MATH 1560 Final Exam Study Guide

MATH:1560 Final Exam Overview and Point Distribution

• The final exam for MATH:1560 is worth a total of $200$ points.

• Exam 1 Content (30 points): Consists of $6$ Multiple Choice Questions covering:     - Convergence Test     - Approximate area under curve     - Definite Integral geometrically     - Geometric Series     - Sequence Convergence     - Taylor polynomials

• Exam 2 Content (30 points): Consists of $6$ Multiple Choice Questions covering:     - Integration by parts     - Partial Fractions     - Trig Sub     - Area Between Curves     - Improper Integral     - Probability

• Chapter 8 Content (60 points): Consists of $12$ Multiple Choice Questions covering:     - Work (2 questions)     - Parametrized curves     - Projections     - Cross Product     - Arc Length (Parametric or Polar)     - Partial Derivative     - Derivatives in Polar Coordinates     - Polar Coordinate Conversion     - Vector Valued Functions     - Surface Area     - Vertical or Horizontal Definite Integral

• Free Response Section (80 points total): There are four free-response questions, each worth $20$ points:     - Power Series ($20$ points)     - Arc Length ($20$ points)     - Center of Mass ($20$ points)     - Area in Polar Coordinates ($20$ points)

Sequence and Series Convergence Analysis

• Direct Comparison Test Application: If the condition $0 \le a_n \le \frac{3}{n^2 + 1}$ holds for all $n$, the Direct Comparison Test is used to determine the convergence of $\sum_{n=1}^{\infty} a_n$. Since $\sum_{n=1}^{\infty} \frac{3}{n^2}$ is a convergent p-series (p = 2 > 1), and \frac{3}{n^2+1} < \frac{3}{n^2}, the series $\sum_{n=1}^{\infty} a_n$ must also converge.

• Sequence Convergence Evaluation: For the sequence $a_n = \frac{5n - 2}{8n + 7}$, convergence is determined by calculating the limit as $n \rightarrow \infty$:     - $\lim_{n \to \infty} \frac{5n - 2}{8n + 7} = \frac{5}{8}$     - The sequence converges to $\frac{5}{8}$.

• Geometric Series Summation: For the series $\sum_{n=0}^{\infty} \frac{7}{3^n}$, it is a geometric series where the first term $a = 7$ and the common ratio $r = \frac{1}{3}$. Since |r| < 1, the series converges. The sum is calculated as:     - S = \frac{a}{1 - r} = \frac{7}{1 -
  \frac{1}{3}} = \frac{7}{
  \frac{2}{3}} = \frac{21}{2}

• Divergence Test / Comparison: To determine the convergence of $\sum_{n=1}^{\infty} \frac{n + 4}{n^2 + 1}$, one can use the Limit Comparison Test with the harmonic series $b_n = \frac{1}{n}$. This series behaves like $\frac{1}{n}$ at infinity and thus diverges.

Power Series and Taylor Polynomials

• Radius of Convergence Calculation: For the power series $\sum_{n=1}^{\infty} n \u0001\frac{x}{4}\u0002^n$, the radius of convergence $R$ is found using the Ratio Test:     - $L = \lim_{n \to \infty} \left| \frac{(n+1)x^{n+1}}{4^{n+1}} \times \frac{4^n}{nx^n} \right| = \frac{|x|}{4}$     - Set L < 1 \rightarrow \frac{|x|}{4} < 1 \rightarrow |x| < 4. Thus, $R = 4$.

• Interval of Convergence Evaluation: For the series $\sum_{n=1}^{\infty} \frac{(x - 2)^n}{n 5^n}$, the interval is found by the Ratio Test:     - $L = \lim_{n \to \infty} \left| \frac{(x-2)^{n+1}}{(n+1)5^{n+1}} \times \frac{n 5^n}{(x-2)^n} \right| = \frac{|x-2|}{5}$     - Centralized at $x=2$ with radius $5$. Boundaries are at $x = -3$ and $x = 7$.     - At $x = 7$, the series is the harmonic series (diverges).     - At $x = -3$, the series is the alternating harmonic series (converges).     - Interval: $[-3, 7)$.

• Power Series Representations:     - For $\frac{1}{(1-x)^2}$, one can differentiate the known series $\frac{1}{1-x} = \sum x^n$ term-by-term.     - For $\int_0^x \frac{1}{1+t^2} dt$, which is $\arctan(x)$, the representation is derived from the integration of the series for $\frac{1}{1+t^2} = \sum_{n=0}^{\infty} (-1)^n t^{2n}$, resulting in $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}$.

