AP Calculus BC

Lesson 1.1 · Unit 1 · CED topic 1.1

Can change occur at an instant?

You already know how to measure change over an interval: pick two points and find the slope between them. Calculus asks a stranger question: how fast is something changing right now, at a single instant, when there's no interval at all?

To get at an instant, shrink the interval. The trouble is that a zero-width interval gives \(\frac{0}{0}\), which is meaningless. The way out is to watch what the average rate approaches as the interval shrinks, and that idea is a limit.

Lesson 1.2 · Unit 1 · CED topic 1.2

Defining limits and limit notation

A limit describes where a function is heading, not where it is. That distinction is the whole point: the value of the function at the target, or whether it even has one, is irrelevant.

Limits also come in one-sided versions. \(\displaystyle\lim_{x \to c^-} f(x)\) approaches from the left, \(\displaystyle\lim_{x \to c^+} f(x)\) from the right. The two-sided limit exists only when both one-sided limits exist and agree.

Lesson 1.3 · Unit 1 · CED topics 1.3–1.4

Estimating limits from graphs and tables

Before you can compute limits algebraically, you need to be able to read them. On the exam, you'll be handed graphs and tables and asked what a function is approaching.

From a graph: trace the curve toward \(x = c\) from each side and read off the \(y\)-value the branches approach. An open circle at \(x = c\) means the function is undefined there: the limit is still the height the branches point to. A filled dot somewhere else is the actual value \(f(c)\); it doesn't affect the limit.

Lesson 1.4 · Unit 1 · CED topics 1.5–1.7

Limit laws and algebraic manipulation

Most limits you'll meet can be found exactly, with algebra. The strategy has two steps: try direct substitution, and if that gives the indeterminate form \(\frac{0}{0}\), rewrite the expression until substitution works.

Lesson 1.5 · Unit 1 · CED topic 1.8

The Squeeze Theorem

Some functions are too wild to handle directly: they oscillate infinitely often near the target. The Squeeze Theorem lets you trap such a function between two tame ones and let the tame ones do the work.

On the exam, the squeeze is usually set up for you: a problem will state inequalities and ask for a limit. Your job is to recognize the theorem and state that the bounds share a limit.

Lesson 1.6 · Unit 1 · CED topics 1.10–1.13

Continuity and types of discontinuities

Informally, a function is continuous if you can draw it without lifting your pencil. The formal version is a three-part checklist, and the exam loves to test each part separately.

When continuity fails, name the failure:

• Removable: the limit exists, but \(f(c)\) is missing or has the wrong value. A hole.
• Jump: the one-sided limits exist but differ. Two branches at different heights.
• Infinite: at least one one-sided limit is \(\pm\infty\). A vertical asymptote.

Lesson 1.7 · Unit 1 · CED topics 1.14–1.15

Infinite limits and asymptotes

Two kinds of "infinity" show up in limits. A function can blow up to \(\pm\infty\) near a finite \(x\), a vertical asymptote, or you can ask where a function settles as \(x\) itself heads to \(\pm\infty\): a horizontal asymptote.

Lesson 1.8 · Unit 1 · CED topic 1.16

The Intermediate Value Theorem

Continuity has a payoff: if a continuous function is below a value at one point and above it at another, it must have hit that value somewhere in between. You can't cross a river without getting wet.

The most common use is proving an equation has a solution without solving it. The exam grades the justification, so learn the sentence structure below word for word.

Three things must appear for full credit: the function is continuous on the closed interval, the target value lies between the endpoint values, and the conclusion names the open interval.

Unit 1 practice · 10 problems

Unit 1 practice: Limits and Continuity

Ten problems covering the whole unit, in roughly exam order. Work each one on paper before revealing the answer: the reveal shows the key steps, not just the number. No calculator needed.

1. Evaluate \(\displaystyle\lim_{x \to 4}\frac{x^2 - 16}{x - 4}\).

   Show answer

   Factor: \(\dfrac{(x-4)(x+4)}{x-4} = x + 4\) for \(x \ne 4\). The limit is \(8\).

2. Evaluate \(\displaystyle\lim_{x \to 0}\frac{\sin 4x}{3x}\).

   Show answer

   \(\dfrac{\sin 4x}{3x} = \dfrac{4}{3}\cdot\dfrac{\sin 4x}{4x} \to \dfrac{4}{3}\cdot 1 = \dfrac{4}{3}\).

3. Evaluate \(\displaystyle\lim_{x \to \infty}\frac{5x^2 - 3x + 1}{2x^2 + 7}\).

   Show answer

   Equal degrees, so the limit is the ratio of leading coefficients: \(\dfrac{5}{2}\).

4. Evaluate \(\displaystyle\lim_{x \to 3^-}\frac{x + 1}{x - 3}\).

   Show answer

   The numerator approaches 4 while the denominator approaches 0 through negative values. The limit is \(-\infty\).

5. Find \(k\) so that \(f(x) = \begin{cases} x^2 + 1 & x < 2 \\ 3x - k & x \ge 2 \end{cases}\) is continuous at \(x = 2\).

   Show answer

   Left limit \(= 5\); right side \(= 6 - k\). Set \(6 - k = 5\), so \(k = 1\).

6. Evaluate \(\displaystyle\lim_{x \to 0}\frac{\sqrt{x + 9} - 3}{x}\).

   Show answer

   Multiply by the conjugate: \(\dfrac{(x + 9) - 9}{x\left(\sqrt{x+9} + 3\right)} = \dfrac{1}{\sqrt{x+9} + 3} \to \dfrac{1}{6}\).

7. Let \(g(x) = \dfrac{x^2 - 5x + 6}{x - 2}\). Classify the discontinuity at \(x = 2\), and give the value of \(g(2)\) that would remove it.

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   \(g(x) = \dfrac{(x-2)(x-3)}{x-2} = x - 3\) for \(x \ne 2\), so the limit at 2 is \(-1\). The discontinuity is removable; defining \(g(2) = -1\) removes it.

8. Evaluate \(\displaystyle\lim_{x \to -\infty}\frac{3x + 2}{\sqrt{x^2 + 1}}\).

   Show answer

   For large negative \(x\), \(\sqrt{x^2 + 1} \approx |x| = -x\), so the expression behaves like \(\dfrac{3x}{-x} = -3\). The limit is \(-3\).

9. \(f\) is continuous on \([1, 5]\) with \(f(1) = -2\) and \(f(5) = 7\). Must there be a \(c\) in \((1, 5)\) with \(f(c) = 3\)? Justify.

   Show answer

   Yes. \(f\) is continuous on \([1, 5]\) and \(f(1) = -2 < 3 < 7 = f(5)\), so by the Intermediate Value Theorem there is a \(c\) in \((1, 5)\) with \(f(c) = 3\).

10. Use the Squeeze Theorem to evaluate \(\displaystyle\lim_{x \to 0} x^4\cos\!\left(\frac{2}{x}\right)\).

    Show answer

    \(-x^4 \le x^4\cos\!\left(\tfrac{2}{x}\right) \le x^4\), and both bounds approach 0. The limit is \(0\).

Lesson 2.1 · Unit 2 · CED topic 2.1

Average and instantaneous rates of change at a point

Unit 1 ended with a promise: the limit idea would let us pin down a rate of change at a single instant. Here's the payoff. The average rate over a shrinking interval has a limit, and that limit is the instantaneous rate.

Notice that the algebra did the work the table did in Lesson 1.1. Instead of guessing where the averages were heading, we canceled the \(h\) and read the limit off exactly.

Lesson 2.2 · Unit 2 · CED topic 2.2

Defining the derivative and derivative notation

Do the instantaneous-rate calculation at a general \(x\) instead of a specific number and you get a new function: one that reports the slope of \(f\) everywhere at once. That function is the derivative.

Lesson 2.3 · Unit 2 · CED topic 2.3

Estimating derivatives from tables and graphs

Real data doesn't come with a formula. When all you have is a table or a graph, you estimate the derivative with the best difference quotient available, and the exam expects you to know which one that is.

