Applications of Calculus

Grade 12 · Math · Study Guide

Explore how derivatives and integrals solve real-world problems including optimization, related rates, and motion analysis.

Calculus becomes powerful when applied to real-world problems. Optimization problems use derivatives to find maximum or minimum values of a function. The procedure involves: (1) identifying the function to optimize, (2) finding its derivative, (3) setting the derivative to zero to find critical points, (4) using the second derivative test to classify them, and (5) checking endpoints if on a closed interval.

Related rates problems involve quantities that change over time and are related by an equation. We differentiate implicitly with respect to time using the Chain Rule. For example, if a balloon's volume V = (4/3)πr³ is increasing, we differentiate to get dV/dt = 4πr² · dr/dt, allowing us to find how fast the radius changes.

In kinematics, if s(t) is position, then velocity v(t) = s'(t) and acceleration a(t) = v'(t) = s''(t). The integral of velocity gives displacement: ∫v(t)dt = s(t). Total distance traveled is ∫|v(t)|dt.

Volumes of solids of revolution use definite integrals. The disk method gives V = π∫ₐᵇ [f(x)]² dx for rotation about the x-axis. These applications demonstrate why calculus is fundamental to physics, engineering, and economics.

Key terms

Optimization
Finding maximum or minimum values of a function using critical points and the second derivative test
Related Rates
Problems where multiple quantities change over time and are related by a differentiable equation
Kinematics
The study of motion using calculus; position, velocity (s'), and acceleration (s'') are related by differentiation
Closed Interval Method
To find absolute extrema on [a,b]: evaluate f at all critical points and endpoints; largest/smallest are the absolute max/min
Disk Method
Volume of solid of revolution: V = π∫ₐᵇ [f(x)]² dx when rotating y=f(x) around the x-axis
Mean Value Theorem
If f is continuous on [a,b] and differentiable on (a,b), then there exists c where f'(c) = [f(b)-f(a)]/(b-a)

Flashcards

Optimization Steps
1) Define variable & function. 2) Find f'(x). 3) Set f'(x)=0. 4) Second Derivative Test. 5) Check endpoints if closed interval.
Related Rates Procedure
1) Draw diagram. 2) Write equation relating variables. 3) Differentiate both sides with respect to t. 4) Substitute known values.
Position, Velocity, Acceleration
v(t) = s'(t), a(t) = v'(t) = s''(t); integrate v(t) to get displacement
Total Distance vs Displacement
Displacement = ∫ₐᵇ v(t)dt; Total distance = ∫ₐᵇ |v(t)| dt
Mean Value Theorem
There exists c in (a,b) where f'(c) = [f(b)-f(a)]/(b-a); the instantaneous rate equals the average rate
Disk Method Volume
V = π∫ₐᵇ [f(x)]² dx for rotation of y=f(x) about the x-axis
Absolute Extrema on [a,b]
Evaluate f at all critical numbers in (a,b) and both endpoints; compare all values
L'Hôpital's Rule
If lim f/g gives 0/0 or ∞/∞, then lim[f(x)/g(x)] = lim[f'(x)/g'(x)]

Questions

  1. A particle's position is s(t) = t³ - 6t² + 9t. At what time t > 0 is the particle at rest?

    1. t = 1 only
    2. t = 3 only
    3. t = 1 and t = 3
    4. t = 2
    Show answer

    Answer: t = 1 and t = 3

    Particle is at rest when v(t)=0. v(t)=s'(t)=3t²-12t+9=3(t-1)(t-3)=0. So t=1 and t=3.

  2. In an optimization problem on a closed interval, the absolute maximum must occur at a critical point, not an endpoint.

    Show answer

    Answer: False

    The absolute maximum on a closed interval can occur at a critical point OR at an endpoint. Always check all three locations.

  3. A ladder 10 feet long leans against a wall. The bottom slides away at 2 ft/s. How fast is the top sliding down when the bottom is 6 feet from the wall?

    1. 1.5 ft/s
    2. 3/2 ft/s
    3. 3/2 ft/s down
    4. All of the above
    Show answer

    Answer: All of the above

    x²+y²=100, 2x(dx/dt)+2y(dy/dt)=0. When x=6: y=8. 2(6)(2)+2(8)(dy/dt)=0, dy/dt=-24/16=-3/2. The top slides down at 3/2 ft/s.

  4. If velocity v(t) = 3t² - 4, then acceleration at t = 2 is ___.

    Show answer

    Answer: 12

    a(t)=v'(t)=6t. At t=2: a(2)=12.

  5. A rectangle has perimeter 40 cm. What dimensions maximize its area?

    1. 10 × 10
    2. 5 × 15
    3. 8 × 12
    4. 2 × 18
    Show answer

    Answer: 10 × 10

    If perimeter=40, then 2l+2w=40, so l+w=20. Area A=lw=l(20-l). A'=20-2l=0 gives l=10, w=10. A 10×10 square maximizes the area.

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