• Maclaurin and Taylor Polynomials:     - For $e^{-x^2}$, the first four nonzero terms of the Maclaurin series are $1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6}$.     - Degree 3 Taylor Polynomial ($T_3(x)$) centered at $0$ for a function where $f(0) = 3$, $f'(0) = 2$, $f''(0) = -4$, and $f^{(3)}(0) = 12$:         - $T_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3$         - $T_3(x) = 3 + 2x - 2x^2 + 2x^3$

Riemann Sums and Work Integrals

• Area Approximations:     - Left-endpoint Riemann sum for $y = x^2 + 1$ on $[0, 4]$ with $n = 4$ ($\Delta x = 1$): $1 \times [f(0) + f(1) + f(2) + f(3)] = 1 \times [1 + 2 + 5 + 10] = 18$.     - Midpoint approximation for $\int_1^5 \sqrt{x} dx$ with $4$ rectangles ($\Delta x = 1$): $\sum_{i=1}^4 \sqrt{i + 0.5}$, specifically $f(1.5) + f(2.5) + f(3.5) + f(4.5)$.

• Work Scenarios:     - Cable and Crate: A crate weighing $120$ lbs is lifted $40$ ft by a cable weighing $1.5$ lb/ft. The work integral is $W = \int_0^{40} (120 + 1.5(40 - y)) dy$.     - Rocket Weigh: Weight decreases linearly from $500$ lbs at $y=0$ to $420$ lbs at $y=800$. The weight function is $w(y) = 500 - \frac{1}{10}y$. The work integral is $W = \int_0^{800} (500 - 0.1y) dy$.     - Pumping Water: A cylindrical tank (radius $3$, height $10$) full of water ($62.4$ lb/ft). The work to pump all water to the top: $W = \int_0^{10} \pi(3)^2 (62.4)(10 - y) dy$.

Vector Operations and Geometry

• Projections and Products:     - Given $\mathbf{u} = \langle 2, -1 \rangle$ and $\mathbf{v} = \langle 3, 4 \rangle$, the projection of $\mathbf{u}$ onto $\mathbf{v}$ is $\text{proj}<em>{\mathbf{v}}(\mathbf{u}) = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|^2} \mathbf{v}$.         - $\mathbf{u} \cdot \mathbf{v} = (2)(3) + (-1)(4) = 2$         - $|\mathbf{v}|^2 = 3^2 + 4^2 = 25$         - $\text{proj}</em>{\mathbf{v}}(\mathbf{u}) = \frac{2}{25} \langle 3, 4 \rangle = \langle \frac{6}{25}, \frac{8}{25} \rangle$     - Dot product of $\langle 1, 2, 3 \rangle$ and $\langle -1, 4, 0 \rangle$: $(1)(-1) + (2)(4) + (3)(0) = 7$.     - Cross product of $\langle 2, 1, -1 \rangle \times \langle 1, 3, 2 \rangle$ results in $\langle 5, -5, 5 \rangle$.

• Lines and Planes:     - Angle between $\langle 1, 1, 0 \rangle$ and $\langle 1, 0, 1 \rangle$ is found using $\cos(\theta) = \frac{1}{\sqrt{2} \sqrt{2}} = \frac{1}{2}$, so $\theta = \frac{\pi}{3}$.     - Vector equation of a line through $(1, -2, 3)$ parallel to $\langle 2, 5, -1 \rangle$: $\mathbf{r}(t) = \langle 1, -2, 3 \rangle + t \langle 2, 5, -1 \rangle$.     - Plane equation through $(1, 0, 2)$ with normal $\langle 3, -1, 4 \rangle$: $3(x - 1) - 1(y - 0) + 4(z - 2) = 0$, which simplifies to $3x - y + 4z = 11$.

Parametric and Polar Calculus

• Parametric Derivatives: For curve $x = t^3 + 1$ and $y = t^2 - 4t$, the derivative is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t - 4}{3t^2}$.

• Arc Length of Parametric Curves: For $x = t^2, y = 2t^3$ for $0 \le t \le 1$, the integral is $L = \int_0^1 \sqrt{(2t)^2 + (6t^2)^2} dt = \int_0^1 \sqrt{4t^2 + 36t^4} dt$.

