Calculus for Business and Social Sciences Final Practice Guide

(1) Proportionality Concepts: Exoplanet Sustainability

Problem Statement: The number of exoplanets $N$ that can sustain life is directly proportional to the square root of the distance $d$ from Earth measured in light-years.
Mathematical Model: * Direct proportionality relationship: $N = k\sqrt{d}$ . * $k$ represents the constant of proportionality.
Determining the Constant of Proportionality: * Given data: The distance to the closest exoplanet that can sustain life is $4.2\,light-years$ . This implies when $d = 4.2$ , the count $N = 1$ . * Substitution: $1 = k\sqrt{4.2}$ . * Solving for $k$ : $k = \frac{1}{\sqrt{4.2}}$ .
Final Formula: * $N = \frac{\sqrt{d}}{\sqrt{4.2}}$ .

(2) Linear Equations: Slope and Line Direction

Problem Statement: Find the equation for a line passing through the points $(-1, 2)$ and $(3, -2)$ and determine if the line is increasing or decreasing.
Slope Calculation: * Let $(x_1, y_1) = (-1, 2)$ and $(x_2, y_2) = (3, -2)$ . * The slope $m$ is calculated as: $m = \frac{-2 - 2}{3 - (-1)} = \frac{-4}{4} = -1$ .
Deriving the Equation: * Using the point-slope form with the point $(-1, 2)$ : $y - 2 = -1(x - (-1))$ . * Simplification: $y - 2 = -1(x + 1)$ . * Distribution: $y - 2 = -x - 1$ . * Final Slope-Intercept Form: $y = -x + 1$ .
Analysis of Line Direction: * The line is decreasing because the slope $m = -1$ is negative (less than zero).

(3) Exponential Functions: Base, Growth Rate, and Evaluation

Model Identification: An exponential function is defined as $Q = g(t) = Q_0 a^t$ , where $Q_0 = g(0)$ is the initial value and a > 0 is the base.
Solving for the Base ( $a$ ): * Given values: $g(8) = 15.3$ and $g(10) = 19.2$ . * Equation for $t=8$ : $Q_0 a^8 = 15.3$ * Equation for $t=10$ : $Q_0 a^{10} = 19.2$ * Dividing the expressions: $\frac{Q_0 a^{10}}{Q_0 a^8} = \frac{19.2}{15.3}$ . * $a^2 = \frac{19.2}{15.3} \approx 1.2549$ . * $a = \sqrt{\frac{19.2}{15.3}} \approx 1.120$ .
Determining Percent Growth Rate ( $r$ ): * The relationship between the base and growth rate is $a = 1 + r$ . * $r = a - 1 = 1.120 - 1 = 0.120$ . * Expressed as a percentage: $r = 12\%$ .
Evaluating Future Values ( $g(12)$ ): * The goal is to find $g(12) = Q_0 a^{12}$ . * Divide the expression for $g(12)$ by the expression for $g(10)$ : $\frac{Q_0 a^{12}}{Q_0 a^{10}} = \frac{g(12)}{19.2}$ . * $a^2 = \frac{g(12)}{19.2}$ . * $g(12) = 19.2 \times a^2 = 19.2 \times (1.120)^2$ . * Result: $g(12) \approx 24.08$ .

(4) Continuous Compounding Interest

Formula: The balance $B$ after time $t$ is given by $B = B_0 e^{rt}$ . * $B_0$ = deposited amount ( $\$5,500$ ). * $r$ = annual interest rate converted to decimal ( $3.22\% = 0.0322$ ).
Goal: Find the time $t_4$ required for the balance to reach a factor of 4 (i.e., $B = 4 \times B_0$ ).
Calculation Steps: * Setting up the equation: $4 \times 5,500 = 5,500 e^{0.0322 t_4}$ . * Divide both sides by $5,500$ : $4 = e^{0.0322 t_4}$ . * Taking the natural logarithm: $0.0322 t_4 = \ln(4)$ . * Isolating $t_4$ : $t_4 = \frac{\ln(4)}{0.0322}$ . * Numerical Result: $t_4 \approx 43.053\,years$ . * Conclusion: It takes approximately $43$ years for the account balance to quadruple.

