Problem 56 - Contemporary Precalculus

Question

(a) Use a calculator to find the average rate of change of \(f(t)=\cos t\) from 5 to \(5+h,\) for each of these values of \(h: 01, .001, .0001,\) and .00001 (b) Compare your answers in part (a) with the number \(-\sin 5 .\) What would you guess that the instantaneous rate of change of \(f(t)=\cos t\) is at \(t=5 ?\)

Step-by-Step Solution

Answer

Question: Given the function \(f(t) = \cos t\), find the average rate of change of the function from 5 to \(5 + h\) for given values of \(h\). Compare these values to \(-\sin 5\) and make a guess of the instantaneous rate of change of the function at \(t = 5\).

Step 1: 1. Find the formula for the average rate of change for f(t)

To find the average rate of change of the function \(f(t) = \cos t\) from 5 to \(5 + h\), we can use the formula: $$\frac{f(5+h) - f(5)}{h}$$

Step 2: 2. Calculate the average rate of change for each given value of h using a calculator

We will now plug in each value of \(h\) into the formula and use a calculator to compute the corresponding average rate of change: For \(h = .01\): $$\frac{\cos(5 + .01) - \cos(5)}{.01}$$ For \(h = .001\): $$\frac{\cos(5 + .001) - \cos(5)}{.001}$$ For \(h = .0001\): $$\frac{\cos(5 + .0001) - \cos(5)}{.0001}$$ For \(h = .00001\): $$\frac{\cos(5 + .00001) - \cos(5)}{.00001}$$ Use a calculator to find the numerical values of these expressions.

Step 3: 3. Compare the obtained values from step 2 to \(-\sin 5\)

Now that you have found the values of the average rate of change for each given \(h\), compare them to the value of \(-\sin 5\). Observe how the values approach \(-\sin 5\) as \(h\) decreases.

Step 4: 4. Guess the instantaneous rate of change at \(t = 5\) based on the comparisons from step 3

Based on the comparisons between the average rate of change and \(-\sin 5\), you can make an educated guess about the instantaneous rate of change at \(t = 5\). The instantaneous rate of change is likely equal to or very close to the value \(-\sin 5\).

Key Concepts

Average Rate of Change

Understanding the average rate of change is essential for studying functions' behavior over specific intervals. Think of it as measuring the steepness of the slope between two points on a graph. To calculate the average rate of change of a function, you take the difference of the function values at two points and divide it by the difference in the input values.

For example, for a trigonometric function like \(f(t) = \cos t\), the average rate of change from \(t = a\) to \(t = a + h\) is given by:\[\frac{f(a + h) - f(a)}{h}\]As \(h\) changes, we get different average rates, giving us a glimpse into the function's local behavior between those two points. Insights gained from examining different values of \(h\) can help inform us about the characteristics of the function across intervals surrounding a point of interest, such as \(t = 5\) in the provided exercise. A smaller \(h\) provides a more precise average, leading us to the concept of instantaneity.

Trigonometric Functions

Trigonometric functions like \(\sin\), \(\cos\), and \(\tan\) are fundamental in various fields, including mathematics, physics, engineering, and even economics. These functions relate the angles of a triangle to the lengths of its sides in a right-angled triangle. But beyond that, they also describe the properties of waves and oscillations.

For our discussion on rates of change, it's important to know that trigonometric functions are periodic and differentiable, meaning they have a repeating pattern and we can calculate their derivative. The derivative of \(\cos t\) is \(-\sin t\), which represents its instantaneous rate of change. This concept is a cornerstone in understanding how we can determine the behavior of trigonometric functions at an exact moment—an application that has implications from signal processing to calculating orbits in astrophysics.

Limits in Calculus

In calculus, a limit is a fundamental concept that describes the value that a function approaches as the input approaches some value. Limits are essential when defining both the continuity of a function and the concept of a derivative, which is the instantaneous rate of change.

The exercise demonstrates the idea of a limit by calculating the average rate of change for progressively smaller values of \(h\). As \(h\) approaches zero, the average rate of change approaches the instantaneous rate of change of the function at that point. This convergence towards the derivative is what we are interested in when we take the limit of the average rate of change as \(h\) approaches zero:\[\lim_{{h \to 0}} \frac{f(a + h) - f(a)}{h}\]Through this process, we're able to capture a snapshot of the function's behavior at a precise instant, which would be equivalent to the function's derivative at the point \(a\). In the context of the original exercise, using limits allows us to predict the exact change in \(\cos t\) at any point, providing a powerful tool for analysis. When applied to \(f(t)=\cos t\) at \(t=5\), the limit will reveal the instantaneous rate of change, hinted to be close to \(-\sin 5\) based on our calculations.