Applications of Functions in Real-Life Contexts
Overview
Functions are an essential part of mathematics and have a wide range of applications in real-life situations. From economics to physics, functions are used to model relationships between variables and solve problems. In this section, we will explore some common real-life applications of functions.
Real-Life Contexts
Functions can describe various real-life scenarios. Here are a few examples:
- Economics: Functions are used to model supply and demand, cost and revenue, and profit functions.
- Physics: Functions describe relationships such as velocity, acceleration, and distance.
- Biology: Functions model population growth, the spread of diseases, and the rate of enzyme reactions.
- Finance: Functions are used in compound interest calculations, loan repayments, and investment growth.
- Engineering: Functions model stress, strain, and electrical circuits.
- Medicine: Functions are used to model drug dosage levels, heart rate, and blood pressure changes over time.
Example 1: Economics - Profit Function
In business, a company’s profit is a function of the number of items sold. A typical profit function can be written as:
\( P(x) = R(x) - C(x) \)
Where:
- \( P(x) \) is the profit from selling \( x \) items
- \( R(x) \) is the revenue from selling \( x \) items
- \( C(x) \) is the cost of producing \( x \) items
For example, if a company sells each item for $10 and the cost to produce each item is $5, the profit function could be written as:
\( P(x) = 10x - 5x \)
Which simplifies to:
\( P(x) = 5x \)
This means the profit increases by $5 for every additional item sold.
Example 2: Physics - Velocity Function
In physics, velocity is often expressed as a function of time. For instance, the velocity of an object in free fall can be modeled by the equation:
\( v(t) = g \cdot t \)
Where:
- \( v(t) \) is the velocity at time \( t \)
- \( g \) is the acceleration due to gravity (approximately 9.8 m/s² on Earth)
- \( t \) is the time the object has been falling
If an object has been falling for 3 seconds, the velocity can be calculated as:
\( v(3) = 9.8 \cdot 3 = 29.4 \, \text{m/s} \)
Example 3: Finance - Compound Interest
The compound interest formula is a function that models the amount of money accumulated over time in an interest-bearing account:
\( A(t) = P(1 + \frac{r}{n})^{nt} \)
Where:
- \( A(t) \) is the amount of money accumulated after \( t \) years, including interest
- \( P \) is the principal amount (initial investment)
- \( r \) is the annual interest rate (decimal)
- \( n \) is the number of times that interest is compounded per year
- \( t \) is the number of years the money is invested or borrowed for
For example, if $1000 is invested at an annual interest rate of 5% compounded quarterly for 3 years, the formula becomes:
\( A(3) = 1000(1 + \frac{0.05}{4})^{4 \cdot 3} \)
Calculating the value of \( A(3) \) will give the total accumulated amount after 3 years.
Practice Questions
- Economics: A company produces and sells t-shirts. The revenue function is \( R(x) = 15x \), where \( x \) is the number of t-shirts sold, and the cost function is \( C(x) = 10x + 500 \), where \( x \) is the number of t-shirts sold. Write the profit function and calculate the profit when 200 t-shirts are sold.
Solution
The profit function is:
\( P(x) = R(x) - C(x) = 15x - (10x + 500) \)
Simplifying:
\( P(x) = 15x - 10x - 500 = 5x - 500 \)
The profit when 200 t-shirts are sold is:
\( P(200) = 5(200) - 500 = 1000 - 500 = 500 \)
The profit is $500.
- Physics: An object is thrown into the air and its height above the ground in meters is modeled by the function \( h(t) = -5t^2 + 20t + 10 \), where \( t \) is the time in seconds. Find the height of the object after 2 seconds and determine when the object hits the ground.
Solution
The height after 2 seconds is:
\( h(2) = -5(2)^2 + 20(2) + 10 = -5(4) + 40 + 10 = -20 + 40 + 10 = 30 \, \text{meters} \)
To find when the object hits the ground, set \( h(t) = 0 \) and solve for \( t \):
\( 0 = -5t^2 + 20t + 10 \)
\( 0 = t^2 - 4t - 2 \)
Using the quadratic formula:
\( t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-2)}}{2(1)} = \frac{4 \pm \sqrt{16 + 8}}{2} = \frac{4 \pm \sqrt{24}}{2} \approx \frac{4 \pm 4.9}{2} \)
So \( t \approx 4.9 \) or \( t \approx -0.9 \). Since time cannot be negative, the object hits the ground at \( t \approx 4.9 \) seconds.
- Finance: A bank account has a principal of $5000 and earns 4% annual interest compounded monthly. Write the compound interest formula and calculate the balance after 5 years.
Solution
The compound interest formula is:
\( A(t) = P(1 + \frac{r}{n})^{nt} \)
Given \( P = 5000 \), \( r = 0.04 \), \( n = 12 \), and \( t = 5 \), the balance after 5 years is:
\( A(5) = 5000(1 + \frac{0.04}{12})^{12 \cdot 5} \approx 5000(1 + 0.00333)^{60} \approx 5000(1.2214) \approx 6107.04 \)
The balance after 5 years is approximately $6107.04.
- Medicine: A doctor is tracking the concentration of a drug in a patient’s bloodstream. The concentration, \( C(t) \), is modeled by the function \( C(t) = 50e^{-0.1t} \), where \( t \) is the time in hours. What is the concentration of the drug after 6 hours?
Solution
The concentration after 6 hours is:
\( C(6) = 50e^{-0.1(6)} = 50e^{-0.6} \approx 50(0.5488) \approx 27.44 \, \text{mg/L} \)