(5) Population Decay Models

Scenario: Fish population in a lake decreasing due to pollution at a continuous rate of $0.15\%$ per year.
Model: $P(t) = P_0 e^{kt}$ . * Continuous Decay Rate ( $k$ ): $-0.15 / 100 = -0.0015$ . * Base Year: 2021 ( $t = 0$ ). * Initial Population ( $P_0$ ): $14,230$ .
Population Function: $P(t) = 14,230 e^{-0.0015 t}$ .
Projection for 2026: Since 2026 is 5 years after 2021, set $t = 5$ . * $P(5) = 14,230 e^{-0.0015(5)}$ . * Calculation: $P(5) = 14,230 e^{-0.0075} \approx 14,124$ .

(6) Business Applications: Cost, Revenue, and Break-Even

Functions Provided: * Cost Function: $C(q) = 1260 + 5q$ . * Revenue Function: $R(q) = 13q$ .
Analysis of Financial components: * Fixed Costs: These are the costs independent of production quantity ( $q=0$ ). For this company, fixed costs are $\$1,260$ . * Marginal Cost: The rate of change of cost, which is the coefficient of $q$ in the linear cost function: $\$5\,per\,item$ . * Marginal Revenue: The rate of change of revenue, which is the coefficient of $q$ in the linear revenue function: $\$13\,per\,item$ . * Price Per Item: Since $R = p \times q$ , and $R = 13q$ , the price charged per item is $p = \$13$ .
Break-Even Point ( $q_0$ ): * Definition: The production level where $R(q) = C(q)$ . * Equation: $13q_0 = 1260 + 5q_0$ . * Subtract $5q_0$ from both sides: $8q_0 = 1260$ . * Solve for $q_0$ : $q_0 = \frac{1260}{8} = 157.5$ . * Conclusion: The company breaks even at approximately $157$ items.

(7) Profit Functions and Derivative Rules

Functions Provided: * $R(q) = \ln(27q) \cdot q^2$ * $C(q) = \frac{\sqrt{2}q}{5} + \frac{3}{\sqrt{5}}$
Profit Function (\pi(q)): * $\pi(q) = R(q) - C(q) = \ln(27q) \cdot q^2 - \left(\frac{\sqrt{2}q}{5} + \frac{3}{\sqrt{5}}\right)$ . * Expanded expression: $\pi(q) = \ln(27q) \cdot q^2 - \frac{\sqrt{2}q}{5} - \frac{3}{\sqrt{5}}$ .
Marginal Profit (\pi'(q)): * Definition: The derivative of the profit function. * Step-by-Step Derivative Process: 1. Product Rule for $\ln(27q) \cdot q^2$ : $\frac{d}{dq}(\ln(27q)) \cdot q^2 + \ln(27q) \cdot \frac{d}{dq}(q^2)$ . 2. Derivative of Natural Log ( $\ln(27q)$ ) with Chain Rule: $\frac{1}{27q} \cdot 27 = \frac{1}{q}$ . 3. Derivative of $q^2$ : $2q$ . 4. Term 1 result: $(\frac{1}{q} \cdot q^2) + \ln(27q) \cdot 2q = q + 2q \ln(27q)$ . 5. Derivative of the linear cost term ( $-\frac{\sqrt{2}q}{5}$ ): $-\frac{\sqrt{2}}{5} \cdot \frac{1}{2}q^{-1/2} = -\frac{\sqrt{2}}{10\sqrt{q}}$ . 6. Derivative of constant term ( $-\frac{3}{\sqrt{5}}$ ): $0$ . * Final Marginal Profit Expression: $\pi'(q) = q + 2q \ln(27q) - \frac{\sqrt{2}}{10\sqrt{q}}$ .

