Problem 55 - Contemporary Precalculus

Question

(a) Use a calculator to find the average rate of change of \(g(t)=\sin t\) from 2 to \(2+h,\) for each of these values of \(h: 01, .001, .0001,\) and .00001 (b) Compare your answers in part (a) with the number cos \(2 .\) What would you guess that the instantaneous rate of change of \(g(t)=\sin t\) is at \(t=2 ?\)

Step-by-Step Solution

Answer

Answer: The guess for the instantaneous rate of change of function g(t) = sin(t) at t = 2 is approximately -0.41614.

Step 1: Find the average rate of change

To calculate the average rate of change of a function, we can use the formula: \(AROC = \frac{g(2+h)-g(2)}{h}\). Use a calculator to substitute the values of \(h\) and compute the average rate of change for each case: - \(h = 0.1\) - \(h = 0.001\) - \(h = 0.0001\) - \(h = 0.00001\)

Step 2: Compute the average rate of change

Using a calculator, compute the average rate of change for each value of \(h\). The results are: - \(AROC(0.1) = \frac{\sin(2+0.1)-\sin 2}{0.1} \approx -0.4024\) - \(AROC(0.001) = \frac{\sin(2+0.001)-\sin 2}{0.001} \approx -0.4160\) - \(AROC(0.0001) = \frac{\sin(2+0.0001)-\sin 2}{0.0001} \approx -0.4162\) - \(AROC(0.00001) = \frac{\sin(2+0.00001)-\sin 2}{0.00001} \approx -0.41614\)

Step 3: Compare the answers in part (a) with the number cos \(2\)

Comparing the values of the average rates of change obtained in step 2 with the number \(\cos 2\): \(\cos 2 \approx -0.41614\) Notice that as \(h\) becomes smaller and closer to 0, \(AROC(h)\) approaches the value of \(\cos 2\). This suggests a relationship between the average rate of change and the instantaneous rate of change.

Step 4: Make a guess on the instantaneous rate of change of \(g(t)\) at \(t=2\)

Given the fact that as \(h\) approaches to 0, the \(AROC(h)\) gets closer to the value of \(\cos 2\), we can guess that the instantaneous rate of change of \(g(t) = \sin t\) at \(t=2\) is equal to the value of \(\cos 2 \approx -0.41614\).

Key Concepts

Instantaneous Rate of Change

The concept of the instantaneous rate of change can be thought of as the speed at which a function is changing at any given point. It represents the slope of the tangent line to a curve at a specific point. One way to estimate the instantaneous rate of change is by looking at the average rate of change over an interval that gets smaller and smaller, as we did in the exercise.

When the interval gets infinitesimally small, the average rate of change approaches the instantaneous rate of change. This concept is fundamental in calculus and is very similar to finding the derivative. In our particular exercise with the function \(g(t) = \sin t\), as the value of \(h\) gets smaller, the average rate of change of \(g(t)\) approaches the instantaneous rate of change.

• As \(h\) approaches zero, the average rate of change approaches \(\cos 2\).
• This suggests that the instantaneous rate of change of \(g(t)\) at \(t=2\) is precisely \(\cos 2\).

This illustration helps in verifying the relationship between the instantaneous rate of change and derivatives.

Trigonometric Functions

Trigonometric functions like \(\sin\), \(\cos\), and \(\tan\) are essential in mathematics, especially when dealing with periodic phenomena. In the given exercise, we examined \(g(t) = \sin t\), highlighting the interplay of trigonometric functions with rates of change and derivatives.

The function \(\sin t\) has known derivative properties:

• The derivative of \(\sin t\) is \(\cos t\).

This means at any point \(t\), the slope of the tangent line to the graph of \(\sin t\) is given by \(\cos t\). Thus, when calculating the instantaneous rate of change at \(t=2\), it is logical that the instantaneous rate of change aligns closely with the value of \(\cos 2\).

Trigonometric functions are pivotal in differential calculus, allowing us to explore how these curves change over time and giving insights into various real-world applications like wave behavior and oscillations.

Derivative Concepts

Understanding derivatives is fundamental in calculus as it is the mathematical tool that measures how a function changes as its input changes. In our exercise, we effectively explored this through the function \(g(t) = \sin t\).

• The derivative of a function at a specific point is an expression of its instantaneous rate of change.

The derivative concept gives us a precise mathematical framework. For trigonometric functions, such as \(\sin t\), the derivative \(\cos t\) tells us how \(\sin t\) is increasing or decreasing at any particular point.

This concept comes into play in real-world scenarios where you need to understand changing rates, such as velocity and acceleration in physics, and in economics to study rates of change in costs and outputs. In our example, as the interval \(h\) decreases to zero, our calculated average rate of change converges to the derivative \(\cos 2\), which is the instantaneous rate of change of the function at \(t = 2\). This illustrates the derivative's critical role in understanding the dynamic behavior of functions.