Problem 135 - College Algebra Essentials

Question

Hurricanes are one of nature's most destructive forces. These low-pressure areas often have diameters of over 500 miles. The function \(f(x)=0.48 \ln (x+1)+27\) models the barometric air pressure, \(f(x),\) in inches of mercury, at a distance of \(x\) miles from the eye of a hurricane. The function \(P(t)=145 e^{-0.092 t}\) models a runner's pulse, \(P(t),\) in beats per minute, \(t\) minutes after a race, where \(0 \leq t \leq 15 .\) Graph the function using a graphing utility. TRACE along the graph and determine after how many minutes the runner's pulse will be 70 beats per minute. Round to the nearest tenth of a minute. Verify your observation algebraically.

Step-by-Step Solution

Answer

To get a short answer, solve the function \(145e^{-0.092t} = 70\) algebraically to obtain \(t\).

Step 1: Visualize the function

The first part involves graphing the function \(P(t) = 145e^{-0.092t}\). This can be achieved by using a graphing utility such as Desmos, or a graphing calculator. One forms the Cartesian plane, and then plots the function on the graph.

Step 2: Trace the graph to estimate the value

Next, one traces along the graph of the function to estimate when the pulse rate is 70 beats per minute. This is done by moving along the curve of the graph from left to right, until the y-coordinate (pulse rate) is approximately 70. The corresponding x-coordinate (time) is noted down.

Step 3: Solve the function algebraically for verification

Now, to verify the graphical answer, the function is solved algebraically to find out when exactly the pulse will be 70 bpm. This involves setting \(P(t) = 70\) and solving the equation 70 = 145e^{-0.092t} for \(t\). After solving it, it will give a time \(t\), rounded to the nearest tenth of a minute.

Key Concepts

Logarithmic Functions

Logarithmic functions play a significant role in mathematical modeling. In the given exercise, the function \(f(x) = 0.48 \ln(x+1) + 27\) is used to describe the barometric air pressure at a distance \(x\) from the eye of a hurricane. This is a classic example of a logarithmic function where \(\ln\) represents the natural logarithm.

A logarithmic function is the inverse of an exponential function and is defined only for positive arguments. This means that \(\ln(x)\) is defined only for \(x > 0\), ensuring realistic modeling within the domain provided.

In practical terms, when graphing a logarithmic function:

• The function increases at a decreasing rate
• Approaches infinity as \(x\) approaches infinity
• Has a vertical asymptote at \(x = -1\) for \(\ln(x+1)\)

The logarithmic model is particularly useful in scenarios like this hurricane example, where the rate of change slows down as one moves farther from the eye of the hurricane.

Exponential Functions

Exponential functions are equally important in modeling dynamic phenomena. The function \(P(t) = 145e^{-0.092t}\) models a runner's pulse after a race. Here, \(e\) is the base of the natural logarithm, approximately equal to 2.71828. This type of function where the variable is in the exponent is called an exponential function.

Exponential decay occurs when the function's exponential factor is less than 1, as seen in this exercise. The graph of \(P(t)\) will decrease rapidly at first and slow down as \(t\) increases. This accurately models a runner's heart rate gradually returning to normal after an intense exercise session.

Key features of graphing exponential functions:

• The curve decreases rapidly
• Never touches the x-axis due to having a horizontal asymptote at \(y=0\)
• Shows a continuous decrease over the given time interval

By understanding the behavior of exponential functions, one can predict changes in real-world scenarios, such as the runner's pulse rate modeled in this exercise.

Algebraic Verification

Algebraic verification is essential to confirm the accuracy of graphical estimation. To verify the time \(t\) when the runner's pulse will be 70 beats per minute, you need to solve the equation \(70 = 145e^{-0.092t}\) algebraically.

First, isolate \(e^{-0.092t}\) by dividing both sides by 145: \[\frac{70}{145} = e^{-0.092t}\].

Next, take the natural logarithm to remove the exponent: \(\ln\left(\frac{70}{145}\right) = -0.092t\).

Solve for \(t\) by dividing both sides by \(-0.092\): \[t = \frac{\ln\left(\frac{70}{145}\right)}{-0.092}\]. This computation will give you the precise time to one decimal place.

Algebraic verification reassures the graphical findings and provides a methodical approach to confirming results. Understanding this process strengthens the ability to derive exact solutions from approximative data visualizations.