Infinity Bounds Slides Infinity Bounds
Exploring Improper Integrals with Infinite Limits
Calculus BC Unit 8
The Paradox of Gabriel's Horn
Consider the function \( f(x) = \frac{1}{x} \) for \( x \ge 1 \) rotated around the x-axis.
The **Volume** is finite.
Wait... the **Surface Area** is infinite?
"You could fill it with paint, but you could never paint the outside."
x → ∞
Defining Improper Integrals (Type I)
An integral is improper if one or both limits of integration are infinite.
Formal Definition
\[ \int_{a}^{\infty} f(x) dx = \lim_{b \to \infty} \int_{a}^{b} f(x) dx \]
\[ \int_{-\infty}^{b} f(x) dx = \lim_{a \to -\infty} \int_{a}^{b} f(x) dx \]
Convergence vs. Divergence: If the limit exists and is finite, the integral **converges**. If not, it **diverges**.
Example 1: The p-test Foundation
\[ \int_{1}^{\infty} \frac{1}{x^2} dx \]
Steps:
Rewrite with a limit: \( \lim_{b \to \infty} \int_{1}^{b} x^{-2} dx \)
Integrate: \( \lim_{b \to \infty} \left[ -x^{-1} \right]_{1}^{b} \)
Evaluate: \( \lim_{b \to \infty} \left( -\frac{1}{b} + 1 \right) \)
Take limit: \( 0 + 1 = 1 \)
Conclusion
The area under \( 1/x^2 \) from 1 to infinity is exactly **1**.
Wait, What about \( 1/x \)?
Try it. It diverges! (ln → ∞)
The L'Hopital Connection
Many improper integrals involve functions that grow at different rates, leading to indeterminate forms in our limits.
Example: \( \int_{1}^{\infty} xe^{-x} dx \)
Requires Integration by Parts + L'Hopital's Rule
Part 1: Parts
\( u = x, dv = e^{-x} dx \)
\( du = dx, v = -e^{-x} \)
Part 2: Limit
Evaluating \( \lim_{b \to \infty} \frac{-b}{e^b} \)
Form: \( \frac{\infty}{\infty} \) → Apply L'H!
Your Turn
\[ \int_{e}^{\infty} \frac{1}{x \ln x} dx \]
Hint 1
Use \( u \)-substitution first where \( u = \ln x \).
Hint 2
Consider the behavior of \( \ln(\ln x) \) as \( x \to \infty \).
Infinity Bounds Worksheet Infinity Bounds Practice
Improper Integrals: Type I
Name:
Date:
Instructions: For each integral below, rewrite the expression using proper limit notation. Determine whether the integral converges or diverges. If it converges, evaluate the exact value. Show all steps, including any use of L'Hopital's Rule or Integration by Parts.
1
\[ \int_{1}^{\infty} \frac{1}{\sqrt[3]{x}} dx \]
2
\[ \int_{0}^{\infty} \frac{1}{x^2 + 1} dx \]
3
Consider: \( \int_{1}^{\infty} xe^{-x^2} dx \)
4
[Advanced] \[ \int_{1}^{\infty} \frac{\ln x}{x^2} dx \]
Hint: Use Integration by Parts. You will likely need L'Hopital's Rule for the limit evaluation.
The Gabriel's Horn Revisited
Recall from the slides that the surface area of the solid formed by rotating \( f(x) = 1/x \) for \( x \ge 1 \) about the x-axis is given by: \[ S = \int_{1}^{\infty} 2\pi \frac{1}{x} \sqrt{1 + \left(-\frac{1}{x^2}\right)^2} dx \] Explain, using the comparison theorem or direct evaluation, why this integral diverges, confirming the "infinite surface area" paradox.