(8) General Differentiation Practice

Problem (a): $f(x) = \frac{1}{\sqrt{2x + 1}} + e^{x^2+2x}$ . * Rewrite as power: $f(x) = (2x+1)^{-1/2} + e^{x^2+2x}$ . * $f'(x) = -\frac{1}{2}(2x+1)^{-3/2}(2) + e^{x^2+2x}(2x + 2)$ .
Problem (b): $g(t) = \sqrt{2t^5 + 4t^3 + 3t + 29}$ . * Rewrite as power: $g(t) = (2t^5 + 4t^3 + 3t + 29)^{1/2}$ . * $g'(t) = \frac{1}{2}(2t^5 + 4t^3 + 3t + 29)^{-1/2}(10t^4 + 12t^2 + 3)$ .
Problem (c): $k(x) = \ln\left(\frac{7x + 4}{x^2 + 9}\right)$ . * Apply log rules or chain rule: $k'(x) = \frac{1}{\left(\frac{7x + 4}{x^2 + 9}\right)} \cdot \left(\frac{7(x^2 + 9) - (7x + 4)(2x)}{(x^2 + 9)^2}\right)$ .
Problem (d): $g(x) = ((11)^{3x} + e^{3x}) \ln\sqrt{8x^2 + 1}$ . * Using Product Rule: $g'(x) = (\ln(11) \cdot 11^{3x} \cdot 3 + 3e^{3x}) \ln\sqrt{8x^2 + 1} + (11^{3x} + e^{3x}) \cdot \frac{1}{\sqrt{8x^2+1}} \cdot \frac{1}{2}(8x^2+1)^{-1/2}(16x)$ .

(9) Tangent Line Calculations

Function: $f(x) = e^{x^3}$ at $x = 3$ .
Find Slope ( $m$ ): * $f'(x) = e^{x^3}(3x^2)$ . * $m = f'(3) = e^{3^3} \cdot 3(3^2) = 27e^{27}$ .
Find Point of Tangency: * The point is $(3, f(3)) = (3, e^{3^3}) = (3, e^{27})$ .
Construct Equation: * Using Point-Slope form: $y - e^{27} = 27e^{27}(x - 3)$ . * Simplification: $y = 27e^{27}x - 81e^{27} + e^{27}$ . * Final Result: $y = 27e^{27}x - 80e^{27}$ .

(10) Curve Analysis: Critical and Inflection Points

Function: $f(x) = (x + 1) e^{-x}$ .
Derivative Calculation: * Using product rule: $f'(x) = (1)e^{-x} + (x+1)(-1)e^{-x}$ . * Factoring out $e^{-x}$ : $f'(x) = (1 - x - 1)e^{-x} = -x e^{-x} = -\frac{x}{e^x}$ .
Identifying Critical Points: * $f'(x) = 0 \rightarrow -\frac{x}{e^x} = 0 \rightarrow x = 0$ . * One critical point exists at $x = 0$ .
First Derivative Test: * Interval $x < 0$ : Test point $x = -1$ . $f'(-1) = -(-1)e^{-(-1)} = e^1$ (Positive, $\nearrow$ ). * Interval $x > 0$ : Test point $x = 1$ . $f'(1) = -(1)e^{-1} = -e^{-1}$ (Negative, $\searrow$ ). * Conclusion: Local Maximum at $x = 0$ .
Second Derivative Test: * $f''(x) = (-1)e^{-x} + (-x)(-e^{-x}) = (x - 1)e^{-x}$ . * Evaluate at CP: f''(0) = (0 - 1)e^{0} = -1 < 0. * Conclusion: Confirms Local Maximum at $x = 0$ .
Determining Inflection Points: * Set $f''(x) = 0 \rightarrow (x - 1)e^{-x} = 0 \rightarrow x = 1$ .
Verifying Inflection Points: * Interval $x < 1$ : Test point $x = 0$ . $f''(0) = -1$ (Negative, Concave Down). * Interval $x > 1$ : Test point $x = 2$ . $f''(2) = (2-1)e^{-2} = e^{-2}$ (Positive, Concave Up). * Conclusion: Concavity changes at $x = 1$ , identifying it as an inflection point.

(11) Global Extrema on a Closed Interval

Function: $f(x) = x^4 - 2x^2$ on the interval $[-2, 2]$ .
Derivatives: * $f'(x) = 4x^3 - 4x$ . * $f''(x) = 12x^2 - 4$ .
Critical Points: * $4x(x^2 - 1) = 0 \rightarrow 4x(x - 1)(x + 1) = 0$ . * Points: $x = -1, x = 0, x = 1$ .
Inflection Points: * $12x^2 = 4 \rightarrow x^2 = \frac{1}{3} \rightarrow x = \pm \sqrt{\frac{1}{3}}$ . * Concavity Table: * $x < -\frac{1}{\sqrt{3}}$ : Sign (+), Concave Up. * $-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$ : Sign (-), Concave Down. * $x > \frac{1}{\sqrt{3}}$ : Sign (+), Concave Up.
Evaluating Global Extrema: * $f(-2) = (-2)^4 - 2(-2)^2 = 16 - 8 = 8$ . * $f(-1) = (-1)^4 - 2(-1)^2 = 1 - 2 = -1$ . * $f(0) = 0^4 - 2(0)^2 = 0$ . * $f(1) = 1^4 - 2(1)^2 = -1$ . * $f(2) = 2^4 - 2(2)^2 = 8$ .
Result: * Global Maximum is $8$ at $x = \pm 2$ . * Global Minimum is $-1$ at $x = \pm 1$ .