From a graph: sketch the tangent line and estimate its slope with two points on it. Sign is often enough: the derivative is positive where the graph rises, negative where it falls, and zero at the tops and bottoms of hills. On the calculator sections, your calculator's numerical derivative does this for you when a formula is given.

Lesson 2.4 · Unit 2 · CED topic 2.4

Connecting differentiability and continuity

Every differentiable function is continuous: you can't have a tangent line at a gap. But continuity is not enough for a derivative. A graph can be unbroken and still have no well-defined slope at a point.

Lesson 2.5 · Unit 2 · CED topics 2.5–2.6

The power rule and the basic derivative rules

Computing every derivative from the limit definition would be unbearable. Fortunately a handful of rules, all provable from the definition, handle every polynomial and most algebraic functions.

Lesson 2.6 · Unit 2 · CED topic 2.7

Derivatives of sin x, cos x, eˣ, and ln x

Four derivatives you need cold. Each can be proved from the limit definition: the sine proof uses the two special limits from Lesson 1.4, but on the exam you simply need to know them.

The fact that \(e^x\) is its own derivative is why \(e\) is the natural base: it's the one exponential whose growth rate equals its current value. You'll lean on that heavily in Unit 7.

Lesson 2.7 · Unit 2 · CED topics 2.8–2.9

The product and quotient rules

The derivative of a sum is the sum of the derivatives. The derivative of a product is not the product of the derivatives, and this is the single most common algebra error in Unit 2.

Lesson 2.8 · Unit 2 · CED topic 2.10

Derivatives of tan, cot, sec, and csc

The remaining four trig derivatives all follow from the quotient rule plus \(\sin\) and \(\cos\). Derive one so you trust them, then memorize all four.

Unit 2 practice · 10 problems

Unit 2 practice, Differentiation: Definition and Fundamental Properties

Ten problems covering the whole unit. Work each one on paper before revealing the answer. No calculator needed.

1. Use the limit definition of the derivative to find \(f'(x)\) for \(f(x) = 3x^2 - x\).

   Show answer

   \(\dfrac{3(x+h)^2 - (x+h) - 3x^2 + x}{h} = \dfrac{6xh + 3h^2 - h}{h} = 6x + 3h - 1 \to 6x - 1\).

2. Evaluate \(\displaystyle\lim_{h \to 0}\frac{\sin\!\left(\frac{\pi}{6} + h\right) - \sin\frac{\pi}{6}}{h}\).

   Show answer

   This is the derivative of \(\sin x\) at \(x = \tfrac{\pi}{6}\): \(\cos\tfrac{\pi}{6} = \dfrac{\sqrt{3}}{2}\).

3. Find \(f'(x)\) for \(f(x) = 2x^5 - 4x^3 + 7x - 12\).

   Show answer

   \(f'(x) = 10x^4 - 12x^2 + 7\).

4. Find \(g'(x)\) for \(g(x) = \dfrac{5}{x^3} - 2\sqrt{x} + e^x\).

   Show answer

   Rewrite as \(5x^{-3} - 2x^{1/2} + e^x\). Then \(g'(x) = -15x^{-4} - x^{-1/2} + e^x = -\dfrac{15}{x^4} - \dfrac{1}{\sqrt{x}} + e^x\).

5. Find the equation of the tangent line to \(y = x^2\ln x\) at \(x = 1\).

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   \(y(1) = 0\). \(y' = 2x\ln x + x\), so \(y'(1) = 1\). Tangent: \(y = x - 1\).

6. Find \(h'(x)\) for \(h(x) = \dfrac{x^2 + 1}{x - 2}\).

   Show answer

   \(h'(x) = \dfrac{2x(x - 2) - (x^2 + 1)}{(x - 2)^2} = \dfrac{x^2 - 4x - 1}{(x - 2)^2}\).

7. \(f\) is differentiable, with \(f(1) = 3\), \(f(2) = 7\), \(f(3) = 13\), \(f(4) = 21\). Estimate \(f'(3)\).

   Show answer

   Use the points straddling \(x = 3\): \(f'(3) \approx \dfrac{21 - 7}{4 - 2} = 7\).

8. Given \(f(1) = 4\), \(f'(1) = -2\), \(g(1) = 3\), \(g'(1) = 5\), find \((fg)'(1)\) and \(\left(\dfrac{f}{g}\right)'(1)\).

   Show answer

   \((fg)'(1) = (-2)(3) + (4)(5) = 14\). \(\left(\dfrac{f}{g}\right)'(1) = \dfrac{(-2)(3) - (4)(5)}{3^2} = -\dfrac{26}{9}\).

9. Is \(f(x) = \begin{cases} x^2 + 2x & x \le 1 \\ 4x - 1 & x > 1 \end{cases}\) differentiable at \(x = 1\)? Justify.

   Show answer

   Continuity: both pieces equal 3 at \(x = 1\). Slopes: \(2x + 2 = 4\) from the left, \(4\) from the right. Both match, so yes, \(f\) is differentiable at 1 with \(f'(1) = 4\).

10. Find \(f'\!\left(\dfrac{\pi}{3}\right)\) for \(f(x) = \sec x\tan x\).

    Show answer

    \(f'(x) = \sec x\tan^2 x + \sec^3 x\). At \(\tfrac{\pi}{3}\), \(\sec = 2\) and \(\tan = \sqrt{3}\): \(2\cdot 3 + 8 = 14\).

Lesson 3.1 · Unit 3 · CED topic 3.1

The chain rule

Most functions worth differentiating are compositions: a function inside another function. The chain rule says: differentiate the outside (leaving the inside alone), then multiply by the derivative of the inside.

Lesson 3.2 · Unit 3 · CED topic 3.2

Implicit differentiation

Not every curve is a function \(y = f(x)\). A circle isn't. But you can still find slopes on it: differentiate the whole equation with respect to \(x\), treating \(y\) as a function of \(x\) and applying the chain rule every time \(y\) appears.

Horizontal tangents occur where the numerator of \(y'\) is zero (and the denominator isn't); vertical tangents where the denominator is zero (and the numerator isn't).

Lesson 3.3 · Unit 3 · CED topic 3.3

Differentiating inverse functions

The graph of \(f^{-1}\) is the graph of \(f\) reflected over \(y = x\). Reflecting a line with slope \(m\) gives a line with slope \(\frac{1}{m}\), so the derivative of the inverse is the reciprocal of the derivative of the original, evaluated at the matching point.

Lesson 3.4 · Unit 3 · CED topic 3.4

Differentiating inverse trigonometric functions

Implicit differentiation gives the derivatives of the inverse trig functions in a few lines. Watch the derivation for \(\arctan x\), then memorize the three you need most.

Lesson 3.5 · Unit 3 · CED topic 3.5

Selecting procedures for calculating derivatives

You now have every derivative rule in the course. The remaining skill is diagnosis: look at a function, identify its outermost structure, and pick the rule that matches. Simplify before differentiating whenever you can: it prevents most errors.

Lesson 3.6 · Unit 3 · CED topic 3.6

Higher-order derivatives

The derivative of a function is a function, so it has a derivative of its own. The second derivative \(f''\) measures how fast the slope is changing: in motion, that's acceleration. In Unit 5 it will tell you which way a graph bends.

Unit 3 practice · 10 problems

Unit 3 practice: Composite, Implicit, and Inverse Functions

Ten problems covering the whole unit. Work each one on paper before revealing the answer. No calculator needed.

1. Differentiate \(y = (2x^3 - 5)^6\).

   Show answer

   \(y' = 6(2x^3 - 5)^5\cdot 6x^2 = 36x^2(2x^3 - 5)^5\).

2. Differentiate \(y = e^{\sin x}\).

   Show answer

   Chain rule with \(u = \sin x\): \(\dfrac{d}{dx}e^{u} = e^{u}\dfrac{du}{dx}\). Here \(\dfrac{du}{dx} = \cos x\), so \(y' = \cos x\, e^{\sin x}\).

3. Differentiate \(y = \ln(\cos x)\).

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   \(y' = \dfrac{-\sin x}{\cos x} = -\tan x\).