Infinity Bounds Answer Key Answer Key: Infinity Bounds
Teacher Reference Only
1
\[ \int_{1}^{\infty} \frac{1}{\sqrt[3]{x}} dx \]
Result: Diverges
Limit notation: \( \lim_{b \to \infty} \int_{1}^{b} x^{-1/3} dx \)
Antiderivative: \( \lim_{b \to \infty} \left[ \frac{3}{2}x^{2/3} \right]_{1}^{b} = \lim_{b \to \infty} \left( \frac{3}{2}b^{2/3} - \frac{3}{2} \right) = \infty \)
2
\[ \int_{0}^{\infty} \frac{1}{x^2 + 1} dx \]
Result: Converges to \( \pi/2 \)
Limit notation: \( \lim_{b \to \infty} [\arctan(x)]_{0}^{b} \)
Evaluation: \( \lim_{b \to \infty} (\arctan(b) - \arctan(0)) = \frac{\pi}{2} - 0 = \frac{\pi}{2} \)
3
\[ \int_{1}^{\infty} xe^{-x^2} dx \]
Result: Converges to \( 1/(2e) \)
Substitution: \( u = -x^2, du = -2x dx \). Integral becomes \( -\frac{1}{2} \int e^u du \).
Limit notation: \( \lim_{b \to \infty} \left[ -\frac{1}{2}e^{-x^2} \right]_{1}^{b} = \lim_{b \to \infty} \left( -\frac{1}{2e^{b^2}} + \frac{1}{2e} \right) = 0 + \frac{1}{2e} \)
4
\[ \int_{1}^{\infty} \frac{\ln x}{x^2} dx \]
Result: Converges to 1
IBP: \( u = \ln x, dv = x^{-2}dx \implies du = \frac{1}{x}dx, v = -\frac{1}{x} \)
Integral: \( \lim_{b \to \infty} \left[ -\frac{\ln x}{x} - \frac{1}{x} \right]_{1}^{b} \)
Limit check: \( \lim_{b \to \infty} \frac{-\ln b}{b} = (\text{L'H}) \lim_{b \to \infty} \frac{-1/b}{1} = 0 \)
Evaluation: \( (0 - 0) - (0 - 1) = 1 \)
Conceptual Solution: Gabriel's Horn
Since \( \sqrt{1 + 1/x^4} > 1 \) for all \( x \ge 1 \), we can compare: \[ \int_{1}^{\infty} \frac{2\pi}{x} \sqrt{1 + \frac{1}{x^4}} dx > \int_{1}^{\infty} \frac{2\pi}{x} dx \] The integral \( \int_{1}^{\infty} \frac{1}{x} dx \) diverges (it's a p-series with \( p=1 \)). By the comparison theorem, the surface area integral also diverges.
Singular Bounds Slides Singular Bounds
Improper Integrals with Infinite Discontinuities
Calculus BC Type II
Journey to the Center of the Earth
Imagine calculating the work done by gravity as an object moves toward a point mass.
\[ W = \int_{0}^{R} \frac{GMm}{r^2} dr \]
There is a **vertical asymptote** at \( r = 0 \).
Can we define the "work" if the force becomes infinite?
Singularity
Defining Improper Integrals (Type II)
These occur when the function has an infinite discontinuity on \([a, b]\).
At the Endpoints
\[ \int_{a}^{b} f(x) dx = \lim_{c \to a^+} \int_{c}^{b} f(x) dx \]
\[ \int_{a}^{b} f(x) dx = \lim_{c \to b^-} \int_{a}^{c} f(x) dx \]
The Hidden Danger
If the asymptote is at \( x = k \) where \( a < k < b \), you **MUST** split the integral:
\[ \int_{a}^{k} f(x) dx + \int_{k}^{b} f(x) dx \]
Example 1: The "Simple" Trap
\[ \int_{-1}^{2} \frac{1}{x^3} dx \]
Wrong Way:
\( [\frac{-1}{2x^2}]_{-1}^{2} = -\frac{1}{8} - (-\frac{1}{2}) = \frac{3}{8} \)
This is false! The function is undefined at \( x = 0 \).
