Applications of Derivatives
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Real-World Applications of Derivatives
Derivatives aren't just abstract math - they're powerful tools for solving practical problems. From finding the best shape for a container to analyzing motion and economics, derivatives help us optimize and understand the world around us.
Optimization Problems
Finding Maximum and Minimum Values
To find maximum or minimum values of a function:
- Find critical points where f'(x) = 0 or f'(x) is undefined
- Use the second derivative test or first derivative test
- Check endpoints if on a closed interval
Second Derivative Test: f''(c) > 0 means minimum; f''(c) < 0 means maximum
Strategy for Optimization Problems
- Understand the problem: What needs to be maximized or minimized?
- Draw a diagram: Visualize the situation
- Assign variables: Identify unknowns
- Write the objective function: Express what you're optimizing
- Write constraints: Express limitations as equations
- Reduce to one variable: Substitute constraints into objective
- Find critical points: Take derivative, set equal to zero
- Verify it's a max/min: Use second derivative test
- Answer the question: State the answer in context
Related Rates
Related Rates Problems
When two or more quantities change over time and are related by an equation, their rates of change are also related. We use implicit differentiation with respect to time.
If x and y are related by an equation, then dx/dt and dy/dt are related
Strategy for Related Rates
- Draw a picture and label quantities that change
- Identify what rates you know and what you need to find
- Write an equation relating the quantities
- Differentiate both sides with respect to time (t)
- Substitute known values and solve
Motion Along a Line
Position, Velocity, and Acceleration
If s(t) represents position at time t:
- Velocity: v(t) = s'(t) = ds/dt
- Acceleration: a(t) = v'(t) = s''(t) = d²s/dt²
- Speed: |v(t)| (absolute value of velocity)
| Condition | Meaning |
|---|---|
| v(t) > 0 | Object moving right/forward |
| v(t) < 0 | Object moving left/backward |
| v(t) = 0 | Object momentarily at rest (may be turning) |
| a(t) > 0 | Velocity increasing (speeding up if v > 0) |
| a(t) < 0 | Velocity decreasing (slowing down if v > 0) |
| v(t) and a(t) same sign | Speeding up |
| v(t) and a(t) opposite signs | Slowing down |
Linear Approximation
Tangent Line Approximation
Near x = a, a function can be approximated by its tangent line:
f(x) ≈ f(a) + f'(a)(x - a)
This is called the linear approximation or linearization of f at a.
Marginal Analysis (Economics)
Economic Applications
- Marginal Cost: C'(x) = additional cost to produce one more unit
- Marginal Revenue: R'(x) = additional revenue from selling one more unit
- Marginal Profit: P'(x) = R'(x) - C'(x)
Maximum profit occurs when R'(x) = C'(x) (marginal revenue = marginal cost)
Curve Sketching
Using Derivatives to Analyze Functions
- f'(x) > 0: Function is increasing
- f'(x) < 0: Function is decreasing
- f''(x) > 0: Function is concave up (smile)
- f''(x) < 0: Function is concave down (frown)
- Inflection point: Where concavity changes (f''(x) = 0)
Examples
Example 1: Optimization - Maximum Area
Problem: A farmer has 400 feet of fencing to enclose a rectangular pasture adjacent to a river (no fence needed on the river side). What dimensions maximize the area?
Solution:
Step 1: Define variables. Let x = width, y = length along river
Step 2: Constraint: 2x + y = 400, so y = 400 - 2x
Step 3: Objective: Maximize A = xy = x(400 - 2x) = 400x - 2x²
Step 4: Find critical points
A'(x) = 400 - 4x = 0
x = 100 feet
Step 5: Verify maximum (A''(x) = -4 < 0, so it's a maximum)
Step 6: Find y: y = 400 - 2(100) = 200 feet
Answer: Width = 100 ft, Length = 200 ft, Maximum Area = 20,000 sq ft
Example 2: Related Rates - Expanding Circle
Problem: A stone dropped in a pond creates a circular ripple. If the radius increases at 3 ft/sec, how fast is the area increasing when the radius is 10 feet?
Solution:
Known: dr/dt = 3 ft/sec, r = 10 ft
Find: dA/dt when r = 10
Equation: A = πr²
Differentiate with respect to t:
dA/dt = 2πr · (dr/dt)
Substitute:
dA/dt = 2π(10)(3) = 60π ≈ 188.5 ft²/sec
Example 3: Motion Analysis
Problem: A particle moves along a line with position s(t) = t³ - 6t² + 9t + 2 (in feet, t in seconds). Find when the particle is at rest and describe its motion.
Solution:
Velocity: v(t) = s'(t) = 3t² - 12t + 9 = 3(t² - 4t + 3) = 3(t - 1)(t - 3)
At rest when v(t) = 0: t = 1 and t = 3 seconds
Motion analysis:
- For t < 1: v(t) > 0, moving right
- At t = 1: at rest, turning point
- For 1 < t < 3: v(t) < 0, moving left
- At t = 3: at rest, turning point
- For t > 3: v(t) > 0, moving right
Positions: s(1) = 6 ft, s(3) = 2 ft
Example 4: Linear Approximation
Problem: Use linear approximation to estimate √4.1
Solution:
Let f(x) = √x, and use a = 4 (where we know the exact value)
f(a) = f(4) = 2
f'(x) = 1/(2√x), so f'(4) = 1/4 = 0.25
Linear approximation:
f(4.1) ≈ f(4) + f'(4)(4.1 - 4)
√4.1 ≈ 2 + 0.25(0.1) = 2 + 0.025 = 2.025
(Actual value: √4.1 ≈ 2.0248, so the estimate is very close!)