4. Find \(\dfrac{dy}{dx}\) for \(x^2 y + y^3 = 10\).

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   \(2xy + x^2 y' + 3y^2 y' = 0\), so \(y' = -\dfrac{2xy}{x^2 + 3y^2}\).

5. Find the slope of the tangent line to \(x^2 - xy + y^2 = 7\) at the point \((2, 3)\).

   Show answer

   \(2x - y - xy' + 2yy' = 0 \Rightarrow y' = \dfrac{y - 2x}{2y - x}\). At \((2, 3)\): \(\dfrac{3 - 4}{6 - 2} = -\dfrac{1}{4}\).

6. Let \(f(x) = x^3 + 2x + 1\) and \(g = f^{-1}\). Find \(g'(4)\).

   Show answer

   \(f(1) = 4\), so \(g(4) = 1\). \(f'(x) = 3x^2 + 2\), \(f'(1) = 5\). \(g'(4) = \dfrac{1}{5}\).

7. Differentiate \(y = \arctan(3x)\).

   Show answer

   Use \(\dfrac{d}{dx}\arctan u = \dfrac{u'}{1 + u^{2}}\) with \(u = 3x\), so \(u' = 3\): \(y' = \dfrac{3}{1 + (3x)^{2}} = \dfrac{3}{1 + 9x^2}\).

8. Differentiate \(y = \arcsin(x^2)\).

   Show answer

   Use \(\dfrac{d}{dx}\arcsin u = \dfrac{u'}{\sqrt{1 - u^{2}}}\) with \(u = x^{2}\), so \(u' = 2x\): \(y' = \dfrac{2x}{\sqrt{1 - (x^{2})^{2}}} = \dfrac{2x}{\sqrt{1 - x^4}}\).

9. Let \(h(x) = f(g(x))\), with \(g(3) = 2\), \(g'(3) = -4\), \(f'(2) = 6\), and \(f'(3) = 10\). Find \(h'(3)\).

   Show answer

   \(h'(3) = f'(g(3))\cdot g'(3) = f'(2)\cdot(-4) = -24\). (The value \(f'(3) = 10\) is a distractor.)

10. Find \(y'''\) for \(y = \sin(2x)\).

    Show answer

    \(y' = 2\cos 2x\), \(y'' = -4\sin 2x\), \(y''' = -8\cos 2x\).

Lesson 4.1 · Unit 4 · CED topic 4.1

Interpreting the meaning of the derivative in context

A derivative is a rate, and a rate has units. Free-response questions routinely ask you to explain what a number like \(f'(5) = -3\) means "in the context of the problem", and they grade the sentence, not the number.

Lesson 4.2 · Unit 4 · CED topic 4.2

Straight-line motion: position, velocity, acceleration

A particle moving along a line has position \(x(t)\). Its velocity is \(v(t) = x'(t)\) and its acceleration is \(a(t) = v'(t) = x''(t)\). Almost every motion question comes down to the signs of \(v\) and \(a\).

Lesson 4.3 · Unit 4 · CED topic 4.3

Rates of change in applied contexts other than motion

Anything that changes has a rate of change, and that rate is a derivative. Populations, costs, areas, temperatures: the calculus is identical; only the units and the story change.

Lesson 4.4 · Unit 4 · CED topics 4.4–4.5

Related rates

Two quantities are linked by an equation, and both change with time. Knowing how fast one changes tells you how fast the other does. Differentiate the linking equation with respect to \(t\): every variable picks up a \(\frac{d}{dt}\) by the chain rule.

Lesson 4.5 · Unit 4 · CED topic 4.6

Local linearity and linearization

Zoom in far enough on any differentiable curve and it looks like a straight line: its tangent. That means the tangent line is a good stand-in for the function near the point of tangency, and lines are easy to evaluate.

Lesson 4.6 · Unit 4 · CED topic 4.7

L'Hospital's Rule

Back to limits, now with derivatives in hand. When a limit has the indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), you may replace the fraction by the ratio of the derivatives: top and bottom differentiated separately, not with the quotient rule.

Unit 4 practice · 10 problems

Unit 4 practice: Contextual Applications of Differentiation

Ten problems covering the whole unit. Work each one on paper before revealing the answer. A calculator is fine for the final arithmetic in problems 4, 5, and 10.

1. \(W(t)\) is the number of gallons of water in a pool \(t\) hours after noon, and \(W'(3) = -15\). Interpret this with units.

   Show answer

   At 3:00 pm, the amount of water in the pool is decreasing at a rate of 15 gallons per hour.

2. A particle's position is \(x(t) = t^3 - 9t^2 + 24t\) for \(t \ge 0\). When is it at rest?

   Show answer

   \(v(t) = 3t^2 - 18t + 24 = 3(t - 2)(t - 4)\). At rest at \(t = 2\) and \(t = 4\).

3. For the same particle, is it speeding up or slowing down at \(t = 1\)?

   Show answer

   \(v(1) = 9\) and \(a(t) = 6t - 18\), so \(a(1) = -12\). Opposite signs: slowing down.

4. The edge of a cube grows at 2 cm/s. How fast is the volume increasing when the edge is 5 cm?

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   \(V = s^3 \Rightarrow \dfrac{dV}{dt} = 3s^2\dfrac{ds}{dt} = 3(25)(2) = 150\) cm³/s.

5. Two cars leave the same point at the same time; one drives north at 60 mph, the other east at 80 mph. How fast is the distance between them increasing after one hour?

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   \(d^2 = x^2 + y^2 \Rightarrow d\,d' = x\,x' + y\,y'\). After 1 h: \(x = 80\), \(y = 60\), \(d = 100\). \(d' = \dfrac{80(80) + 60(60)}{100} = 100\) mph.

6. Use a linearization to approximate \(\sqrt[3]{8.06}\).

   Show answer

   \(f(x) = x^{1/3}\) at \(a = 8\): \(f(8) = 2\), \(f'(8) = \tfrac{1}{3}(8)^{-2/3} = \tfrac{1}{12}\). \(L(8.06) = 2 + \tfrac{0.06}{12} = 2.005\).

7. Evaluate \(\displaystyle\lim_{x \to 0}\frac{e^{2x} - 1}{\sin x}\).

   Show answer

   Form \(\frac{0}{0}\). L'Hospital: \(\displaystyle\lim_{x \to 0}\frac{2e^{2x}}{\cos x} = 2\).

8. Evaluate \(\displaystyle\lim_{x \to \infty}\frac{x^3}{e^x}\).

   Show answer

   Form \(\frac{\infty}{\infty}\); apply L'Hospital three times: \(\dfrac{3x^2}{e^x} \to \dfrac{6x}{e^x} \to \dfrac{6}{e^x} \to 0\).

9. Evaluate \(\displaystyle\lim_{x \to 1}\frac{\ln x}{x^2 - 1}\).

   Show answer

   Form \(\frac{0}{0}\). \(\displaystyle\lim_{x \to 1}\frac{1/x}{2x} = \frac{1}{2}\).

10. A 13 ft ladder leans against a wall. The top slides down at 2 ft/s. How fast is the base moving away from the wall when the base is 5 ft from it?

    Show answer

    \(x^2 + y^2 = 169\); when \(x = 5\), \(y = 12\). \(2x\,x' + 2y\,y' = 0 \Rightarrow 5x' + 12(-2) = 0\), so \(x' = \dfrac{24}{5} = 4.8\) ft/s.

Lesson 5.1 · Unit 5 · CED topic 5.1

The Mean Value Theorem

If you drive 120 miles in 2 hours, at some instant your speedometer read exactly 60. That's the Mean Value Theorem: somewhere on a smooth curve, the tangent line is parallel to the secant line through the endpoints.

Lesson 5.2 · Unit 5 · CED topic 5.2

Extreme Value Theorem, extrema, and critical points

Where can a function be biggest or smallest? Only in a few places: at points where the derivative is zero, where it doesn't exist, or at the ends of the interval. Everything in this unit is built on that fact.

