AP Calculus BC Review 2022: Session 4

Hey guys! Welcome back to another AP Calculus BC review session. In this fourth session, we're diving deep into some crucial topics to help you ace that exam. Let's get started!

Understanding Series Convergence and Divergence

Series convergence and divergence are foundational concepts in Calculus BC. You absolutely need to nail these down. So, what's the deal? A series converges if the sequence of its partial sums approaches a finite limit. Otherwise, it diverges. Easy enough, right? Well, the devil's in the details.

Convergence Tests: The Bread and Butter

To determine whether a series converges or diverges, we rely on a variety of tests. Here are some of the most important ones:

1. The Integral Test: This test compares a series to an integral. If ${\int_{1}^{\infty} f(x) dx}$ converges, then ${\sum_{n=1}^{\infty} a_n}$ converges, provided that ${f(x)}$ is continuous, positive, and decreasing. Basically, if the area under the curve is finite, the sum is finite too.

2. The Comparison Test: This one's all about, well, comparison! If ${0 \leq a_n \leq b_n}$ for all ${n}$ and ${\sum b_n}$ converges, then ${\sum a_n}$ converges. Similarly, if ${a_n \geq b_n \geq 0}$ and ${\sum b_n}$ diverges, then ${\sum a_n}$ diverges. It's like saying, "If something smaller converges, the bigger thing might too, and if something smaller diverges, so does the bigger thing."

3. The Limit Comparison Test: A more robust version of the comparison test. If ${\lim_{n \to \infty} \frac{a_n}{b_n} = c}$, where ${0 < c < \infty}$, then ${\sum a_n}$ and ${\sum b_n}$ either both converge or both diverge. This test is super handy when you have series that behave similarly but aren't directly comparable.

4. The Ratio Test: This test is fantastic for series involving factorials or exponentials. Calculate ${L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|}$. If ${L < 1}$, the series converges absolutely. If ${L > 1}$, the series diverges. If ${L = 1}$, the test is inconclusive. Think of it as checking if the terms are shrinking fast enough to converge.

5. The Root Test: Similar to the ratio test, but you take the nth root. Compute ${L = \lim_{n \to \infty} \sqrt[n]{|a_n|}}$. Again, if ${L < 1}$, the series converges absolutely; if ${L > 1}$, the series diverges; and if ${L = 1}$, the test is inconclusive. This is great for series where the entire term is raised to the nth power.

6. The Alternating Series Test: This test applies to alternating series (where the signs alternate). An alternating series ${\sum_{n=1}^{\infty} (-1)^n b_n}$ converges if ${b_n}$ is decreasing and ${\lim_{n \to \infty} b_n = 0}$. Basically, the terms must shrink to zero to ensure convergence.

Absolute vs. Conditional Convergence

Now, let's talk about absolute and conditional convergence. A series ${\sum a_n}$ converges absolutely if ${\sum |a_n|}$ converges. If ${\sum a_n}$ converges but ${\sum |a_n|}$ diverges, then ${\sum a_n}$ converges conditionally. Absolute convergence is stronger; it implies convergence regardless of the signs of the terms.

Examples to Solidify Understanding

Let's solidify these concepts with examples:

• Example 1: Determine whether ${\sum_{n=1}^{\infty} \frac{1}{n^2}}$ converges or diverges.

  Solution: This is a p-series with ${p = 2 > 1}$, so it converges by the p-series test.

• Example 2: Determine whether ${\sum_{n=1}^{\infty} \frac{1}{n}}$ converges or diverges.

  Solution: This is a p-series with ${p = 1}$, so it diverges (harmonic series).

• Example 3: Determine whether ${\sum_{n=1}^{\infty} \frac{(-1)^n}{n}}$ converges or diverges.

  Solution: This is an alternating series with ${b_n = \frac{1}{n}}$. Since ${\frac{1}{n}}$ is decreasing and ${\lim_{n \to \infty} \frac{1}{n} = 0}$, it converges by the alternating series test. However, ${\sum_{n=1}^{\infty} \left| \frac{(-1)^n}{n} \right| = \sum_{n=1}^{\infty} \frac{1}{n}}$ diverges, so it converges conditionally.

Understanding these tests and being able to apply them correctly is crucial for the AP exam. Practice makes perfect, so work through as many problems as you can!

