June 22, 2026 - Calculus 2- Exam Review Applications of Integration and Differential Equations

Differential Equations: Sugar Mixing and Concentration Analysis

• Initial Setup of the Sugar Mixing Problem:

  • The problem defines $y(t)$ as the amount of sugar in a tank at time $t$, measured in kilograms ($kg$).

  • The inflow rate of water/solution is $2.4\,L/min$.

  • To find the concentration, the amount of sugar $y(t)$ is divided by the total volume. In this specific scenario, a divisor of $255$ is identified (likely derived from volume calculations or specific tank parameters mentioned in the problem context).

  • The rate of change of sugar is established by setting up a differential equation representing inflow minus outflow.

• Solving the Differential Equation:

  • The differential equation must be rewritten to be separable (multiplication/division rather than addition/subtraction).

  • A common denominator is required to combine terms. For example, using $0.24 \times 2 = 61.2$ to find a common numerator/denominator structure.

  • The expression is rearranged to divide both sides by $(61.2 - y)$, facilitating integration.

  • The integration involves the anti-derivative of the left side. Applying the chain rule to the natural log leads to a negative sign: $- \ln|61.2 - y|$.

  • The right side integrates to $\frac{t}{255} + C$.

• Exponential Form and Constants:

  • The negative sign is moved to the other side: $\ln|61.2 - y| = -(\frac{t}{255} + C)$.

  • The equation is converted to exponential form: $61.2 - y = C e^{-\frac{t}{255}}$. The constant $C$ absorbs the sign associated with dropping the absolute value bars.

  • Rearranging for $y(t)$: $y(t) = C e^{-\frac{1}{255}t} + 61.2$.

• Applying Initial Conditions:

  • The initial condition provided is that at $t = 0$, the amount of sugar $y(0) = 0\,kg$.

  • Plugging in values: $0 = C e^0 + 61.2 \rightarrow 0 = C + 61.2$, therefore $C = -61.2$.

  • Final function: $y(t) = -61.2 e^{-\frac{1}{255}t} + 61.2$. This shows that as time passes, the term with the negative exponent gets smaller, meaning less is subtracted from $61.2$, so the total amount of sugar increases.

• Limiting Behavior:

  • The "limiting amount" of sugar is found by taking the limit as $t \to \infty$.

  • As $t$ approaches infinity, the term $-61.2 e^{-\frac{1}{255}t}$ goes to zero.

  • The limit is $61.2\,kg$, which is the maximum amount of sugar the tank will ever contain.

Salt Concentration Mixing Problems

• Parameters for the Salt Problem:

  • Initial volume of solution: $1,000\,L$.

  • Initial salt amount: $100\,kg$.

  • Initial concentration: $\frac{100\,kg}{1,000\,L} = 0.1\,kg/L$.

  • Variable definition: Let $s(t)$ be the amount of salt in the tank at time $t$.

• Inflow and Outflow Rates:

  • Inflow: A concentration of $0.05\,kg/L$ flows in at a rate of $9\,L/min$.

  • Calculation: $0.05\,kg/L \times 9\,L/min = 0.45\,kg/min$.

  • Outflow: The solution leaves the tank at the same rate of $9\,L/min$.

  • Outflow concentration: $\frac{s(t)}{1,000}\,kg/L$.

  • Outflow rate: $\frac{s(t)}{1,000} \times 9 = \frac{9s(t)}{1,000}$.

• Differential Equation Construction:

  • $\frac{ds}{dt} = 0.45 - \frac{9s}{1,000}$.

  • This is solved using the same separation and integration techniques as the sugar problem by finding a common denominator (e.g., $1,000$).

Work: Pumping Water Out of a Container

• Problem Setup (Inverted Cone):

  • Shape: Inverted cone with height $H = 12\,ft$ and base radius $R = 4\,ft$.

  • Scenario: The tank is initially full. Water is pumped out to the top until only $4\,ft$ of water remains in the tank.

  • Work Definition: Work ($W$) is $\int \text{Force} \times d(\text{displacement})$.

  • Coordinate System: The x-axis is vertical, with the origin $(0,0)$ at the top of the cone. The cone extends down to $x = 12$. Because we leave $4\,ft$ at the bottom, we pump water from $x = 0$ to $x = 8$.

• Constants and Geometry:

  • Weight density of water in US units: $62.4\,lb/ft^3$.

  • To find the radius $r$ of a water slice at any depth $x$, use the equation of a line or similar triangles.

  • Points for the line: $(0, 4)$ (top radius) and $(12, 0)$ (bottom vertex).

  • Slope ($m$) calculation: $\frac{0 - 4}{12 - 0} = -\frac{4}{12} = -\frac{1}{3}$.

  • Radius equation: $r = -\frac{1}{3}x + 4$.

• Force and Volume of a Slice:

  • Volume of a thin cylindrical disc: $V = \pi r^2 dx = \pi (-\frac{1}{3}x + 4)^2 dx$.

  • Force: $\text{Weight Density} \times \text{Volume} = 62.4 \pi (-\frac{1}{3}x + 4)^2 dx$.

  • Distance lifted: Since the origin is at the top, the distance each slice at position $x$ must be lifted is simply $x$.

• Integration for Work:

  • $W = \int_0^8 62.4 \pi x (-\frac{1}{3}x + 4)^2 dx$.

  • Substitution ($u$-sub): Let $u = -\frac{1}{3}x + 4$. Then $du = -\frac{1}{3}dx$ (or $dx = -3du$). Solve for $x$: $x = 3(4 - u) = 12 - 3u$.