The Correct Path
Evaluate two limits:
Limit as \( x \to 0^- \)
Limit as \( x \to 0^+ \)
Result: Divergent!
If either half diverges, the whole integral diverges.
Workflow for Type II Integrals
1
Scan
Check the interval for vertical asymptotes. Are they at the ends? In the middle?
2
Limit
Rewrite with limits. Use \( c \to k^+ \) or \( c \to k^- \) as appropriate.
3
Check
Integrate and take the limit. If result is \(\pm\infty\), the integral diverges.
Your Turn
\[ \int_{0}^{1} \frac{1}{\sqrt{1-x}} dx \]
Identifying the Problem
Where is the asymptote? How will you set up your limit?
Singular Bounds Worksheet Singular Bounds Worksheet
Improper Integrals: Type II (Discontinuities)
Name:
Date:
Danger Zone: Unlike Type I integrals, the "impropriety" of Type II integrals is often hidden. Always check for vertical asymptotes within the limits of integration before applying the Fundamental Theorem of Calculus.
1
\[ \int_{0}^{8} \frac{1}{\sqrt[3]{x}} dx \]
Identify the discontinuity and evaluate using limits.
2
\[ \int_{-1}^{1} \frac{1}{x^2} dx \]
Be careful! Split the integral if necessary.
3
\[ \int_{0}^{1} \ln x dx \]
Case Study: The Gravitational Trap
The force of gravity between a point mass \( M \) at the origin and a particle of mass \( m \) is \( F(r) = GMm/r^2 \). Theoretically, the work done to move the particle from a distance \( R \) into the center (distance \( 0 \)) is: \[ W = \int_{0}^{R} \frac{GMm}{r^2} dr \]
A) Evaluate this integral to determine if the work is finite or infinite.
B) Physical Interpretation: Based on your result, why do physicists say this "point mass" model fails at small scales (singularities)?
Singular Bounds Answer Key Answer Key: Singular Bounds
Teacher Reference Only
1
\[ \int_{0}^{8} \frac{1}{\sqrt[3]{x}} dx \]
Result: Converges to 6
Limit: \( \lim_{c \to 0^+} \int_{c}^{8} x^{-1/3} dx \)
Antiderivative: \( \lim_{c \to 0^+} \left[ \frac{3}{2}x^{2/3} \right]_{c}^{8} \)
Evaluation: \( \frac{3}{2}(8)^{2/3} - \frac{3}{2}(0)^{2/3} = \frac{3}{2}(4) - 0 = 6 \)
2
\[ \int_{-1}^{1} \frac{1}{x^2} dx \]
Result: Divergent
Split: \( \int_{-1}^{0} \frac{1}{x^2} dx + \int_{0}^{1} \frac{1}{x^2} dx \)
Right side: \( \lim_{c \to 0^+} \int_{c}^{1} x^{-2} dx = \lim_{c \to 0^+} \left[ -\frac{1}{x} \right]_{c}^{1} = \lim_{c \to 0^+} (-1 + \frac{1}{c}) = \infty \)
Since one part diverges, the whole integral diverges.
3
\[ \int_{0}^{1} \ln x dx \]
Result: Converges to -1
IBP: \( \int \ln x dx = x \ln x - x \)
Limit: \( \lim_{c \to 0^+} [x \ln x - x]_{c}^{1} = (0 - 1) - \lim_{c \to 0^+} (c \ln c - c) \)
L'Hopital: \( \lim_{c \to 0^+} \frac{\ln c}{1/c} = \lim_{c \to 0^+} \frac{1/c}{-1/c^2} = \lim_{c \to 0^+} (-c) = 0 \)
Total: \( -1 - (0 - 0) = -1 \)
Case Study Solution
A) \( W = \lim_{c \to 0^+} \int_{c}^{R} \frac{GMm}{r^2} dr = GMm \lim_{c \to 0^+} [-\frac{1}{r}]_{c}^{R} = GMm (-\frac{1}{R} + \lim_{c \to 0^+} \frac{1}{c}) = \infty \).