Power Series and Taylor/Maclaurin Series

Power series and Taylor/Maclaurin series are essential tools in Calculus BC, allowing us to represent functions as infinite polynomials. This representation can be incredibly useful for approximating function values, solving differential equations, and more.

What is a Power Series?

A power series is an infinite series of the form:

${\sum_{n=0}^{\infty} c_n (x - a)^n = c_0 + c_1(x - a) + c_2(x - a)^2 + c_3(x - a)^3 + \cdots}$

where:

• ${x}$ is a variable.
• ${c_n}$ are the coefficients.
• ${a}$ is the center of the series.

The power series converges for certain values of ${x}$ and diverges for others. The set of ${x}$-values for which the series converges is called the interval of convergence.

Radius and Interval of Convergence

To find the radius of convergence ${R}$, we typically use the ratio test:

${L = \lim_{n \to \infty} \left| \frac{c_{n+1}(x - a)^{n+1}}{c_n(x - a)^n} \right| = |x - a| \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right|}$

For the series to converge, we need ${L < 1}$, so:

${|x - a| < \frac{1}{\lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right|}}$

Thus, the radius of convergence is:

${R = \frac{1}{\lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right|}}$

(If the limit is 0, then ${R = \infty}$, and if the limit is ${\infty}$, then ${R = 0}$).

The interval of convergence is the interval ${(a - R, a + R)}$, but you must also check the endpoints ${x = a - R}$ and ${x = a + R}$ to see if the series converges at these points.

Taylor and Maclaurin Series: Representing Functions

A Taylor series represents a function ${f(x)}$ as a power series centered at ${x = a}$:

${f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n}$

where ${f^{(n)}(a)}$ is the nth derivative of ${f(x)}$ evaluated at ${x = a}$.

A Maclaurin series is a special case of the Taylor series where the center is at ${x = 0}$:

${f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n}$

Common Maclaurin Series to Know

Memorizing these common Maclaurin series will save you time on the AP exam:

• ${e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots}$
• ${\sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots}$
• ${\cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots}$
• ${\frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots}$ (geometric series, ${|x| < 1}$)

Manipulating Power Series

You can manipulate power series to find new series. Common techniques include:

• Substitution: Replace ${x}$ with another expression (e.g., ${x^2}$ or ${-x}$).
• Differentiation: Differentiate a known series term-by-term.
• Integration: Integrate a known series term-by-term.
• Multiplication: Multiply a series by a constant or another series.

Examples to Illustrate the Concepts

Let's look at some examples:

• Example 1: Find the Maclaurin series for ${f(x) = e^{2x}}$.

  Solution: Using the Maclaurin series for ${e^x}$, substitute ${2x}$ for ${x}$:

  ${e^{2x} = \sum_{n=0}^{\infty} \frac{(2x)^n}{n!} = \sum_{n=0}^{\infty} \frac{2^n x^n}{n!}}$

• Example 2: Find the Taylor series for ${f(x) = \sin(x)}$ centered at ${a = \frac{\pi}{2}}$.

  Solution: We need to find the derivatives of ${\sin(x)}$ and evaluate them at ${x = \frac{\pi}{2}}$:

  ${f(x) = \sin(x), \quad f(\frac{\pi}{2}) = 1}$

  ${f'(x) = \cos(x), \quad f'(\frac{\pi}{2}) = 0}$

  ${f''(x) = -\sin(x), \quad f''(\frac{\pi}{2}) = -1}$

  ${f'''(x) = -\cos(x), \quad f'''(\frac{\pi}{2}) = 0}$

  The Taylor series is:

  ${\sin(x) = 1 - \frac{(x - \frac{\pi}{2})^2}{2!} + \frac{(x - \frac{\pi}{2})^4}{4!} - \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n (x - \frac{\pi}{2})^{2n}}{(2n)!}}$

Understanding and manipulating power series, along with recognizing common Maclaurin series, is crucial for success on the AP Calculus BC exam.

Vector-Valued Functions

Vector-valued functions are functions that map real numbers to vectors. These functions are particularly useful for describing motion in space. Let's dive into the details.

Definition and Representation

A vector-valued function ${\mathbf{r}(t)}$ in two dimensions is given by:

${\mathbf{r}(t) = \langle x(t), y(t) \rangle = x(t) \mathbf{i} + y(t) \mathbf{j}}$

and in three dimensions:

${\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle = x(t) \mathbf{i} + y(t) \mathbf{j} + z(t) \mathbf{k}}$

where ${x(t)}$, ${y(t)}$, and ${z(t)}$ are scalar functions of the parameter ${t}$.

