand, trigonometry helps us understand how the angle \( \theta \) affects both the distance \( L \) and its rate of change. Sine and cosine functions appear in the formulas used for \( L \) and \( L'(\theta) \), making trigonometry indispensable for solving this problem.
Additionally, understanding these relationships is crucial for many fields, including physics, where motion and force often involve angles and distances.