  • Limits change: When $x = 0, u = 4$. When $x = 8, u = \frac{4}{3}$.

  • The integral becomes: $\int_{4}^{4/3} 62.4 \pi (12 - 3u) u^2 (-3) du$.

  • Simplify and evaluate: Distribution leads to a polynomial that is integrated using the power rule.

Radial Mass Density

• Concept Analysis:

  • Linear mass density is the integral of the density over length.

  • Radial mass density involves a circular object where density changes as you move from the center (radius $0$) outward to a fixed radius $R$.

  • The mass of a thin ring (washer) at radius $x$ is the density at that radius multiplied by the circumference ($2\pi x$) and the width ($dx$).

  • Mass Formula: $M = \int_0^R 2\pi x \cdot \rho(x) dx$.

• Example Calculation:

  • Given density $\rho(x) = e^{-x^2}$. mass $= \int_0^R 2\pi x e^{-x^2} dx$.

  • Using substitution $u = -x^2$, $du = -2x dx$.

  • The integral becomes $\int -\pi e^u du$, which evaluates to $\pi(1 - e^{-R^2})$.

Arc Length and Surface Area of Revolution

• Arc Length:

  • Formula: $L = \int_a^b \sqrt{1 + (\frac{dy}{dx})^2} dx$.

  • Example: Find length of $y = 3x + 5$ from $0$ to $4$.

  • Derivative: $\frac{dy}{dx} = 3$.

  • Integral: $\int_0^4 \sqrt{1 + 3^2} dx = \int_0^4 \sqrt{10} dx$. Result: $4\sqrt{10}$.

• Surface Area of Revolution:

  • Formula: $S = \int_a^b 2\pi f(x) \sqrt{1 + (f'(x))^2} dx$.

  • Example: Rotate $y = 2x + 2$ around the x-axis from $x = 1$ to $x = 2$.

  • $f'(x) = 2$. The radical part is $\sqrt{1 + 2^2} = \sqrt{5}$.

  • Integral: $2\pi \sqrt{5} \int_1^2 (2x + 2) dx = 2\pi \sqrt{5} [x^2 + 2x]_1^2 = 2\pi \sqrt{5} (8 - 3) = 10\pi \sqrt{5}$.

Volume of Solids (Disk and Washer Methods)

• Disk Method:

  • Rotational solids around the x-axis: Volume $V = \int_a^b \pi [f(x)]^2 dx$.

  • Example: $y = 3x^2$ from $0$ to $1$ around x-axis.

  • $V = \int_0^1 \pi (3x^2)^2 dx = \int_0^1 9\pi x^4 dx = [\frac{9\pi x^5}{5}]_0^1 = \frac{9\pi}{5}$.

• Washer Method:

  • Used when rotating the area between two curves, resulting in a hollow center.

  • Volume $V = \int_a^b \pi ([R(x)]^2 - [r(x)]^2) dx$, where $R(x)$ is the outer radius (top function) and $r(x)$ is the inner radius (bottom function).

  • Example: Two lines $y = x + 3$ and $y = x$ from $x = 0$ to $x = 4$ around x-axis.

  • Identify Outer/Inner: $R = x + 3$, $r = x$.

  • Integral: $\pi \int_0^4 ((x+3)^2 - x^2) dx = \pi \int_0^4 (x^2 + 6x + 9 - x^2) dx = \pi \int_0^4 (6x + 9) dx$.

  • Result: $\pi [3x^2 + 9x]_0^4 = \pi (48 + 36) = 84\pi$.

Advanced Separable Equations and Partial Fraction Decomposition

• General Technique:

  • Separable equations take the form $\frac{dy}{dx} = g(x)h(y)$. Move all $y$ terms to one side and $x$ terms to the other.

  • Example Equation: $\frac{dy}{dx} = (x^2 + 1)(y - 4)(y + 2)$.

  • Rearrange: $\int \frac{1}{(y - 4)(y + 2)} dy = \int (x^2 + 1) dx$.

  • Partial Fraction Decomposition (PFD):

    • $\frac{1}{(y-4)(y+2)} = \frac{A}{y-4} + \frac{B}{y+2}$.

    • Multiply by denominator: $1 = A(y+2) + B(y-4)$.

    • If $y = 4, 1 = 6A \rightarrow A = 1/6$.

    • If $y = -2, 1 = -6B \rightarrow B = -1/6$.

    • Integrant: $\frac{1}{6} \ln|y-4| - \frac{1}{6} \ln|y+2| = \frac{1}{6} \ln|\frac{y-4}{y+2}|$.

• Solving for Initial Value:

  • Condition: $y(0) = 8$.

  • Integrate right side: $\frac{1}{3}x^3 + x + C$.

  • Resulting equation: $\frac{1}{6} \ln|\frac{y-4}{y+2}| = \frac{x^3}{3} + x + C$.

  • Algebraic manipulation (multiplying by 6, exponentiating) allows one to solve explicitly for $y$.

Questions & Discussion

• Question: Is the test on Wednesday or Thursday?

• Response: The test is on Wednesday. The first review was Thursday, which might have caused the confusion.

• Question: Will there be hydrostatic pressure or spring problems on the test?

• Response: There will be no hydrostatic pressure problems. There will likely be no spring math on the test either.

• Question: On the pumping water problem, why is the radius term not $2x$?

• Response: $2x$ is typically used when looking for surface area in certain contexts, but for volume discs in an inverted cone, the radius is calculated based on the linear slope of the cone wall.

• Question: Are the books linked in the syllabus always free?

• Response: Yes, the books linked are free. One student noted they previously had to buy an expensive book for Economics, which contrasts with this course's free resources.