Not every critical point is an extremum: \(f(x) = x^3\) has \(f'(0) = 0\) but just flattens out momentarily at the origin and keeps rising.

Lesson 5.3 · Unit 5 · CED topics 5.3–5.4

Increasing and decreasing intervals; the first derivative test

The sign of \(f'\) tells you which way \(f\) is going. Where the sign changes, \(f\) turns around, and that's a local extremum.

Lesson 5.4 · Unit 5 · CED topic 5.5

Absolute extrema on a closed interval: the candidates test

On a closed interval, the absolute maximum and minimum are guaranteed to exist (Extreme Value Theorem) and guaranteed to be at a critical point or an endpoint. So list the candidates, evaluate \(f\) at each, and compare. No sign charts needed.

Lesson 5.5 · Unit 5 · CED topics 5.6–5.7

Concavity, points of inflection, and the second derivative test

The first derivative says whether a graph rises or falls. The second derivative says which way it bends, like a cup holding water (concave up) or spilling it (concave down).

Lesson 5.6 · Unit 5 · CED topics 5.8–5.9

Sketching graphs and connecting f, f′, and f″

The exam's favorite Unit 5 question hands you the graph of \(f'\), not \(f\), and asks about \(f\). Every feature of \(f\) can be read off the derivative's graph if you know the translation.

Lesson 5.7 · Unit 5 · CED topics 5.10–5.11

Optimization

Largest area, least material, shortest distance: optimization problems turn a word problem into a function of one variable and then find its extreme value with the tools of this unit.

Lesson 5.8 · Unit 5 · CED topic 5.12

Exploring behaviors of implicit relations

The tools of this unit (critical points, the second derivative) work on implicitly defined curves too. The difference is that \(y'\) is a fraction in \(x\) and \(y\), so horizontal tangents come from its numerator and vertical tangents from its denominator.

The logic is the same as for functions; the only extra step is substituting the point's \(y\)-coordinate as well as its \(x\).

Unit 5 practice · 10 problems

Unit 5 practice: Analytical Applications of Differentiation

Ten problems covering the whole unit: one of the two heaviest-weighted units on the exam. Work each one on paper before revealing the answer. No calculator needed.

1. Find the absolute maximum and minimum of \(f(x) = x^3 - 12x\) on \([-3, 3]\).

   Show answer

   \(f'(x) = 3x^2 - 12 = 0\) at \(x = \pm 2\). Candidates: \(f(-3) = 9\), \(f(-2) = 16\), \(f(2) = -16\), \(f(3) = -9\). Absolute max 16 at \(x = -2\); absolute min \(-16\) at \(x = 2\).

2. Find the value of \(c\) guaranteed by the Mean Value Theorem for \(f(x) = \sqrt{x}\) on \([1, 9]\).

   Show answer

   Average rate \(= \dfrac{3 - 1}{8} = \dfrac{1}{4}\). \(f'(c) = \dfrac{1}{2\sqrt{c}} = \dfrac{1}{4} \Rightarrow c = 4\).

3. Find the intervals of concavity and the points of inflection of \(g(x) = x^4 - 8x^2 + 3\).

   Show answer

   \(g''(x) = 12x^2 - 16 = 0 \Rightarrow x = \pm\dfrac{2}{\sqrt{3}}\). Concave up for \(|x| > \dfrac{2}{\sqrt{3}}\), concave down on \(\left(-\dfrac{2}{\sqrt{3}}, \dfrac{2}{\sqrt{3}}\right)\); inflection points at \(x = \pm\dfrac{2}{\sqrt{3}} \approx \pm 1.155\).

4. \(f'(x) = x^2(x - 4)(x + 2)\). At which \(x\) does \(f\) have a local maximum? A local minimum?

   Show answer

   \(x^2 \ge 0\) never changes sign. \((x - 4)(x + 2)\) is positive for \(x < -2\), negative on \((-2, 4)\), positive for \(x > 4\). Local max at \(x = -2\) (\(+\) to \(-\)); local min at \(x = 4\) (\(-\) to \(+\)); nothing at \(x = 0\).

5. Use the second derivative test to classify the critical points of \(h(x) = 2x^3 - 3x^2 - 36x + 5\).

   Show answer

   \(h'(x) = 6(x - 3)(x + 2)\); critical points \(3\) and \(-2\). \(h''(x) = 12x - 6\): \(h''(3) = 30 > 0\) ⇒ local min at 3; \(h''(-2) = -30 < 0\) ⇒ local max at \(-2\).

6. A rectangle has its base on the \(x\)-axis and its upper corners on \(y = 12 - x^2\). Find the maximum possible area.

   Show answer

   \(A(x) = 2x(12 - x^2) = 24x - 2x^3\); \(A'(x) = 24 - 6x^2 = 0 \Rightarrow x = 2\); \(A''(2) < 0\). Maximum area \(= 4 \cdot 8 = 32\) (a 4 by 8 rectangle).

7. An open-top box with a square base must hold 32 in³. Find the dimensions that minimize the surface area.

   Show answer

   \(x^2 h = 32 \Rightarrow h = \dfrac{32}{x^2}\). \(S = x^2 + 4xh = x^2 + \dfrac{128}{x}\); \(S' = 2x - \dfrac{128}{x^2} = 0 \Rightarrow x = 4\), \(h = 2\). Minimum surface area \(= 48\) in².

8. \(f\) is twice differentiable on \([0, 6]\), with \(f' > 0\) on \((0, 2)\), \(f' < 0\) on \((2, 6)\), \(f'' < 0\) on \((0, 4)\), and \(f'' > 0\) on \((4, 6)\). Describe the graph of \(f\).

   Show answer

   \(f\) increases to a local maximum at \(x = 2\), then decreases. It is concave down on \((0, 4)\) and concave up on \((4, 6)\), with a point of inflection at \(x = 4\).

9. Find all points on \(x^2 + 4y^2 = 16\) where the tangent line is horizontal, and all points where it is vertical.

   Show answer

   \(2x + 8yy' = 0 \Rightarrow y' = -\dfrac{x}{4y}\). Horizontal where \(x = 0\): \((0, \pm 2)\). Vertical where \(y = 0\): \((\pm 4, 0)\).

10. Find and classify the local extrema of \(f(x) = \dfrac{x}{x^2 + 1}\).

    Show answer

    \(f'(x) = \dfrac{1 - x^2}{(x^2 + 1)^2}\), zero at \(x = \pm 1\); positive on \((-1, 1)\), negative outside. Local minimum \(f(-1) = -\tfrac{1}{2}\); local maximum \(f(1) = \tfrac{1}{2}\).

Lesson 6.1 · Unit 6 · CED topic 6.1

Exploring accumulation of change

Integration begins with a picture, not a formula. If you know how fast something changes, the area under the rate graph is how much it changed. That single idea is the whole of Unit 6.

Lesson 6.2 · Unit 6 · CED topic 6.2

Approximating areas with Riemann sums

Most rate graphs aren't rectangles and triangles. Riemann sums estimate the area with rectangles anyway, many thin ones, and the exam tests four flavors plus a judgment about whether each over- or underestimates.

Lesson 6.3 · Unit 6 · CED topic 6.3

Summation notation and the definite integral

Use more and more rectangles and the Riemann sum approaches the exact area. That limit is the definite integral. The exam mostly asks you to translate between the two notations.

Lesson 6.4 · Unit 6 · CED topics 6.4–6.5

The Fundamental Theorem of Calculus and accumulation functions

Here is the theorem the whole course has been building toward. An accumulation function adds up \(f\) from a fixed start to a moving endpoint \(x\). Its rate of change is just \(f(x)\): integration and differentiation undo each other.

Lesson 6.5 · Unit 6 · CED topic 6.6

Properties of definite integrals

Because a definite integral is a signed area, it obeys the rules areas do: you can split it, reverse it, scale it, and add it. These properties let you compute integrals from given values without any formula for \(f\).