B) An infinite amount of work to move to a center implies infinite energy, which is unphysical. This indicates that real objects aren't "point masses" and that general relativity or quantum mechanics must take over at these scales.
Growth Limits Slides Growth Limits
Partial Fractions & Logistic Growth
Calculus BC Modeling
The Outbreak Scenario
In a town of 5,000 people, a "zombie" virus begins to spread.
Growth depends on the number of infected (y).
Growth slows down as everyone gets infected (5,000 - y).
\[ \frac{dy}{dt} = ky(5000 - y) \]
How do we solve for y(t)?
We need Partial Fractions.
Step 1: Separate the Variables
Differential Form
\[ \frac{dy}{y(5000 - y)} = k dt \]
To integrate the left side, we must break the complex fraction into simpler parts that are easy to integrate (natural logs).
The Goal
\[ \frac{1}{y(5000 - y)} = \frac{A}{y} + \frac{B}{5000 - y} \]
The Mechanics of PFD
1. Clear Denominators
\[ 1 = A(5000 - y) + By \]
2. Pick "Smart" Values
If \( y = 0 \): \( 1 = 5000A \implies A = \frac{1}{5000} \)
If \( y = 5000 \): \( 1 = 5000B \implies B = \frac{1}{5000} \)
Integrated Form:
\[ \frac{1}{5000} \int \left( \frac{1}{y} + \frac{1}{5000-y} \right) dy = \int k dt \]
This leads to the familiar logistic solution involving exponentials.
The Solution Profile
Solving for \( y \) gives the Logistic Function:
\[ y(t) = \frac{L}{1 + Ce^{-Lkt}} \]
\( L \) = Carrying Capacity
Max growth rate occurs at \( y = L/2 \)
Limit (L) Rapid Growth
Mental Check
If a forest can support 2,000 deer, and the population is currently 100, at what population size will the deer be multiplying at their fastest rate?
1,000
(\( L/2 \))
Outbreak Logic Worksheet Outbreak Logic Worksheet
Subject: Modeling Logistic Growth via PFD
Name:
Date:
Part 1: Fraction Deconstruction
Decompose the following rational expressions into partial fractions. Show all algebraic steps to find constants \( A \) and \( B \).
\[ \frac{5x - 4}{x(x - 2)} \]
\[ \frac{1}{y(10 - y)} \]
Part 2: Outbreak Simulation
A population of 10,000 residents is exposed to a virus. Let \( P(t) \) be the number of infected individuals. The rate of infection is proportional to both the number of infected people and the number of uninfected people: \[ \frac{dP}{dt} = 0.0001 P(10000 - P) \] Initially, 100 people are infected (\( P(0) = 100 \)).
A) Separate the variables and use partial fraction decomposition to rewrite the integral for \( P \).
B) Integrate both sides and solve for the general solution \( P(t) \).
C) Apply the initial condition \( P(0) = 100 \) to find the constant of integration and the particular solution.
