Problem 68 - A Graphical Approach to College Algebra

Question

Sprinter's Speed and Time During the 100 -meter dash, the elapsed time \(T\) for a sprinter to reach a speed of x meters per second is given by the following function. $$T(x)=-1.2 \ln \left(1-\frac{x}{11}\right)$$ (a) How much time had elapsed when the sprinter was running 0 meters per second? Interpret your answer. (b) At the end of the race, the sprinter was moving at 10.998 meters per second. What was the sprinter's time for this 100 -meter dash? (c) Find T^{-1}(x) and interpret its meaning.

Step-by-Step Solution

Answer

(a) 0 seconds. (b) Approx. 12.56 seconds. (c) \( T^{-1}(x) = 11(1-e^{-\frac{x}{1.2}}) \) gives speed for a time.

Step 1: Understand the Function and Variables

The given function for the time, \( T(x) = -1.2 \ln \left(1-\frac{x}{11}\right) \), represents the time it takes for a sprinter to reach a speed of \( x \) meters per second. \( T(x) \) is the elapsed time, and \( x \) is the sprinter's speed.

Step 2: Calculate elapse time at 0 m/s speed

For part (a), plug \( x = 0 \) m/s into the function; \( T(0) = -1.2 \ln(1 - \frac{0}{11}) = -1.2 \ln(1) \). Since \( \ln(1) = 0 \), the calculation becomes \( T(0) = 0 \). This means no time has elapsed when the speed is 0 m/s, which is logical because it signifies the start of the race.

Step 3: Determine time when speed is 10.998 m/s

For part (b), plug \( x = 10.998 \) m/s into the function to find the elapsed time: \( T(10.998) = -1.2 \ln\left(1-\frac{10.998}{11}\right) \). Simplifying \( 1-\frac{10.998}{11} \approx 0.0001818 \), so \( T(10.998) = -1.2 \ln(0.0001818) \approx 12.56 \) seconds. This is the sprinter's time for completing the 100 meters.

Step 4: Find the inverse function

To find the inverse function \( T^{-1}(x) \), start by setting \( y = -1.2 \ln(1-\frac{x}{11}) \). Solve for \( x \) in terms of \( y \):1. Divide both sides by -1.2: \( \frac{y}{-1.2} = \ln(1-\frac{x}{11}) \).2. Exponentiate both sides: \( e^{\frac{y}{-1.2}} = 1-\frac{x}{11} \).3. Solve for \( \frac{x}{11} \): \( \frac{x}{11} = 1 - e^{\frac{y}{-1.2}} \).4. Finally, \( x = 11(1 - e^{\frac{y}{-1.2}}) \).Therefore, \( T^{-1}(x) = 11(1-e^{-\frac{x}{1.2}}) \). This function gives the speed of the sprinter \( x \) for a given elapsed time \( y \).

Key Concepts

Inverse Functions

Inverse functions reverse the process of a given function. For a function like our sprinter's, where elapsed time is calculated based on speed, the inverse function finds the speed when a certain amount of time has passed.
In our example with the function \( T(x) = -1.2 \ln\left(1-\frac{x}{11}\right) \), the inverse \( T^{-1}(x) \) tells us what speed the sprinter reaches after a certain time \( x \).
Here’s how it works:

• Start with the relationship given by the function.
• Swap the roles of the dependent and independent variables. This is what solving for the inverse entails.
• Finally, perform algebraic manipulations to solve for the original variable in terms of the other.

For our function, the process involved isolating \( x \) gives \( T^{-1}(x) = 11(1-e^{-\frac{x}{1.2}}) \).
The interpretation is straightforward: if you know how long the sprinter has been running, this inverse function tells you their current speed.

Logarithmic Functions

Logarithmic functions are the inverse of exponential functions. They help handle processes that change rapidly at first but slow down over time.
In our sprinter’s function, \( \ln \left(1 - \frac{x}{11}\right) \), the logarithm serves to moderate how time is calculated based on speed.
Here's why logarithms are essential in this context:

• When \( x \) is small, the speed is low, and the time is less affected.
• As \( x \) increases, potential changes strongly influence the logarithm, thus affecting the time calculated.
• The negative sign \(-1.2 \) inversely adjusts the time according to the sprinter's increasing speed.

Think of logarithms as functions that help modulate rapid changes, like how time changes when speed reaches its peak in a race. They make it possible to graph smooth, interpretable functions, rather than erratic and wild logarithmic changes.

Graphing Functions

Graphing is a key skill to understand functions like our sprinter’s and their behaviors visually.
When graphing the function \( T(x) = -1.2 \ln\left(1-\frac{x}{11}\right) \), here is what to consider:

• The x-axis represents the speed \( x \), while the y-axis represents the time \( T(x) \).
• Start by plotting points for known values, like \( T(0) \), where time is zero at the beginning.
• Plot the endpoint, such as \( x = 10.998 \), where the graph takes significant values close to completion.
• Notice the graph's curve: it starts flat (as time changes little from increasing speed initially) and then drops steeply as speed nears the maximal value (11 m/s).

Graphing helps illuminate how the sprinter’s time varies with their increasing speed during the dash. It reveals the pace and acceleration as visual trends on a coordinate plane.