Lesson 6.6 · Unit 6 · CED topic 6.7

The Fundamental Theorem, Part 2: evaluating definite integrals

Part 1 said accumulation functions have derivative \(f\). Part 2 turns that around into a computing tool: to find the exact area under \(f\), find any function whose derivative is \(f\) and subtract its values at the endpoints.

Lesson 6.7 · Unit 6 · CED topic 6.8

Finding antiderivatives and indefinite integrals

FTC Part 2 makes antiderivatives the central skill. Every derivative rule from Units 2–3, read backwards, is an integration rule, plus a constant, since any constant differentiates to zero.

Lesson 6.8 · Unit 6 · CED topic 6.9

Integrating using substitution

Substitution is the chain rule run backwards. If you can spot an inner function whose derivative also appears (up to a constant), rename the inner function \(u\) and the integral collapses to a basic one.

Lesson 6.9 · Unit 6 · CED topics 6.10, 6.14

Long division, completing the square, and selecting a technique

Two algebraic rewrites unlock rational integrands that substitution can't touch. Then, with every AB technique in hand, the remaining skill is choosing the right one quickly.

Lesson 6.10 · Unit 6 · CED topic 6.11 · BC only

Integration by parts

Substitution undoes the chain rule. Integration by parts undoes the product rule: it trades an integral you can't do for one you can, by differentiating one factor and integrating the other. It handles products like \(x e^x\) and \(x\cos x\), and anything containing \(\ln x\).

Exam tip: check for a substitution first. \(\int x e^{x^2}\,dx\) is a \(u\)-sub, not parts.

Lesson 6.11 · Unit 6 · CED topic 6.12 · BC only

Integrating with linear partial fractions

A fraction whose denominator factors into distinct linear pieces can be split into simpler fractions, each of which integrates to a logarithm. On the BC exam the denominator is always a product of linear, non-repeating factors, so the setup is always the same shape.

Exam tip: look before you decompose. \(\displaystyle\int\frac{2x}{x^2 - 4}\,dx\) is a plain \(u\)-substitution, and a denominator that doesn't factor calls for completing the square.

Lesson 6.12 · Unit 6 · CED topic 6.13 · BC only

Evaluating improper integrals

An integral is improper when an endpoint is infinite or the integrand blows up somewhere on the interval. The FTC needs a finite interval and a continuous integrand, so you never apply it directly. Instead, you integrate to a variable endpoint and take a limit, and the limit is what you write on the exam.

Caution: \(\displaystyle\int_{-1}^{1}\frac{dx}{x^2}\) is improper at \(x = 0\), inside the interval. Blindly applying the FTC gives \(\big[-1/x\big]_{-1}^{1} = -2\), a negative "area" for a positive function. Split at 0; each half diverges, so the integral diverges.

Unit 6 practice · 10 problems

Unit 6 practice: Integration and Accumulation of Change

Ten problems covering the whole unit, in roughly exam order: the AB core first, then the BC techniques (parts, partial fractions, improper integrals) in problems 7–10. Work each one on paper before revealing the answer: the reveal shows the key steps, not just the number. No calculator needed.

1. Water flows into a tank at a rate of \(R(t)\) liters per minute. Using the table, estimate the total amount of water that enters during \(0 \le t \le 5\) with a trapezoidal sum on the three subintervals given: \(t = 0, 2, 3, 5\); \(R(t) = 4, 10, 12, 20\).

   Show answer

   \(\tfrac{4 + 10}{2}(2) + \tfrac{10 + 12}{2}(1) + \tfrac{12 + 20}{2}(2) = 14 + 11 + 32 = 57\) liters. (The widths are 2, 1, 2: read them from the table, don't assume they are equal.)

2. Express \(\displaystyle\lim_{n\to\infty}\sum_{k=1}^{n}\frac{3}{n}\sqrt{1 + \frac{3k}{n}}\) as a definite integral, then evaluate it.

   Show answer

   \(\Delta x = \tfrac{3}{n}\) means an interval of length 3, and \(x_k = 0 + \tfrac{3k}{n}\) means it starts at 0: \(\displaystyle\int_0^3\sqrt{1 + x}\,dx = \Big[\tfrac{2}{3}(1 + x)^{3/2}\Big]_0^3 = \tfrac{2}{3}(8 - 1) = \dfrac{14}{3}\).

3. Find \(\dfrac{d}{dx}\displaystyle\int_x^{3} e^{t^2}\,dt\).

   Show answer

   Flip the limits first: \(\displaystyle\int_x^3 e^{t^2}\,dt = -\int_3^x e^{t^2}\,dt\). By FTC Part 1 the derivative is \(-e^{x^2}\).

4. Evaluate \(\displaystyle\int_0^2\frac{x}{x^2 + 1}\,dx\).

   Show answer

   \(u = x^2 + 1\), \(du = 2x\,dx\), limits \(1 \to 5\): \(\tfrac{1}{2}\displaystyle\int_1^5\frac{du}{u} = \tfrac{1}{2}\Big[\ln u\Big]_1^5 = \dfrac{\ln 5}{2} \approx 0.805\).

5. The graph of \(f\) is a horizontal segment at height 3 on \([0, 2]\), followed by a line segment from \((2, 3)\) to \((6, -3)\). Let \(g(x) = \displaystyle\int_0^x f(t)\,dt\). Find \(g(6)\) and \(g'(4)\), and find the value of \(x\) in \([0, 6]\) at which \(g\) attains its maximum. Justify.

   Show answer

   \(g(2) = 6\) (rectangle). The segment crosses zero at \(t = 4\): the triangle on \([2, 4]\) has area \(\tfrac{1}{2}(2)(3) = 3\) and the one on \([4, 6]\) contributes \(-3\), so \(g(6) = 6 + 3 - 3 = 6\). \(g'(4) = f(4) = 0\). Since \(g' = f\) is positive on \((0, 4)\) and negative on \((4, 6)\), \(g\) increases then decreases, so its maximum is at \(x = 4\), where \(g(4) = 9\).

6. Find \(\displaystyle\int\frac{dx}{x^2 + 6x + 13}\).

   Show answer

   The denominator doesn't factor; complete the square: \(x^2 + 6x + 13 = (x + 3)^2 + 4\). With \(u = x + 3\): \(\displaystyle\int\frac{du}{u^2 + 2^2} = \frac{1}{2}\arctan\frac{x + 3}{2} + C\).

7. Find \(\displaystyle\int x\cos 2x\,dx\).

   Show answer

   Parts with \(u = x\), \(dv = \cos 2x\,dx\), so \(du = dx\) and \(v = \tfrac{1}{2}\sin 2x\): \(\dfrac{x}{2}\sin 2x - \displaystyle\int\tfrac{1}{2}\sin 2x\,dx = \dfrac{x}{2}\sin 2x + \dfrac{1}{4}\cos 2x + C\). Check by differentiating: \(\tfrac{1}{2}\sin 2x + x\cos 2x - \tfrac{1}{2}\sin 2x = x\cos 2x\).

8. Evaluate \(\displaystyle\int_3^5\frac{5x - 1}{(x - 2)(x + 1)}\,dx\). Give an exact answer and a decimal approximation.

   Show answer

   \(\dfrac{5x - 1}{(x - 2)(x + 1)} = \dfrac{A}{x - 2} + \dfrac{B}{x + 1}\), so \(5x - 1 = A(x + 1) + B(x - 2)\). At \(x = 2\): \(9 = 3A\), \(A = 3\). At \(x = -1\): \(-6 = -3B\), \(B = 2\). Then \(\Big[3\ln|x - 2| + 2\ln|x + 1|\Big]_3^5 = (3\ln 3 + 2\ln 6) - (0 + 2\ln 4) = 3\ln 3 + 2\ln\tfrac{3}{2} = \ln\dfrac{243}{4} \approx 4.107\).