Outbreak Logic Answer Key Answer Key: Outbreak Logic
Teacher Reference Only
Part 1: Fraction Deconstruction
\[ \frac{5x - 4}{x(x - 2)} = \frac{A}{x} + \frac{B}{x-2} \]
Step 1: \( 5x - 4 = A(x-2) + Bx \)
If \( x=0 \): \( -4 = -2A \implies A = 2 \)
If \( x=2 \): \( 6 = 2B \implies B = 3 \)
Final: \( \frac{2}{x} + \frac{3}{x-2} \)
\[ \frac{1}{y(10 - y)} = \frac{A}{y} + \frac{B}{10-y} \]
Step 1: \( 1 = A(10-y) + By \)
If \( y=0 \): \( 1 = 10A \implies A = 1/10 \)
If \( y=10 \): \( 1 = 10B \implies B = 1/10 \)
Final: \( \frac{1}{10y} + \frac{1}{10(10-y)} \)
Part 2: Outbreak Simulation
A) Separation & PFD
\[ \frac{dP}{P(10000-P)} = 0.0001 dt \]
Using PFD constants: \( A=1/10000, B=1/10000 \)
\[ \frac{1}{10000} \int \left( \frac{1}{P} + \frac{1}{10000-P} \right) dP = \int 0.0001 dt \]
\[ \int \left( \frac{1}{P} + \frac{1}{10000-P} \right) dP = \int 1 dt \]
B) Integration
\[ \ln|P| - \ln|10000-P| = t + C \implies \ln\left| \frac{P}{10000-P} \right| = t + C \]
Exponentiate: \( \frac{P}{10000-P} = Ce^t \)
Solving for P: \( P = (10000-P)Ce^t \implies P = 10000Ce^t - PCe^t \)
\[ P(1 + Ce^t) = 10000Ce^t \implies P(t) = \frac{10000Ce^t}{1 + Ce^t} = \frac{10000}{1 + Ke^{-t}} \]
C) Initial Condition \( P(0) = 100 \)
\[ 100 = \frac{10000}{1 + K} \implies 1 + K = 100 \implies K = 99 \]
Particular Solution: \( P(t) = \frac{10000}{1 + 99e^{-t}} \)
Curved Paths Slides Curved Paths
Arc Length & Surface Area of Revolution
Calculus BC Engineering
The Bridge Builder's Problem
A suspension bridge cable follows a curve \( y = \cosh(x) \). How do we find the **exact length** of the cable between two towers?
Arc Length is the accumulation of infinite tiny hypotenuses:
\[ dL = \sqrt{dx^2 + dy^2} \]
Length = ?
The Arc Length Formula
Standard Form
\[ S = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} dx \]
Requires \( f'(x) \) to be continuous on \( [a, b] \).
Note: This formula almost always leads to a square root of a quadratic. This is why we need **Trig Substitution**.
Surface Area of Revolution
Imagine a curve rotated about an axis. The surface area is the sum of the circumferences of thin "bands".
The Formula:
\[ SA = \int_{a}^{b} 2\pi r \cdot dL \]
where \( r = f(x) \) (rotating about x-axis).
Visualizing the Band
Band: 2πr ΔL
The Essential Tool: Trig Sub
Form: \( \sqrt{a^2 - x^2} \)
Use \( x = a \sin \theta \)
Id: \( 1 - \sin^2 = \cos^2 \)
Form: \( \sqrt{a^2 + x^2} \)
Use \( x = a \tan \theta \)
Id: \( 1 + \tan^2 = \sec^2 \)
Form: \( \sqrt{x^2 - a^2} \)
Use \( x = a \sec \theta \)
Id: \( \sec^2 - 1 = \tan^2 \)
The Derivation Challenge
"Derive the surface area of a sphere of radius R by rotating the curve \( y = \sqrt{R^2 - x^2} \) about the x-axis."
Setup Checklist:
1. Find \( dy/dx \)
2. Build the \( \sqrt{1 + (dy/dx)^2} \) radical
3. Set up the \( 2\pi y \cdot dL \) integral
Blueprint Curves Worksheet Blueprint Curves Practice
Arc Length & Surface Area of Revolution
Project:
Engineer:
Arc Length Formula
\[ L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} dx \]
Surface Area (x-axis)
\[ SA = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} dx \]
1
Find the length of the curve \( y = \frac{1}{2}x^2 \) from \( x = 0 \) to \( x = 1 \).
Hint: This will lead to an integral requiring the \( x = \tan \theta \) substitution.
2
Derive the surface area of a sphere of radius \( R \) by rotating the semi-circle \( y = \sqrt{R^2 - x^2} \) from \( x = -R \) to \( x = R \) about the x-axis.
