Applications of Derivatives

Finding Extrema (Max and Min)

Critical points: Where f'(x) = 0 or undefined

To find absolute max/min on [a, b]:

1. Find all critical points in (a, b)
2. Evaluate f at critical points and endpoints
3. Largest value = absolute max, smallest = absolute min

Example: Find Maximum

f(x) = x³ - 3x² + 1 on [0, 3]

Find f'(x):

f'(x) = 3x² - 6x = 3x(x - 2)

Critical points: x = 0, x = 2

Evaluate:

• f(0) = 1
• f(2) = 8 - 12 + 1 = -3
• f(3) = 27 - 27 + 1 = 1

Absolute max: 1 (at x = 0 and x = 3) Absolute min: -3 (at x = 2)

First Derivative Test

Determines if critical point is max, min, or neither

At critical point c:

• If f'(x) changes from + to -: local maximum
• If f'(x) changes from - to +: local minimum
• If f'(x) doesn't change sign: neither (inflection point)

Example: Classify Critical Points

f(x) = x³ - 3x + 1

f'(x) = 3x² - 3 = 3(x² - 1) = 3(x-1)(x+1)

Critical points: x = -1, x = 1

Test intervals:

• x < -1: f'(x) = 3(-)(-) = + → increasing
• -1 < x < 1: f'(x) = 3(+)(-) = - → decreasing
• x > 1: f'(x) = 3(+)(+) = + → increasing

At x = -1: + to - → local max At x = 1: - to + → local min

Second Derivative Test

Alternative test using concavity

At critical point c where f'(c) = 0:

• If f''(c) > 0: local minimum (concave up)
• If f''(c) < 0: local maximum (concave down)
• If f''(c) = 0: test inconclusive (use first derivative test)

Example: Second Derivative Test

f(x) = x³ - 3x + 1

f'(x) = 3x² - 3 f''(x) = 6x

At x = 1: f''(1) = 6 > 0 → local min ✓ At x = -1: f''(-1) = -6 < 0 → local max ✓

Optimization Problems

Real-world problems finding maximum or minimum values

Strategy:

1. Draw diagram, identify variables
2. Write equation for quantity to optimize
3. Use constraints to express as function of one variable
4. Find derivative, set equal to 0
5. Solve for critical points
6. Verify max or min
7. Answer in context

Example: Maximize Area

Problem: 200 m of fencing for rectangular enclosure. Maximize area.**

Variables: x = width, y = length

Constraint: 2x + 2y = 200 → y = 100 - x

Objective: A = xy = x(100 - x) = 100x - x²

Find maximum:

A'(x) = 100 - 2x

Set equal to 0:

100 - 2x = 0
x = 50

Then y = 100 - 50 = 50

Check: A''(x) = -2 < 0 → maximum ✓

Maximum area: 50 × 50 = 2500 m²

(Square shape!)

Example: Minimize Cost

Problem: Cylindrical can holds 500 cm³. Minimize material (surface area).**

Volume: πr²h = 500 → h = 500/(πr²)

Surface area: SA = 2πr² + 2πrh

Substitute h:

SA = 2πr² + 2πr(500/(πr²))
SA = 2πr² + 1000/r

Find minimum:

SA' = 4πr - 1000/r²

Set equal to 0:

4πr = 1000/r²
4πr³ = 1000
r³ = 250/π
r ≈ 4.3 cm

Then h ≈ 8.6 cm

Minimum material when h = 2r (height = diameter)

Related Rates

Problems where multiple quantities change with respect to time

Strategy:

1. Draw diagram, identify variables
2. Write equation relating variables
3. Differentiate both sides with respect to time (t)
4. Substitute known rates and values
5. Solve for unknown rate

Example: Ladder Sliding

Problem: 10 m ladder leans against wall. Bottom slides away at 0.5 m/s. How fast is top sliding down when bottom is 6 m from wall?**

Relationship: x² + y² = 100 (Pythagorean theorem)

Given: dx/dt = 0.5 m/s when x = 6

Find: dy/dt when x = 6

Differentiate:

2x(dx/dt) + 2y(dy/dt) = 0

When x = 6: y² = 100 - 36 = 64, so y = 8

Substitute:

2(6)(0.5) + 2(8)(dy/dt) = 0
6 + 16(dy/dt) = 0
dy/dt = -6/16 = -0.375 m/s

Top sliding down at 0.375 m/s

Example: Balloon Inflating

Problem: Spherical balloon inflated at 10 cm³/s. How fast is radius increasing when r = 5 cm?**

Volume: V = (4/3)πr³

Given: dV/dt = 10 cm³/s

Find: dr/dt when r = 5

Differentiate:

dV/dt = 4πr²(dr/dt)

Substitute:

10 = 4π(25)(dr/dt)
dr/dt = 10/(100π)
dr/dt = 1/(10π) ≈ 0.032 cm/s

Motion Problems

Position function: s(t)

Velocity: v(t) = s'(t)

Acceleration: a(t) = v'(t) = s''(t)

Speed: |v(t)|

Particle moving right: v(t) > 0 Particle moving left: v(t) < 0 Particle at rest: v(t) = 0

Example: Analyze Motion

s(t) = t³ - 6t² + 9t (position in meters, t in seconds)

Velocity:

v(t) = 3t² - 12t + 9 = 3(t² - 4t + 3)
     = 3(t - 1)(t - 3)

Acceleration:

a(t) = 6t - 12 = 6(t - 2)

When at rest: v(t) = 0 → t = 1 or t = 3

Direction:

• 0 < t < 1: v > 0 (moving right)
• 1 < t < 3: v < 0 (moving left)
• t > 3: v > 0 (moving right)

At t = 2: a(2) = 0 (acceleration is zero)

Curve Sketching

Use derivatives to sketch accurate graphs

Information from f:

• f(x) > 0: above x-axis
• f(x) < 0: below x-axis
• f(x) = 0: x-intercepts

Information from f':

• f'(x) > 0: increasing
• f'(x) < 0: decreasing
• f'(x) = 0: horizontal tangent (critical points)

Information from f'':

• f''(x) > 0: concave up
• f''(x) < 0: concave down
• f''(x) = 0: possible inflection point

Example: Sketch Curve

f(x) = x³ - 3x

f'(x) = 3x² - 3 = 3(x² - 1) Critical points: x = ±1

f''(x) = 6x Inflection point: x = 0

Analysis:

• f'(x) = 0 at x = -1 (local max) and x = 1 (local min)
• f''(0) = 0 (inflection point)
• Concave down for x < 0, concave up for x > 0

Mean Value Theorem

If f continuous on [a, b] and differentiable on (a, b):

There exists c in (a, b) where:

f'(c) = [f(b) - f(a)] / (b - a)

Meaning: Somewhere, instantaneous rate equals average rate

Example: MVT

f(x) = x² on [1, 3]

Average rate: [9 - 1]/[3 - 1] = 4

f'(x) = 2x

Find c where f'(c) = 4:

2c = 4
c = 2

At x = 2, instantaneous rate = average rate ✓

Practice