Unit 4 brings calculus to life with real-world applications. You'll learn how to:
In real-world contexts, derivatives represent rates of change:
Units: (units of f) per (units of x)
Always include units when interpreting derivatives in context. The derivative tells us how fast something is changing at a specific moment.
If P(t) represents the population of a city at time t (in years), then P'(t) represents:
| Context | Function | Derivative Meaning | Units |
|---|---|---|---|
| Position | s(t) | Velocity | m/s or ft/s |
| Velocity | v(t) | Acceleration | m/s² or ft/s² |
| Volume | V(r) | Rate of volume change | cm³/cm or in³/in |
| Temperature | T(t) | Rate of temperature change | °C/min or °F/hour |
| Cost | C(x) | Marginal cost | $ per item |
Related rates problems involve finding how one quantity changes with respect to time when you know how another related quantity changes:
Always draw a diagram and label all known and unknown quantities. This helps you see the relationships between variables.
A circle's radius is increasing at 3 cm/s. How fast is the area increasing when the radius is 5 cm?
Solution:
1) We want dA/dt when r = 5
2) A = πr² (area of circle)
3) dA/dt = 2πr · dr/dt (chain rule)
4) dA/dt = 2π(5)(3) = 30π cm²/s
Critical points occur where f'(x) = 0 or f'(x) is undefined:
Find all critical points of f(x) = x³ - 3x² + 2 on [-1, 3]
Solution:
1) f'(x) = 3x² - 6x = 3x(x - 2)
2) Set f'(x) = 0: 3x(x - 2) = 0
3) So x = 0 or x = 2
4) Check endpoints: x = -1 and x = 3
5) Critical points: x = -1, 0, 2, 3
For a critical point x = c:
For a critical point x = c:
The Mean Value Theorem (MVT) states:
The MVT guarantees that somewhere between a and b, the instantaneous rate of change equals the average rate of change.
For f(x) = x² on [1, 3], find the value of c guaranteed by MVT
Solution:
1) f'(x) = 2x
2) f(1) = 1, f(3) = 9
3) Average rate: [f(3) - f(1)]/(3 - 1) = (9 - 1)/2 = 4
4) Set f'(c) = 4: 2c = 4, so c = 2
5) Check: 2 is in (1, 3) ✓
To solve optimization problems:
Find the dimensions of a rectangular box with volume 1000 cm³ that has minimum surface area.
Solution:
1) Let x, y, z be the dimensions
2) Volume: V = xyz = 1000, so z = 1000/(xy)
3) Surface area: S = 2xy + 2xz + 2yz = 2xy + 2000/y + 2000/x
4) Find critical points: ∂S/∂x = 2y - 2000/x² = 0
5) This gives x = y = 10, so z = 10
6) The optimal box is a cube: 10 × 10 × 10 cm
A spherical balloon is being inflated at a rate of 10 cm³/s. How fast is the radius increasing when the radius is 5 cm?
Solution:
1) We want dr/dt when r = 5
2) Volume: V = (4/3)πr³
3) Differentiate: dV/dt = 4πr² · dr/dt
4) Substitute: 10 = 4π(5)² · dr/dt
5) Solve: dr/dt = 10/(100π) = 1/(10π) cm/s
Find the maximum area of a rectangle inscribed in a semicircle of radius 2.
Solution:
1) Let the rectangle have width 2x and height y
2) From the semicircle: x² + y² = 4, so y = √(4 - x²)
3) Area: A = 2xy = 2x√(4 - x²)
4) Find critical points: A' = 2√(4 - x²) + 2x(-x)/√(4 - x²) = 0
5) This gives x = √2, so y = √2
6) Maximum area: A = 2(√2)(√2) = 4
A ladder 10 feet long rests against a vertical wall. If the bottom of the ladder slides away from the wall at 1 ft/s, how fast is the top sliding down when the bottom is 6 feet from the wall?
Solution:
1) We want dy/dt when x = 6
2) From Pythagorean theorem: x² + y² = 100
3) When x = 6: y = 8
4) Differentiate: 2x(dx/dt) + 2y(dy/dt) = 0
5) Substitute: 2(6)(1) + 2(8)(dy/dt) = 0
6) Solve: dy/dt = -12/16 = -3/4 ft/s
7) The top is sliding down at 3/4 ft/s
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In Unit 5, you'll learn about analytical applications of differentiation including L'Hôpital's rule and advanced curve sketching.
Go to Unit 5: Analytical Applications →