Design Challenge: The Satellite Dish
A satellite dish is formed by rotating the curve \( y = x^2 \) from \( x = 0 \) to \( x = 2 \) about the y-axis. Calculate the surface area of the reflective material needed for the dish.
Note: When rotating about the y-axis, the formula becomes \( SA = \int 2\pi x \sqrt{1 + (dy/dx)^2} dx \).
Blueprint Curves Answer Key Answer Key: Blueprint Curves
Teacher Reference Only
1
Arc Length of \( y = \frac{1}{2}x^2 \) on \( [0, 1] \)
Step 1: \( y' = x \)
Step 2: \( L = \int_{0}^{1} \sqrt{1 + x^2} dx \)
Step 3: Trig Sub \( x = \tan\theta \implies dx = \sec^2\theta d\theta \)
Integral: \( \int \sec^3\theta d\theta = \frac{1}{2}(\sec\theta\tan\theta + \ln|\sec\theta + \tan\theta|) \)
Back-sub: \( \left[ \frac{1}{2}(x\sqrt{1+x^2} + \ln|x + \sqrt{1+x^2}|) \right]_{0}^{1} \)
Final: \( \frac{1}{2}(\sqrt{2} + \ln(1 + \sqrt{2})) \approx 1.148 \)
2
Sphere Surface Area Derivation
Curve: \( y = \sqrt{R^2 - x^2} \). Derivative: \( y' = \frac{-x}{\sqrt{R^2 - x^2}} \)
Radical: \( \sqrt{1 + (y')^2} = \sqrt{1 + \frac{x^2}{R^2-x^2}} = \sqrt{\frac{R^2-x^2+x^2}{R^2-x^2}} = \frac{R}{\sqrt{R^2-x^2}} \)
Setup: \( SA = \int_{-R}^{R} 2\pi y \sqrt{1+(y')^2} dx = \int_{-R}^{R} 2\pi \sqrt{R^2-x^2} \cdot \frac{R}{\sqrt{R^2-x^2}} dx \)
Simplify: \( \int_{-R}^{R} 2\pi R dx = [2\pi R x]_{-R}^{R} = 2\pi R^2 - (-2\pi R^2) \)
Final: \( 4\pi R^2 \)
3
Satellite Dish Surface Area
Curve: \( y = x^2 \), so \( y' = 2x \).
Setup: \( SA = \int_{0}^{2} 2\pi x \sqrt{1 + (2x)^2} dx = \int_{0}^{2} 2\pi x \sqrt{1 + 4x^2} dx \)
u-sub: \( u = 1 + 4x^2, du = 8x dx \implies x dx = \frac{1}{8}du \)
Limits: \( x=0 \to u=1, x=2 \to u=17 \)
Integral: \( 2\pi \cdot \frac{1}{8} \int_{1}^{17} u^{1/2} du = \frac{\pi}{4} [ \frac{2}{3}u^{3/2} ]_{1}^{17} \)
Final: \( \frac{\pi}{6}(17\sqrt{17} - 1) \approx 36.177 \text{ units}^2 \)
Mission Control Slides Mission Control
Case Study: Advanced Synthesis
Classified Calculus Mastery
Anomalous Data Detected
"The thruster fuel tank was designed using an arc length approximation. However, at high pressure, the curve became sharper. Our models diverged, and the tank failed."
Today, we are the forensic mathematicians. We find where the models failed by applying perfect integration.
Scenario A
Calculating Escape Velocity (Improper Integrals + Limits)
Scenario B
Fuel Tank Volume vs. Surface Tension (Trig Sub + Arc Length)
Technique Selection Matrix
Improper
Use when bounds are infinite or vertical asymptotes exist.
Partial Frac
Use for rational functions with factorable denominators.
Trig Sub
Use for radicals of quadratic sums or differences.
Parts
Use for products of transcendental functions.
Which technique would you use for: \( \int \frac{1}{\sqrt{x^2 - 9}} dx \)?