Example 5: Economic Optimization
Problem: A company's cost function is C(x) = 0.01x² + 5x + 100 and revenue is R(x) = 15x - 0.01x², where x is units. Find the production level that maximizes profit.
Solution:
Profit function: P(x) = R(x) - C(x)
P(x) = (15x - 0.01x²) - (0.01x² + 5x + 100)
P(x) = 15x - 0.01x² - 0.01x² - 5x - 100
P(x) = -0.02x² + 10x - 100
Find critical point:
P'(x) = -0.04x + 10 = 0
x = 10/0.04 = 250 units
Verify maximum: P''(x) = -0.04 < 0, so it's a maximum
Maximum profit: P(250) = -0.02(62500) + 10(250) - 100 = $1,150
Practice
Apply derivatives to solve real-world problems.
1. A rectangle has perimeter 40 cm. What dimensions maximize the area?
A) 5 × 15 B) 8 × 12 C) 10 × 10 D) 4 × 16
2. If s(t) = t² - 4t + 3 is position, when is velocity zero?
A) t = 1 B) t = 2 C) t = 3 D) t = 4
3. A sphere's radius increases at 2 cm/sec. How fast is volume increasing when r = 3 cm?
A) 24π B) 36π C) 72π D) 108π cm³/sec
4. The function f(x) = x³ - 3x² has an inflection point at:
A) x = 0 B) x = 1 C) x = 2 D) x = 3
5. Use linear approximation: (1.02)³ ≈ ?
A) 1.04 B) 1.06 C) 1.08 D) 1.10
6. If C(x) = 100 + 5x + 0.1x², marginal cost at x = 50 is:
A) 5 B) 10 C) 15 D) 20
7. A particle has v(t) = 4 and a(t) = -2 at time t. The particle is:
A) Speeding up B) Slowing down C) At rest D) Moving backward
8. Where is f(x) = x⁴ - 4x³ concave up?
A) x < 0 B) 0 < x < 2 C) x < 0 or x > 2 D) All x
9. A ladder 10 ft long leans against a wall. If bottom slides away at 1 ft/sec, how fast is top sliding down when bottom is 6 ft from wall?
A) 0.5 ft/sec B) 0.75 ft/sec C) 1 ft/sec D) 1.25 ft/sec
10. To maximize revenue R(p) = 1000p - 2p², what price should be charged?
A) $125 B) $250 C) $500 D) $1000
Click to reveal answers
- C) 10 × 10 - Square maximizes area for given perimeter
- B) t = 2 - v(t) = 2t - 4 = 0 when t = 2
- C) 72π - dV/dt = 4πr²·(dr/dt) = 4π(9)(2) = 72π
- B) x = 1 - f''(x) = 6x - 6 = 0 when x = 1
- B) 1.06 - f(x) = x³, f(1) + f'(1)(0.02) = 1 + 3(0.02) = 1.06
- C) 15 - C'(x) = 5 + 0.2x, C'(50) = 5 + 10 = 15
- B) Slowing down - v and a have opposite signs
- C) x < 0 or x > 2 - f''(x) = 12x² - 24x = 12x(x-2) > 0
- B) 0.75 ft/sec - Using x² + y² = 100, 2x(dx/dt) + 2y(dy/dt) = 0
- B) $250 - R'(p) = 1000 - 4p = 0, p = 250
Check Your Understanding
1. Explain the general strategy for solving optimization problems.
Show answer
First, identify what needs to be optimized (maximized or minimized) and express it as a function. Then identify constraints that limit the variables and use them to reduce to one variable. Take the derivative of the objective function, set it to zero, and solve for critical points. Use the second derivative test to verify whether it's a max or min. Finally, check endpoints if applicable and answer in context.
2. In related rates problems, why do we differentiate with respect to time?
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We differentiate with respect to time because all the quantities in the problem are changing over time. The relationship between the quantities stays the same (like the Pythagorean theorem for a ladder), but the values change as time passes. Differentiating with respect to t gives us relationships between the rates of change (like dx/dt and dy/dt), which is what we need to solve the problem.
3. How do you determine when an object is speeding up versus slowing down?
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An object speeds up when velocity and acceleration have the same sign (both positive or both negative). It slows down when they have opposite signs. This is because acceleration tells you how velocity is changing - if acceleration pushes velocity further from zero (same sign), the object speeds up; if it pushes velocity toward zero (opposite sign), the object slows down.
4. Why is marginal analysis useful in economics?
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Marginal analysis helps businesses make decisions about whether to produce one more unit. The marginal cost tells you the additional cost of producing that unit, while marginal revenue tells you the additional income. If marginal revenue exceeds marginal cost, producing more increases profit. Maximum profit occurs where they're equal - producing beyond that point means the extra cost exceeds the extra revenue.
🚀 Next Steps
- Review any concepts that felt challenging
- Move on to the next lesson when ready
- Return to practice problems periodically for review