Problem 62 - Single Variable Calculus

Question

Average Speed According to data from a study by General Motors, the average speed of a trip, \(A\) (in miles per hour), is related to the number of stops per mile made on that trip, \(x\), by the equation $$A=\frac{26.5}{x^{0.45}}$$ Compute \(d A / d x\) for \(x=0.25\) and \(x=2\), and interpret your results.

Step-by-Step Solution

Answer

For $x=0.25$, the derivative $\frac{dA}{dx} \approx -10.336$, indicating a decrease in average speed as the number of stops per mile increases. For $x=2$, the derivative $\frac{dA}{dx} \approx -1.226$, also indicating a decrease in average speed but at a slower rate than for $x=0.25$. Thus, the average speed decreases less rapidly as the number of stops per mile increases.

Step 1: Find the derivative of A with respect to x

To find the derivative of \(A\) with respect to \(x\), use the power rule. For an equation of the form \(y=ax^b\), the derivative is given by \(y'=abx^{b-1}\). In our case, we need to find the derivative of the function: $$A= \frac{26.5}{x^{0.45}} = 26.5x^{-0.45}$$ Now, apply the power rule to find the derivative, \(\frac{dA}{dx}\): $$\frac{dA}{dx} = -0.45\cdot 26.5x^{-0.45-1}$$

Step 2: Find the value of the derivative for x=0.25

Now, compute the value of the derivative \(\frac{dA}{dx}\) for \(x=0.25\). Plug \(x=0.25\) into the equation for the derivative: $$\frac{dA}{dx} = -0.45\cdot 26.5(0.25)^{-0.45-1} \approx -10.336$$

Step 3: Find the value of the derivative for x=2

Next, compute the value of the derivative \(\frac{dA}{dx}\) for \(x=2\). Plug \(x=2\) into the equation for the derivative: $$\frac{dA}{dx} = -0.45\cdot 26.5(2)^{-0.45-1} \approx -1.226$$

Step 4: Interpret the results

The first result, \(\frac{dA}{dx} \approx -10.336\), corresponds to an average speed that is decreasing as the number of stops per mile (\(x\)) increases. This makes sense since more stops per mile would reasonably decrease the overall average speed. The second result, \(\frac{dA}{dx} \approx -1.226\), also corresponds to a decreasing average speed as the number of stops per mile increases. However, the magnitude of the derivative is smaller than for the case when \(x=0.25\). This means that the average speed decreases less rapidly when the number of stops per mile is higher. In other words, the rate at which the average speed decreases slows down as the number of stops per mile increases.

Key Concepts

Understanding the Derivative

In calculus, a derivative represents the rate of change of a function with respect to one of its variables. Think of the derivative as a way to say how fast something is changing. In this exercise, we are looking at how the average speed, denoted as \(A\), changes as the number of stops per mile, \(x\), changes.
The derivative here, \(\frac{dA}{dx}\), tells us how the average speed changes for tiny changes in the number of stops per mile. When you compute a derivative, you are essentially finding a formula that describes this rate of change at any given point.
In our case, the derivative tells us how sensitive the average speed is to variations in the number of stops.

Calculating Average Speed

Average speed is a simple concept but understanding how it reacts to factors like stops can be intricate. Here, average speed \(A\), measured in miles per hour, depends on how often the car stops per mile, represented by \(x\). The formula given is \(A = \frac{26.5}{x^{0.45}}\). This formula shows that as the number of stops increases, average speed decreases.
Why does this happen? Because more stops mean more time spent not moving, thus reducing the average speed over your trip. Hence, fewer stops allow for more continuous driving, increasing the average speed.

The Power Rule and Its Application

The power rule is a handy rule in calculus for taking derivatives of functions of the form \(y = ax^b\). It simplifies this to \(y' = abx^{b-1}\). In our problem, we rewrite the original function \(A = \frac{26.5}{x^{0.45}}\) as \(A = 26.5x^{-0.45}\) to use the power rule effectively.
By applying the power rule, you differentiate \(A\) to find \(\frac{dA}{dx} = -0.45 \times 26.5x^{-0.45 - 1}\). This computation gives us a formula that we can use to find how fast the average speed decreases with additional stops per mile.
The beauty of the power rule is in its ability to quickly provide the derivative, a crucial tool for understanding the behavior of complex equations.

Interpreting the Results

After finding the derivative, we evaluate it at specific values: \(x = 0.25\) and \(x = 2\). At \(x = 0.25\), \(\frac{dA}{dx} \approx -10.336\). This means that a small increase in stops per mile will lead to a significant drop in average speed, illustrating high sensitivity to changes in stops.
When \(x = 2\), \(\frac{dA}{dx} \approx -1.226\). Here, the impact of additional stops is less drastic, indicating that as stops per mile increase, average speed decreases, but at a slower rate.
This shows a diminishing sensitivity—initially, additional stops lower speeds quickly, but the effect tapers off as stops become more frequent. This is valuable for practical scenarios where understanding the rate change helps in decision-making for efficient travel planning.