Word Problems
Translate real-world scenarios into calculus problems and use derivatives to find rates of change, velocities, and optimal values.
Learn
Derivatives appear throughout science, economics, and engineering. The key to solving word problems is translating the scenario into a mathematical function, then applying derivative rules to answer the question.
Common Types of Word Problems
- Rate of Change: How fast is something changing at a specific moment?
- Velocity and Acceleration: Given position, find velocity (first derivative) or acceleration (second derivative)
- Marginal Analysis: In economics, the derivative of cost/revenue/profit functions
- Related Rates: How are two changing quantities connected?
- Optimization: Finding maximum or minimum values
Problem-Solving Framework
- Read carefully: Identify what quantity is changing and what you need to find
- Define variables: Let x, t, or other variables represent the quantities
- Write the function: Express the relationship mathematically
- Differentiate: Find the derivative
- Solve: Substitute values and answer the question
- Interpret: State your answer in context with units
Examples
Example 1: Motion Problem
A ball is thrown upward with position s(t) = -16t^2 + 64t + 80 feet after t seconds. Find the velocity at t = 2 seconds.
Step 1: Velocity is the derivative of position
Step 2: v(t) = s'(t) = -32t + 64
Step 3: v(2) = -32(2) + 64 = -64 + 64 = 0
Answer: At t = 2 seconds, the velocity is 0 ft/s (the ball is at its highest point)
Example 2: Business Application
A company's profit function is P(x) = -2x^2 + 100x - 800 dollars, where x is units sold. Find the marginal profit when 20 units are sold.
Step 1: Marginal profit is P'(x)
Step 2: P'(x) = -4x + 100
Step 3: P'(20) = -4(20) + 100 = -80 + 100 = 20
Answer: Marginal profit at 20 units is $20 per additional unit
Example 3: Rate of Change
A circular oil slick has area A = pi*r^2. If the radius is increasing at 0.5 m/min when r = 4 m, how fast is the area increasing?
Step 1: Find dA/dr = 2*pi*r
Step 2: Use chain rule: dA/dt = (dA/dr)(dr/dt)
Step 3: dA/dt = 2*pi*r * (0.5) = pi*r
Step 4: At r = 4: dA/dt = 4*pi m^2/min
Answer: The area is increasing at 4*pi (approximately 12.57) square meters per minute
Practice
Solve these word problems using derivatives. Remember to include units in your answers.
Problem 1
A particle moves along a line with position s(t) = t^3 - 6t^2 + 9t meters at time t seconds. Find the velocity at t = 1 second.
Show Solution
v(t) = s'(t) = 3t^2 - 12t + 9
v(1) = 3 - 12 + 9 = 0 m/s
Problem 2
The height of a plant after t weeks is h(t) = 20 + 8t - 0.5t^2 cm. How fast is it growing at t = 4 weeks?
Show Solution
h'(t) = 8 - t
h'(4) = 8 - 4 = 4 cm/week
Problem 3
A manufacturer's cost function is C(x) = 0.01x^3 - 0.6x^2 + 15x + 100 dollars. Find the marginal cost when x = 30 units.
Show Solution
C'(x) = 0.03x^2 - 1.2x + 15
C'(30) = 0.03(900) - 1.2(30) + 15 = 27 - 36 + 15 = $6 per unit
Problem 4
A rocket's altitude is h(t) = 200t - 5t^2 meters. At what time does the rocket reach its maximum height?
Show Solution
h'(t) = 200 - 10t = 0
t = 20 seconds
Problem 5
A spherical balloon has volume V = (4/3)*pi*r^3. Find the rate of change of volume with respect to radius when r = 5 cm.
Show Solution
dV/dr = 4*pi*r^2
At r = 5: dV/dr = 4*pi*(25) = 100*pi cubic cm per cm (approximately 314.16)
Problem 6
Water is being drained from a tank. The volume after t minutes is V(t) = 100 - 4t - 0.1t^2 liters. How fast is water draining at t = 10 minutes?
Show Solution
V'(t) = -4 - 0.2t
V'(10) = -4 - 2 = -6 liters per minute (negative indicates draining)
Problem 7
The revenue from selling x items is R(x) = 50x - 0.02x^2 dollars. Find the number of items that maximizes revenue.
Show Solution
R'(x) = 50 - 0.04x = 0
x = 1250 items
Problem 8
A car's position on a highway is s(t) = 60t + 0.5t^2 miles after t hours. Find its acceleration.
Show Solution
v(t) = s'(t) = 60 + t miles per hour
a(t) = v'(t) = 1 mile per hour per hour (constant acceleration)
Problem 9
Population of bacteria is P(t) = 1000 * e^(0.05t). Find the growth rate at t = 20 hours. (Use the fact that d/dx[e^(kx)] = k*e^(kx))
Show Solution
P'(t) = 1000 * 0.05 * e^(0.05t) = 50 * e^(0.05t)
P'(20) = 50 * e^1 = 50e (approximately 136 bacteria per hour)
Problem 10
A rectangle has perimeter 40 cm. Express the area as a function of width w, then find the width that maximizes area.
Show Solution
If perimeter = 40, then 2l + 2w = 40, so l = 20 - w
A(w) = w(20 - w) = 20w - w^2
A'(w) = 20 - 2w = 0
w = 10 cm (the rectangle is actually a square)
Check Your Understanding
Reflect on these conceptual questions:
- What does a negative derivative mean in a motion problem?
- Why is "marginal" cost/revenue the derivative rather than the average?
- How do you know when a function reaches its maximum or minimum value?
- What are the units of the derivative if f(t) has units of meters and t has units of seconds?
Next Steps
- Practice identifying the type of problem before solving
- Always check that your answer makes sense in context
- Move on to Common Mistakes to avoid typical errors
- Return to this lesson when you encounter word problems in other subjects