• Polar Coordinates:     - Polar to Cartesian: $r = 4 \sin(\theta)$ becomes $r^2 = 4r \sin(\theta) \rightarrow x^2 + y^2 = 4y$.     - Slope of polar curve $r = 1 + 2 \cos(\theta)$ at $\theta = \frac{\pi}{3}$: Requires \frac{dy}{dx} = \frac{
  dr/d\theta \sin(\theta) + r \cos(\theta)}{
  dr/d\theta \cos(\theta) - r \sin(\theta)}.

• Polar Area Integration:     - Area inside $r = 3 - 3 \cos(\theta)$: $A = \int_0^{2\pi} \frac{1}{2} (3 - 3 \cos(\theta))^2 d\theta$.     - Shared area between circles $r = 2$ and $r = 2 \cos(\theta)$, finding intersection points to establish integration limits.

Advanced Integration and Multivariable Calculus

• Integration Methods:     - Integration by Parts (IBP): $\int x e^{2x} dx$ and $\int x \cos(4x) dx$.     - Partial Fractions: $\int \frac{1}{x^2 - 4} dx$ and $\int \frac{3x + 5}{x^2 + x - 2} dx$.     - Trigonometric Substitution: For $\int \frac{x^2}{\sqrt{1 - x^2}} dx$, use $x = \sin(\theta)$, $dx = \cos(\theta) d\theta$, resulting in $\int \frac{\sin^2(\theta)}{\cos(\theta)} \cos(\theta) d\theta = \int \sin^2(\theta) d\theta$.

• Improper Integrals:     - $\int_1^{\infty} \frac{1}{x^{3/2}} dx$: Converges to $2$.     - $\int_0^1 \frac{1}{\sqrt{x}} dx$: Converges to $2$.

• Partial Derivatives:     - For $f(x, y) = x^3 y^2 + e^x \cos(y)$, the partial derivative with respect to $x$ is $f_x = 3x^2 y^2 + e^x \cos(y)$.     - For $f(x, y) = x^2 e^y \sin(y)$, finding $f_{xy}$ requires differentiating $f_x = 2x e^y \sin(y)$ with respect to $y$: $f_{xy} = 2x (e^y \sin(y) + e^y \cos(y))$.

Area, Surface Area, Mass, and Probability

• Geometric Area:     - Between $y = x^2$ and $y = 2x$: Integate $2x - x^2$ from $0$ to $2$.     - Between $x = y^3$ and $y = x$: Integrate $y^{1/3} - y$ or $x - x^3$ depending on intersection boundaries.

• Surface Area of Rotation:     - For $y = \sqrt{x + 1}$ on $[0, 3]$ around x-axis: $S = \int_0^3 2\pi \sqrt{x+1} \sqrt{1 + \u0001\frac{1}{2\sqrt{x+1}}\u0002^2} dx$.     - For $y = \ln(x)$ on $[1, e^2]$ about x-axis (with respect to $y$): Since $x = e^y$, the range is $0 \le y \le 2$; $S = \int_0^2 2\pi y \sqrt{1 + (e^y)^2} dy$.

• Mass and Moments:     - For region bounded by $y = 3x$ and $y = x^2$ on $[0, 3]$ with density $\rho(x) = 2x$:         - Mass $M = \int_0^3 2x(3x - x^2) dx$.         - Moment about y-axis $M_y = \int_0^3 x(2x)(3x - x^2) dx$.         - Moment about x-axis $M_x = \int_0^3 \frac{1}{2} (2x) ((3x)^2 - (x^2)^2) dx = \int_0^3 x (9x^2 - x^4) dx$ .

• Probability Distributions:     - Verification of PDF: For $f(x) = 6x(1 - x)$ on $[0, 1]$, checking $\int_0^1 f(x) dx = 1$ is necessary.     - Variance: For density $f(x) = \frac{3}{8}x^2$ on $[0, 2]$, variance is calculated as $V(X) = E[X^2] - (E[X])^2$.         - $E[X] = \int_0^2 x \cdot \frac{3}{8}x^2 dx = \int_0^2 \frac{3}{8}x^3 dx = \left[ \frac{3}{32}x^4 \right]_0^2 = \frac{48}{32} = 1.5$.         - $E[X^2] = \int_0^2 x^2 \cdot \frac{3}{8}x^2 dx = \int_0^2 \frac{3}{8}x^4 dx = \left[ \frac{3}{40}x^5 \right]_0^2 = \frac{96}{40} = 2.4$.         - Variance $V(x) = 2.4 - (1.5)^2 = 2.4 - 2.25 = 0.15$.