(12) Estimating Totals with Riemann Sums

Scenario: Tickets sold per minute follows $s(t) = 100 \ln(t + 1)$ . Estimate total for first 10 minutes.
Partitioning with $\Delta t = 2$ : * Endpoints: $(0, 2, 4, 6, 8, 10)$ . * Left-hand sum: * Sum: $s(0)\Delta t + s(2)\Delta t + s(4)\Delta t + s(6)\Delta t + s(8)\Delta t$ . * Calculation: $2 \times (100\ln(1) + 100\ln(3) + 100\ln(5) + 100\ln(7) + 100\ln(9)) \approx 1,370$ . * Right-hand sum: * Sum: $s(2)\Delta t + s(4)\Delta t + s(6)\Delta t + s(8)\Delta t + s(10)\Delta t$ . * Calculation: $2 \times (100\ln(3) + 100\ln(5) + 100\ln(7) + 100\ln(9) + 100\ln(11)) \approx 1,849$ .
Total Estimate: * Average the sums: $\frac{1370 + 1849}{2} = 1609.5$ . * Final estimation: Approximately $1,609$ tickets.

(13) Integration Techniques

Problem (a): Algebraic Indefinite Integral * $\int \left(4t^3 + \frac{1}{\sqrt{3}t} + \frac{t^{1/3}}{5}\right) dt = \int \left(4t^3 + \frac{t^{-1/2}}{\sqrt{3}} + \frac{t^{1/3}}{5}\right) dt$ . * Result: $t^4 + \frac{2}{\sqrt{3}} t^{1/2} + \frac{3}{20} t^{4/3} + C$ .
Problem (b): Logarithmic and Exponential Integration * $\int \left(\frac{13}{17x} + e^{4x+3}\right) dx = \int \left(\frac{13}{17} \cdot \frac{1}{x} + e^3 e^{4x}\right) dx$ . * Result: $\frac{13}{17} \ln(x) + \frac{e^3}{4} e^{4x} + C$ .
Problem (c): Definite Integral * $\int_1^5 \left(e^{2t} + \frac{1}{7t} + \sqrt{\pi} t^{1/2}\right) dt$ . * Antiderivative: $\left(\frac{1}{2} e^{2t} + \frac{1}{7} \ln(t) + \frac{2\sqrt{\pi}}{3} t^{3/2}\right) \bigg|_1^5$ . * Evaluation: $\left(\frac{1}{2} e^{10} + \frac{1}{7} \ln(5) + \frac{2\sqrt{\pi}}{3} 5^{3/2}\right) - \left(\frac{1}{2} e^2 + \frac{1}{7} \ln(1) + \frac{2\sqrt{\pi}}{3} 1^{3/2}\right)$ .
Problem (d): Definite Integral Over Symmetrical Range * $\int_{-1}^1 \left(t^5 - t^7 - 2t^{-3}\right) dt$ . * Antiderivative: $\left(\frac{1}{6} t^6 - \frac{1}{8} t^8 + t^{-2}\right) \bigg|_{-1}^1$ . * Result: Because all terms result in even powers, evaluated limits cancel out: $0$ .
Problem (e): Substitution Method * $\int \left(5x^{-1/5} - 2xe^{x^2}\right) dx$ . * Part 1: $\int 5x^{-1/5} dx = \frac{25}{4} x^{4/5} + C_1$ . * Part 2 (Substitution): Let $u = x^2$ , then $du = 2x\,dx$ . * $\int e^u du = e^u + C_2 = e^{x^2} + C_2$ . * Combined Result: $\frac{25}{4} x^{4/5} - e^{x^2} + C$ .