9. Which of the following improper integrals converges?

   \(\displaystyle\int_1^{\infty}\frac{dx}{\sqrt{x}}\) \(\displaystyle\int_1^{\infty}\frac{dx}{x}\) \(\displaystyle\int_1^{\infty}\frac{dx}{x^{3/2}}\) \(\displaystyle\int_0^{1}\frac{dx}{x^2}\)

   This is \(\displaystyle\int_1^{\infty}x^{-1/2}\,dx\) with \(p = \tfrac{1}{2}\); the p-test for an infinite upper limit needs \(p \gt 1\), and indeed \(\big[2\sqrt{x}\big]_1^b = 2\sqrt{b} - 2 \to \infty\). Picking this usually means remembering the p-test in the wrong direction.

   \(p = 1\) is the borderline case, and it diverges: the antiderivative is \(\ln x\), and \(\ln b \to \infty\) as \(b \to \infty\), just very slowly. The \(1/x\) integrand is the classic looks-like-it-should-converge trap.

   By the p-test with \(p = \tfrac{3}{2} \gt 1\) the integral converges, and its value is \(\displaystyle\lim_{b\to\infty}\Big[-2x^{-1/2}\Big]_1^b = \lim_{b\to\infty}\left(2 - \tfrac{2}{\sqrt{b}}\right) = 2\).

   The bounds are finite, but the integrand blows up at \(x = 0\), so this is improper too, and near a vertical asymptote the p-test reverses: \(\displaystyle\int_0^1 x^{-p}\,dx\) converges only for \(p \lt 1\). With \(p = 2\), \(\displaystyle\lim_{t\to 0^+}\Big[-\tfrac{1}{x}\Big]_t^1 = \lim_{t\to 0^+}\left(\tfrac{1}{t} - 1\right) = \infty\).

10. Determine whether \(\displaystyle\int_0^{\infty} x e^{-x}\,dx\) converges, and find its value if it does.

    Show answer

    Parts with \(u = x\), \(dv = e^{-x}\,dx\), \(v = -e^{-x}\): \(\displaystyle\int_0^b x e^{-x}\,dx = \Big[-x e^{-x}\Big]_0^b + \int_0^b e^{-x}\,dx = -\dfrac{b}{e^b} + 1 - e^{-b}\). As \(b \to \infty\), \(\dfrac{b}{e^b}\) is an \(\tfrac{\infty}{\infty}\) form; by L'Hospital it behaves like \(\dfrac{1}{e^b} \to 0\), and \(e^{-b} \to 0\). The integral converges to \(1\).

Lesson 7.1 · Unit 7 · CED topic 7.1

Modeling situations with differential equations

A differential equation is an equation involving a derivative. Most real laws are stated this way, not "here is the population" but "here is how fast the population is changing." The first skill is translating a sentence about rates into symbols.

Lesson 7.2 · Unit 7 · CED topic 7.2

Verifying solutions to differential equations

A solution to a differential equation is a function that makes the equation true. Checking a proposed solution needs no solving at all: differentiate it, substitute, and see whether both sides agree.

A general solution contains an arbitrary constant (\(y = Ce^{2x}\)); a particular solution has the constant pinned down by an initial condition (\(y = 3e^{2x}\)).

Lesson 7.3 · Unit 7 · CED topics 7.3–7.4

Sketching and reasoning with slope fields

A differential equation \(\dfrac{dy}{dx} = F(x, y)\) tells you the slope at every point in the plane. Draw a short segment with that slope at a grid of points and you have a slope field: a picture of every solution at once, before you've solved anything.

To sketch a particular solution through a given point, start at that point and follow the segments in both directions, staying tangent to the field. The curve never crosses the segments.

Lesson 7.4 · Unit 7 · CED topic 7.6

Finding general solutions using separation of variables

The one solving technique in AB Calculus: if the right side factors into a function of \(x\) times a function of \(y\), move everything involving \(y\) to the left with \(dy\), everything involving \(x\) to the right with \(dx\), and integrate both sides.

Lesson 7.5 · Unit 7 · CED topic 7.7

Finding particular solutions using initial conditions

An initial condition, one point the solution passes through, selects a single curve from the family of general solutions. Solve in general, substitute the point, and solve for \(C\). Then check the domain: the solution must be a continuous curve through the given point.

Lesson 7.6 · Unit 7 · CED topic 7.8

Exponential models with differential equations

The most important differential equation in the course is \(\dfrac{dy}{dt} = ky\): a quantity changing at a rate proportional to itself. Separation of variables solves it once and for all, and the solution describes populations, radioactive decay, interest, and cooling.

Lesson 7.7 · Unit 7 · CED topic 7.5 · BC only

Approximating solutions using Euler's method

Most differential equations can't be solved by separation. Euler's method sidesteps solving altogether: starting at the initial point, follow the tangent line for a short step, recompute the slope from the differential equation at the new point, and repeat. It is the slope field turned into arithmetic.

Exam tip: show the table. Graders award points for the correct slope at each step and the correct update, even when a later arithmetic slip spoils the final number.

Lesson 7.8 · Unit 7 · CED topic 7.9 · BC only

Logistic models with differential equations

Exponential growth never stops, but real populations run into limits: food, space, competition. The logistic model keeps the "proportional to \(P\)" growth for small populations and throttles it as \(P\) approaches a ceiling. Nearly every exam question about it can be answered from the equation alone, without solving it.

Separating variables (with partial fractions, Lesson 6.11) gives the explicit solution \(P = \dfrac{L}{1 + Ae^{-kt}}\), but the exam almost never asks for it. It asks for \(L\), the limit, the fastest-growth population, and concavity.

Unit 7 practice · 10 problems

Unit 7 practice: Differential Equations

Ten problems covering the whole unit: the AB core in problems 1–6, then Euler's method and logistic models in 7–10. Work each one on paper before revealing the answer. A calculator is fine for the final arithmetic in problems 6 and 8.

1. A rumor spreads through a school of 1200 students at a rate proportional to the number of students who have not yet heard it. Write a differential equation for \(N(t)\), the number of students who have heard the rumor after \(t\) days.

   Show answer

   \(\dfrac{dN}{dt} = k(1200 - N)\) with \(k \gt 0\). (Since \(1200 - N \gt 0\), the rate is positive and \(N\) increases toward 1200.)

2. Verify that \(y = x e^{x}\) is a solution of \(y' = y + e^{x}\).

   Show answer

   Product rule: \(y' = e^x + x e^x\). Right side: \(y + e^x = x e^x + e^x\). Equal. ✓

3. For \(\dfrac{dy}{dx} = x - y\), find the slopes of the slope-field segments at \((2, -1)\) and \((1, 3)\), and describe where the segments are horizontal.

   Show answer

   At \((2, -1)\): \(2 - (-1) = 3\). At \((1, 3)\): \(1 - 3 = -2\). Horizontal where \(x - y = 0\), i.e. along the line \(y = x\).

4. Find the general solution of \(\dfrac{dy}{dx} = x y^2\).

   Show answer

   Separate: \(\displaystyle\int y^{-2}\,dy = \int x\,dx\), so \(-\dfrac{1}{y} = \dfrac{x^2}{2} + C\). Solving, \(y = -\dfrac{1}{x^2/2 + C} = -\dfrac{2}{x^2 + K}\) where \(K = 2C\) is an arbitrary constant. (\(y = 0\) is also a solution.)

5. Solve \(\dfrac{dy}{dx} = y^2\sin x\) with \(y(0) = 1\).

   Show answer

   \(\displaystyle\int y^{-2}\,dy = \int\sin x\,dx \Rightarrow -\dfrac{1}{y} = -\cos x + C\). At \((0, 1)\): \(-1 = -1 + C\), so \(C = 0\). Then \(\dfrac{1}{y} = \cos x\), so \(y = \sec x\) (valid on \(-\tfrac{\pi}{2} \lt x \lt \tfrac{\pi}{2}\)).

6. Coffee at 90°C is left in a 20°C room, and its temperature satisfies \(\dfrac{dT}{dt} = k(T - 20)\). After 10 minutes it is 60°C. Find \(k\), and find when the coffee reaches 40°C.

