Lesson 1-3: Applications in Physics & Economics

Applications in Physics & Economics

Apply calculus concepts to solve real-world problems in physics and economics. Learn to calculate displacement, work, average values, and perform marginal analysis using integration and differentiation techniques with practical examples and applications.

Learning Objectives

Physics Applications

Calculate displacement, work, and other physical quantities using integration

Average Value

Compute average values of functions over intervals using definite integrals

Economic Analysis

Apply marginal analysis and cost/revenue optimization using calculus

Real-World Modeling

Create mathematical models for practical problems in various fields

Core Knowledge Points

Physics Applications

Displacement from Velocity

$s = \int_{t_1}^{t_2} v(t) \, dt$

Net displacement over time interval [t₁, t₂]

Work by Variable Force

$W = \int_a^b F(x) \, dx$

Work done by force F(x) over distance [a, b]

Average Value

$\bar{f} = \frac{1}{b-a} \int_a^b f(x) \, dx$

Average value of f(x) over interval [a, b]

Economic Applications

Total Cost from Marginal Cost

$C(Q) = C_0 + \int_0^Q MC(q) \, dq$

Total cost as integral of marginal cost

Revenue from Marginal Revenue

$\Delta R = \int_{Q_1}^{Q_2} MR(q) \, dq$

Revenue change from marginal revenue

Consumer Surplus

$CS = \int_0^{Q^*} [D(q) - P^*] \, dq$

Area between demand curve and price line

Real-World Applications

Physics: Projectile Motion

Problem: Ball Thrown Upward

A ball is thrown upward with velocity $v(t) = 20 - 9.8t$ m/s. Find the displacement and average velocity over the first 3 seconds.

Displacement: $s = \int_0^3 (20 - 9.8t) \, dt = [20t - 4.9t^2]_0^3 = 15.9$ m

Average velocity: $\bar{v} = \frac{15.9}{3} = 5.3$ m/s

Economics: Cost Analysis

Problem: Manufacturing Cost

A company's marginal cost is $MC(q) = 0.1q^2 - 2q + 10$ . Fixed cost is $1000. Find the total cost of producing 50 units.

Total Cost: $C(50) = 1000 + \int_0^{50} (0.1q^2 - 2q + 10) \, dq$

$= 1000 + [\frac{0.1q^3}{3} - q^2 + 10q]_0^{50} = 1000 + 2083.33 = 3083.33$

Comprehensive Example Analysis

Physics: Variable Force Work

Problem: Spring Compression

A spring follows Hooke's law: $F(x) = kx$ where k = 200 N/m. Find the work done compressing the spring from 0 to 0.1 m.

1. Set up the integral

$W = \int_0^{0.1} 200x \, dx$

2. Integrate

$W = 200 \left[ \frac{x^2}{2} \right]_0^{0.1} = 200 \cdot \frac{(0.1)^2}{2} = 1$ J

3. Interpretation

The work done is 1 Joule, representing the energy stored in the compressed spring.

Economic Analysis Example

Consumer Surplus Calculation

Problem: Market Analysis

Demand function: $D(q) = 100 - 2q$ , Supply function: $S(q) = 20 + 3q$ . Find consumer surplus at equilibrium.

1. Find equilibrium

$100 - 2q = 20 + 3q \Rightarrow q^* = 16, P^* = 68$

2. Calculate consumer surplus

$CS = \int_0^{16} [(100 - 2q) - 68] \, dq = \int_0^{16} (32 - 2q) \, dq$

3. Evaluate integral

$CS = [32q - q^2]_0^{16} = 512 - 256 = 256$

Practice Problems

Problem 1: Physics Applications

A particle moves with velocity $v(t) = t^2 - 4t + 3$ m/s. Find:

a) Displacement from t = 0 to t = 4 seconds

b) Average velocity over this time interval

c) Total distance traveled

Problem 2: Economic Analysis

A company's marginal revenue is $MR(q) = 50 - 0.5q$ and marginal cost is $MC(q) = 10 + 0.3q$ . Find:

a) The optimal production level

b) Total profit at optimal level

c) Producer surplus

Problem 3: Average Value

Find the average value of $f(x) = x^2 + 1$ over the interval [0, 3]:

a) Calculate the average value

b) Find the x-value where f(x) equals the average value

c) Interpret the result geometrically

Key Takeaways

Physics Applications

Use integration to calculate displacement, work, and other physical quantities

Average Value

Calculate average values using definite integrals and interpret geometrically

Economic Analysis

Apply marginal analysis and calculate consumer/producer surplus

Real-World Modeling

Create mathematical models for practical problems across disciplines

Advanced Insights

Fundamental Theorem Applications: The connection between derivatives and integrals is crucial for understanding rates of change and accumulation in real-world problems.

Optimization in Economics: Marginal analysis using derivatives helps find optimal production levels, while integration helps calculate total costs and benefits.

Geometric Interpretation: Definite integrals represent areas under curves, which often correspond to meaningful physical or economic quantities.

Common Pitfalls

• Confusing displacement (net change) with total distance traveled
• Forgetting to account for fixed costs when calculating total cost from marginal cost
• Not setting up the correct limits of integration for the problem context
• Misinterpreting the sign of the integral in economic applications

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