Advanced Case: Escape Velocity
Work required to escape a planet's gravity:
\[ W = \int_{R}^{\infty} \frac{GMm}{r^2} dr \]
Is the integral convergent?
What happens if \( R \to 0 \)?
The Synthesis Point
This connects Lesson 1 (Infinite Bounds) with Lesson 2 (Discontinuities) and Lesson 4 (Physical Application).
The Master Challenge
"A rocket is burning fuel at a rate of \( \frac{dm}{dt} = \frac{-k}{t^2 + a^2} \). Determine the total mass lost as \( t \to \infty \)."
Identify
What kind of integral is this? Which technique is best?
Execute
Does the total mass converge to a finite value?
Mastery Achieved
You have moved from solving textbook problems to modeling the physical universe. Integration is the tool that lets us tame infinity and predict growth.
Mission Control Worksheet Mission Analysis Report
Project: Infinity & Growth Synthesis
Classification: Mastery
Agent ID:
Phase 1: Strategic Selection
Match each physical scenario to the most efficient integration technique. Briefly justify your choice.
Calculating total heat dissipated over infinite time.
Method: ______________________
Reason: _____________________
Finding the surface area of a parabolic thruster nozzle.
Method: ______________________
Reason: _____________________
Modeling the spread of a digital virus in a network of 10k nodes.
Method: ______________________
Reason: _____________________
Phase 2: The Escape Velocity Crisis
A probe of mass \( m \) is launched from the surface of Planet X (radius \( R \)). The work required to move it to a distance \( L \) is \( W = \int_{R}^{L} \frac{GMm}{r^2} dr \).
Calculate the limit of the work as \( L \to \infty \) (Total Escape Work). Does it converge?
Critical Inquiry:
If the energy supply for the probe is finite, is it possible to actually "escape" the planet's gravity? Justify using your result.
Phase 3: The Thruster Optimization
A new thruster nozzle is shaped like the curve \( y = e^{2x} \) from \( x = 0 \) to \( x = 1 \) (units in meters). To ensure thermal integrity, we must calculate the **arc length** of this curve.
Task: Setup the integral and determine which substitution method is required.
"The accumulation of finite data from infinite potential is the hallmark of the master mathematician." - Mission Debrief
Mission Debrief Answer Key Answer Key: Mission Analysis
Teacher Reference Only
Phase 1: Technique Matrix
Heat over infinite time: Improper Integral (Type I). Reason: Upper limit of integration is \(\infty\).
Parabolic thruster surface area: Surface Area Formula + Trig Sub. Reason: Radical of form \(\sqrt{1+4x^2}\).
Network virus spread: Logistic Growth / Partial Fractions. Reason: Growth depends on both infected and uninfected populations.
Phase 2: Escape Velocity
Calculation:
\[ W = \lim_{L \to \infty} \int_{R}^{L} \frac{GMm}{r^2} dr = GMm \lim_{L \to \infty} \left[ -\frac{1}{r} \right]_{R}^{L} \]
\[ = GMm \lim_{L \to \infty} \left( -\frac{1}{L} + \frac{1}{R} \right) = GMm(0 + \frac{1}{R}) = \frac{GMm}{R} \]
Conclusion: It converges!
Yes, it is possible to escape. Even though gravity extends to infinity, the total amount of energy (work) required to leave the field is finite. If the probe has kinetic energy \( > GMm/R \), it escapes.
Phase 3: Thruster Optimization
Setup:
Curve: \( y = e^{2x} \implies y' = 2e^{2x} \)
\[ L = \int_{0}^{1} \sqrt{1 + (2e^{2x})^2} dx = \int_{0}^{1} \sqrt{1 + 4e^{4x}} dx \]
Substitution Strategy:
Let \( u = 2e^{2x} \). Then \( du = 4e^{2x} dx \). This is complex but eventually leads to a **Trig Substitution** where \( u = \tan\theta \), resulting in a \(\sec^3\theta\) form integral.