   Show answer

   Separating gives \(T - 20 = 70e^{kt}\). At \(t = 10\): \(40 = 70e^{10k}\), so \(k = \tfrac{1}{10}\ln\tfrac{4}{7} \approx -0.0560\) per minute. For \(T = 40\): \(20 = 70e^{kt} \Rightarrow t = \dfrac{\ln(2/7)}{k} = \dfrac{10\ln(2/7)}{\ln(4/7)} \approx 22.4\) minutes.

7. Let \(\dfrac{dy}{dx} = 2x + y\) with \(y(0) = 1\). Use Euler's method with three steps of size \(h = 0.2\) to approximate \(y(0.6)\). Then decide whether the approximation is an overestimate or an underestimate, and justify.

   Show answer

   xₙ	yₙ	slope = 2xₙ + yₙ	h × slope	yₙ₊₁
   0	1	1	0.2	1.2
   0.2	1.2	1.6	0.32	1.52
   0.4	1.52	2.32	0.464	1.984

   So \(y(0.6) \approx 1.984\). Differentiating the equation, \(y'' = 2 + y' = 2 + 2x + y \gt 0\) near the solution (\(x, y \gt 0\)), so the solution is concave up, the tangent lines lie below the curve, and Euler's method underestimates. (Exact solution \(y = 3e^x - 2x - 2\), so \(y(0.6) \approx 2.266\).)

8. Let \(\dfrac{dy}{dx} = 1 + y^2\) with \(y(0) = 0\). (a) Use Euler's method with two steps of \(h = 0.25\) to approximate \(y(0.5)\). (b) Solve the equation exactly and compare.

   Show answer

   (a) Step 1: slope \(1 + 0 = 1\), \(y(0.25) \approx 0 + 0.25(1) = 0.25\). Step 2: slope \(1 + 0.25^2 = 1.0625\), \(y(0.5) \approx 0.25 + 0.25(1.0625) = 0.5156\). (b) \(\displaystyle\int\frac{dy}{1 + y^2} = \int dx \Rightarrow \arctan y = x + C\); \(y(0) = 0\) gives \(C = 0\), so \(y = \tan x\) and \(y(0.5) = \tan 0.5 \approx 0.5463\). Euler's estimate is low, as expected: \(y'' = 2y\,y' \gt 0\) for \(y \gt 0\), so the curve is concave up.

9. A fish population satisfies \(\dfrac{dP}{dt} = 0.4P\left(1 - \dfrac{P}{1200}\right)\) with \(P(0) = 200\), where \(t\) is in years. Find \(\displaystyle\lim_{t\to\infty}P(t)\), the population at which \(P\) is growing fastest and the growth rate then, and describe the concavity of the graph of \(P\) for \(t \gt 0\).

   Show answer

   Carrying capacity \(L = 1200\), so \(\displaystyle\lim_{t\to\infty}P(t) = 1200\). Growth is fastest at \(P = \tfrac{L}{2} = 600\), where \(\dfrac{dP}{dt} = 0.4(600)\left(\tfrac{1}{2}\right) = 120\) fish per year. Since \(P(0) = 200 \lt 600\), the graph is concave up until \(P = 600\) (the inflection point) and concave down afterward as it levels off at 1200.

10. A population is modeled by \(\dfrac{dP}{dt} = 3P - 0.006P^2\). What is the carrying capacity?

    3 500 250 0.006

    3 is the growth constant \(k\), the per-capita growth rate when \(P\) is tiny, not the carrying capacity. In the logistic form \(kP\left(1 - \tfrac{P}{L}\right)\), \(k\) is the coefficient of \(P\), while \(L\) is the population at which the rate returns to zero.

    Factor: \(3P - 0.006P^2 = 3P(1 - 0.002P) = 3P\left(1 - \dfrac{P}{500}\right)\), so \(k = 3\) and \(L = 500\). Equivalently, set \(dP/dt = 0\) and take the nonzero root, \(P = 3/0.006 = 500\).

    250 is \(L/2\), the population at which \(dP/dt\) is largest (the inflection point of the logistic curve). It is a favorite exam distractor because it is the other special value of \(P\) that appears in logistic questions.

    0.006 is just the coefficient of \(P^2\); the carrying capacity is the ratio \(3/0.006\), not the coefficient itself. This choice comes from not rewriting the equation in the standard \(kP(1 - P/L)\) form.

Lesson 8.1 · Unit 8 · CED topic 8.1

Finding the average value of a function

The average of a list of numbers is their sum divided by how many there are. The average of a function on an interval is the same idea with an integral in place of the sum: total accumulation divided by the length of the interval.

Lesson 8.2 · Unit 8 · CED topic 8.2

Connecting position, velocity, and acceleration using integrals

Unit 4 went from position to velocity to acceleration by differentiating. Integrals go the other way, and they make one distinction sharp: displacement versus total distance.

Lesson 8.3 · Unit 8 · CED topic 8.3

Using accumulation functions and definite integrals in applied contexts

The signature AB free-response problem: something flows in at one rate and out at another, and you're asked how much there is at a given time and when there's the most. The tools are net change and the first derivative test, applied to an integral.

Lesson 8.4 · Unit 8 · CED topic 8.4

Finding the area between curves expressed as functions of x

The area under one curve is an integral. The area between two curves is the integral of the gap between them: top minus bottom, over the \(x\)-interval where the region lives. Finding that interval usually means finding where the curves intersect.

Lesson 8.5 · Unit 8 · CED topics 8.5–8.6

Areas with functions of y, and regions with multiple intersections

Sometimes a region is awkward to slice vertically: a curve doubles back, or the "top" changes partway through. Slice horizontally instead, integrating right minus left with respect to \(y\). And when curves cross more than twice, split the integral wherever top and bottom trade places.

Lesson 8.6 · Unit 8 · CED topics 8.7–8.8

Volumes with cross sections

Slice a solid into thin slabs perpendicular to an axis. Each slab has volume (cross-sectional area) × (thickness). Add them up with an integral and you have the volume, no revolving required.

Lesson 8.7 · Unit 8 · CED topics 8.9–8.10

Volumes of revolution: the disc method

Revolve a region around a line and every cross section perpendicular to that line is a circle. The cross-section formula becomes \(A = \pi R^2\), where \(R\) is the distance from the axis of rotation to the curve, so the only new work is writing that radius correctly.

Lesson 8.8 · Unit 8 · CED topics 8.11–8.12

Volumes of revolution: the washer method

If the region doesn't touch the axis of rotation, revolving it leaves a hole. Each cross section is a washer, a disc with a smaller disc removed, and its area is \(\pi R^2 - \pi r^2\).

Unit 8 practice · 10 problems

Unit 8 practice: Applications of Integration

Ten problems covering the whole unit. Work each one on paper before revealing the answer. No calculator needed.

1. Find the average value of \(f(x) = 6x^2\) on \([0, 2]\).

   Show answer

   \(\dfrac{1}{2}\displaystyle\int_0^2 6x^2\,dx = \dfrac{1}{2}\Big[2x^3\Big]_0^2 = 8\).

2. \(v(t) = 3t^2 - 12\) on \([0, 3]\). Find the displacement and the total distance traveled.

   Show answer

   Displacement: \(\Big[t^3 - 12t\Big]_0^3 = -9\). \(v = 0\) at \(t = 2\): \(\displaystyle\int_0^2 v = -16\), \(\displaystyle\int_2^3 v = 7\). Total distance \(= 16 + 7 = 23\).

3. \(a(t) = 2t - 4\) and \(v(0) = 3\). Find \(v(t)\) and the displacement on \([0, 3]\).

   Show answer

   \(v(t) = t^2 - 4t + 3\). Displacement \(= \Big[\tfrac{t^3}{3} - 2t^2 + 3t\Big]_0^3 = 9 - 18 + 9 = 0\).

4. Water enters a tank at \(12 - t\) gal/min and leaves at 4 gal/min for \(0 \le t \le 12\); the tank holds 50 gal at \(t = 0\). When is the amount of water greatest, and how much is there then?

   Show answer

   Net rate \(8 - t\), positive then negative, zero at \(t = 8\). Amount \(= 50 + \displaystyle\int_0^8 (8 - t)\,dt = 50 + 32 = 82\) gallons at \(t = 8\).