Calculus with Vector-Valued Functions

Derivatives

The derivative of a vector-valued function ${\mathbf{r}(t)}$ is found by differentiating each component:

${\mathbf{r}'(t) = \langle x'(t), y'(t), z'(t) \rangle}$

The derivative ${\mathbf{r}'(t)}$ represents the tangent vector to the curve traced by ${\mathbf{r}(t)}$. The magnitude of the derivative, ${|\mathbf{r}'(t)|}$, gives the speed of the particle along the curve.

Integrals

The integral of a vector-valued function ${\mathbf{r}(t)}$ is found by integrating each component:

${\int \mathbf{r}(t) dt = \langle \int x(t) dt, \int y(t) dt, \int z(t) dt \rangle}$

Velocity, Acceleration, and Speed

If ${\mathbf{r}(t)}$ represents the position of a particle at time ${t}$, then:

• Velocity is ${\mathbf{v}(t) = \mathbf{r}'(t)}$.
• Acceleration is ${\mathbf{a}(t) = \mathbf{v}'(t) = \mathbf{r}''(t)}$.
• Speed is ${|\mathbf{v}(t)| = |\mathbf{r}'(t)|}$.

Arc Length

The arc length ${L}$ of a curve traced by ${\mathbf{r}(t)}$ from ${t = a}$ to ${t = b}$ is given by:

${L = \int_a^b |\mathbf{r}'(t)| dt = \int_a^b \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2} dt}$

Examples to Illustrate Concepts

Let's consider some examples:

• Example 1: Find the velocity, acceleration, and speed of a particle whose position is given by ${\mathbf{r}(t) = \langle t^2, \sin(t), \cos(t) \rangle}$.

  Solution:

  • Velocity: ${\mathbf{v}(t) = \mathbf{r}'(t) = \langle 2t, \cos(t), -\sin(t) \rangle}$
  • Acceleration: ${\mathbf{a}(t) = \mathbf{v}'(t) = \langle 2, -\sin(t), -\cos(t) \rangle}$
  • Speed: ${|\mathbf{v}(t)| = \sqrt{(2t)^2 + (\cos(t))^2 + (-\sin(t))^2} = \sqrt{4t^2 + 1}}$

• Example 2: Find the arc length of the curve ${\mathbf{r}(t) = \langle \cos(t), \sin(t), t \rangle}$ from ${t = 0}$ to ${t = 2\pi}$.

  Solution:

  • ${\mathbf{r}'(t) = \langle -\sin(t), \cos(t), 1 \rangle}$
  • ${|\mathbf{r}'(t)| = \sqrt{(-\sin(t))^2 + (\cos(t))^2 + 1^2} = \sqrt{\sin^2(t) + \cos^2(t) + 1} = \sqrt{2}}$
  • ${L = \int_0^{2\pi} \sqrt{2} dt = \sqrt{2} \int_0^{2\pi} dt = \sqrt{2} [t]_0^{2\pi} = 2\pi \sqrt{2}}$

Understanding vector-valued functions is essential for problems involving motion and curves in space. Pay close attention to the derivatives, integrals, and their applications.

Polar Coordinates and Polar Curves

Polar coordinates and polar curves provide an alternative way to describe points and curves in the plane, especially useful when dealing with circular symmetry. Let's explore these concepts in detail.

Introduction to Polar Coordinates

In polar coordinates, a point ${P}$ in the plane is represented by ${(r, \theta)}$, where:

• ${r}$ is the distance from the origin to the point ${P}$ (the radius).
• ${\theta}$ is the angle measured counterclockwise from the positive x-axis to the line segment connecting the origin to ${P}$.

Conversion Between Polar and Cartesian Coordinates

To convert between polar coordinates ${(r, \theta)}$ and Cartesian coordinates ${(x, y)}$, we use the following relationships:

• ${x = r \cos(\theta)}$
• ${y = r \sin(\theta)}$
• ${r^2 = x^2 + y^2}$
• ${\tan(\theta) = \frac{y}{x}}$ (if ${x \neq 0}$)

Polar Curves

A polar curve is a curve defined by an equation in polar coordinates, typically in the form ${r = f(\theta)}$.

Common Polar Curves

• Circles: ${r = a}$ (circle centered at the origin with radius ${a}$)
• Lines: ${\theta = c}$ (line through the origin making an angle ${c}$ with the x-axis)
• Cardioids: ${r = a \pm a \cos(\theta)}$ or ${r = a \pm a \sin(\theta)}$ (heart-shaped curves)
• Limacons: ${r = a \pm b \cos(\theta)}$ or ${r = a \pm b \sin(\theta)}$ (curves with an inner loop if ${a < b}$)
• Roses: ${r = a \cos(n\theta)}$ or ${r = a \sin(n\theta)}$ (petal-shaped curves)

Calculus with Polar Curves

Slope of a Polar Curve

To find the slope ${\frac{dy}{dx}}$ of a polar curve ${r = f(\theta)}$, we use the parametric equations:

${x = r \cos(\theta) = f(\theta) \cos(\theta)}$

${y = r \sin(\theta) = f(\theta) \sin(\theta)}$

Then,

${\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{f'(\theta) \sin(\theta) + f(\theta) \cos(\theta)}{f'(\theta) \cos(\theta) - f(\theta) \sin(\theta)}}$

Area Enclosed by a Polar Curve

The area ${A}$ enclosed by a polar curve ${r = f(\theta)}$ from ${\theta = a}$ to ${\theta = b}$ is given by:

${A = \frac{1}{2} \int_a^b [f(\theta)]^2 d\theta = \frac{1}{2} \int_a^b r^2 d\theta}$

Arc Length of a Polar Curve

The arc length ${L}$ of a polar curve ${r = f(\theta)}$ from ${\theta = a}$ to ${\theta = b}$ is given by:

${L = \int_a^b \sqrt{r^2 + (\frac{dr}{d\theta})^2} d\theta = \int_a^b \sqrt{[f(\theta)]^2 + [f'(\theta)]^2} d\theta}$

Examples to Illustrate Concepts

Let's consider a couple of examples:

• Example 1: Find the slope of the polar curve ${r = 2 \cos(\theta)}$ at ${\theta = \frac{\pi}{4}}$.

  Solution:

  • ${f(\theta) = 2 \cos(\theta)}$, so ${f'(\theta) = -2 \sin(\theta)}$

  • ${\frac{dy}{dx} = \frac{f'(\theta) \sin(\theta) + f(\theta) \cos(\theta)}{f'(\theta) \cos(\theta) - f(\theta) \sin(\theta)} = \frac{-2 \sin(\theta) \sin(\theta) + 2 \cos(\theta) \cos(\theta)}{-2 \sin(\theta) \cos(\theta) - 2 \cos(\theta) \sin(\theta)}}$

  • At ${\theta = \frac{\pi}{4}}$, ${\sin(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}}$, so

    ${\frac{dy}{dx} = \frac{-2(\frac{\sqrt{2}}{2})^2 + 2(\frac{\sqrt{2}}{2})^2}{-2(\frac{\sqrt{2}}{2})(\frac{\sqrt{2}}{2}) - 2(\frac{\sqrt{2}}{2})(\frac{\sqrt{2}}{2})} = \frac{-1 + 1}{-1 - 1} = \frac{0}{-2} = 0}$

• Example 2: Find the area enclosed by the polar curve ${r = 2 + 2 \cos(\theta)}$.

  Solution:

  • ${A = \frac{1}{2} \int_0^{2\pi} [2 + 2 \cos(\theta)]^2 d\theta = \frac{1}{2} \int_0^{2\pi} (4 + 8 \cos(\theta) + 4 \cos^2(\theta)) d\theta}$

  • Using the identity ${\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}}$:

    ${A = \frac{1}{2} \int_0^{2\pi} (4 + 8 \cos(\theta) + 4(\frac{1 + \cos(2\theta)}{2})) d\theta = \frac{1}{2} \int_0^{2\pi} (6 + 8 \cos(\theta) + 2 \cos(2\theta)) d\theta}$

    ${A = \frac{1}{2} [6\theta + 8 \sin(\theta) + \sin(2\theta)]_0^{2\pi} = \frac{1}{2} [6(2\pi) + 0 + 0 - (0 + 0 + 0)] = 6\pi}$

Understanding polar coordinates and polar curves is essential for certain types of problems, especially those involving symmetry about the origin.

Okay, guys, that wraps up this review session! Make sure to practice these concepts, and you'll be well on your way to acing that AP Calculus BC exam. Keep up the great work, and I'll catch you in the next session!