5. Find the area between \(y = x^2 - 4\) and \(y = 2x - 1\).

   Show answer

   Intersections at \(x = -1, 3\); the line is on top. \(\displaystyle\int_{-1}^{3}\big(-x^2 + 2x + 3\big)\,dx = \Big[-\tfrac{x^3}{3} + x^2 + 3x\Big]_{-1}^{3} = 9 + \tfrac{5}{3} = \tfrac{32}{3}\).

6. Set up and evaluate an integral with respect to \(y\) for the area bounded by \(y = \sqrt{x}\), \(y = 0\), and \(x = 4\).

   Show answer

   Horizontal strips run from \(x = y^2\) to \(x = 4\) for \(0 \le y \le 2\): \(\displaystyle\int_0^2 (4 - y^2)\,dy = 8 - \tfrac{8}{3} = \tfrac{16}{3}\). (Check: \(\displaystyle\int_0^4\sqrt{x}\,dx = \tfrac{16}{3}\).)

7. The region bounded by \(y = x^2\), \(y = 0\), and \(x = 2\) is revolved about the \(x\)-axis. Find the volume.

   Show answer

   \(\pi\displaystyle\int_0^2 (x^2)^2\,dx = \pi\Big[\tfrac{x^5}{5}\Big]_0^2 = \dfrac{32\pi}{5}\).

8. The region between \(y = \sqrt{x}\) and \(y = \dfrac{x}{2}\) is revolved about the \(x\)-axis. Find the volume.

   Show answer

   They meet at \(x = 0\) and \(4\), with \(\sqrt{x}\) on top. \(\pi\displaystyle\int_0^4\left(x - \tfrac{x^2}{4}\right)dx = \pi\Big[\tfrac{x^2}{2} - \tfrac{x^3}{12}\Big]_0^4 = \pi\left(8 - \tfrac{16}{3}\right) = \dfrac{8\pi}{3}\).

9. The base of a solid is the region between \(y = x\) and \(y = x^2\). Cross sections perpendicular to the \(x\)-axis are equilateral triangles. Find the volume.

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   Side \(s = x - x^2\), \(A = \tfrac{\sqrt{3}}{4}s^2\). \(V = \tfrac{\sqrt{3}}{4}\displaystyle\int_0^1 (x - x^2)^2\,dx = \tfrac{\sqrt{3}}{4}\left(\tfrac{1}{3} - \tfrac{1}{2} + \tfrac{1}{5}\right) = \tfrac{\sqrt{3}}{4}\cdot\tfrac{1}{30} = \dfrac{\sqrt{3}}{120}\).

10. The region between \(y = x^2\) and \(y = 1\) is revolved about the line \(y = 2\). Find the volume.

    Show answer

    Outer radius \(2 - x^2\), inner radius \(1\). \(V = \pi\displaystyle\int_{-1}^{1}\Big[(2 - x^2)^2 - 1\Big]dx = 2\pi\displaystyle\int_0^1\big(3 - 4x^2 + x^4\big)\,dx = 2\pi\cdot\tfrac{28}{15} = \dfrac{56\pi}{15}\).

Lesson 9.1 · Unit 9 · CED topic 9.1

Defining and differentiating parametric equations

A parametric curve gives \(x\) and \(y\) separately as functions of a third variable \(t\), usually time. The curve can loop, cross itself, and fail the vertical line test: none of which a function \(y = f(x)\) can do. The slope of the curve still makes sense, and the chain rule delivers it.

Exam tip: the slope is a function of \(t\). Write \(\dfrac{dy}{dx}\) in terms of \(t\), substitute the \(t\)-value, and report the point in \((x, y)\) form. Never try to differentiate \(y\) "with respect to \(x\)" directly.

Lesson 9.2 · Unit 9 · CED topic 9.2

Second derivatives of parametric equations

Concavity of a parametric curve needs \(\dfrac{d^2y}{dx^2}\), and the most common BC error is to divide the second \(t\)-derivatives: \(\dfrac{y''(t)}{x''(t)}\) is wrong. The second derivative is the derivative of the slope with respect to \(x\), so apply the same chain-rule quotient again, this time to \(\dfrac{dy}{dx}\).

Exam tip: on a calculator-free question you'll usually be asked for the value at one \(t\). Compute \(\dfrac{dy}{dx}\) as a function of \(t\), take \(\dfrac{d}{dt}\) of it, then divide by \(\dfrac{dx}{dt}\): three labeled lines earn the points.

Lesson 9.3 · Unit 9 · CED topic 9.3

Arc lengths of curves given by parametric equations

Zoom in on a tiny piece of a parametric curve and it's a straight segment with legs \(dx\) and \(dy\); by the Pythagorean theorem its length is \(\sqrt{dx^2 + dy^2}\). Add those pieces up with an integral and you have arc length.

Arc length equals distance traveled only if the curve isn't retraced on \([a, b]\). If the particle backs up along the same path, the integral still counts every bit of motion, which is what "distance traveled" means anyway (Lesson 9.4).

Lesson 9.4 · Unit 9 · CED topics 9.4–9.6

Vector-valued functions: velocity, speed, and position

Package the parametric equations as one position vector and Unit 4's straight-line motion becomes motion in the plane. Everything differentiates and integrates component by component; the one new idea is that speed is the length of the velocity vector.

Lesson 9.5 · Unit 9 · CED topic 9.7

Defining polar coordinates and differentiating in polar form

Polar coordinates locate a point by its distance \(r\) from the origin and the angle \(\theta\) from the positive \(x\)-axis. Circles, spirals, and flower shapes that are awkward in \(x\) and \(y\) become one-line equations \(r = f(\theta)\). A polar curve is really a parametric curve with parameter \(\theta\), so the slope formula you already know applies.

Lesson 9.6 · Unit 9 · CED topic 9.8

Finding the area of a polar region

Rectangles are the wrong shape for polar regions. The natural slice is a thin wedge from the origin, which is a sector of a circle with area \(\tfrac{1}{2}r^2\,d\theta\). Summing wedges gives the polar area formula, and the \(\tfrac{1}{2}\) is the part students drop.

Exam tip: you will need \(\cos^2\theta = \dfrac{1 + \cos 2\theta}{2}\) and \(\sin^2\theta = \dfrac{1 - \cos 2\theta}{2}\) on the no-calculator section. On the calculator section, set up the integral with correct limits and evaluate numerically.

Lesson 9.7 · Unit 9 · CED topic 9.9

Finding the area of a region bounded by two polar curves

The region between two polar curves is a ring-shaped wedge: outer radius \(R\), inner radius \(r\). Subtract the inner sector from the outer one before integrating, and do it on exactly the \(\theta\)-interval where the region lives. Finding that interval, the intersection angles, is most of the work.

Exam tip: setting \(R = r\) misses intersections at the pole, where the curves arrive at different \(\theta\)-values. Sketch first; check whether each curve reaches \(r = 0\). On the calculator section, find intersection angles numerically and store them.

Unit 9 practice · 10 problems

Unit 9 practice: Parametric Equations, Polar Coordinates, and Vector-Valued Functions

Ten problems covering the whole unit, in roughly exam order: parametric curves, then motion in the plane, then polar coordinates. Work each one on paper before revealing the answer. A calculator is fine for problems 4 and 6.

1. A curve is given by \(x = t^3\), \(y = t^2 - 4t\). (a) Find \(\dfrac{dy}{dx}\) at \(t = 1\) and write the equation of the tangent line there. (b) Find the points where the tangent is horizontal and where it is vertical.

   Show answer

   \(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{2t - 4}{3t^2}\). (a) At \(t = 1\): slope \(-\tfrac{2}{3}\) at the point \((1, -3)\), so \(y + 3 = -\tfrac{2}{3}(x - 1)\). (b) Horizontal where \(2t - 4 = 0\), \(t = 2\) (and \(dx/dt = 12 \ne 0\)): the point \((8, -4)\). Vertical where \(3t^2 = 0\), \(t = 0\) (and \(dy/dt = -4 \ne 0\)): the point \((0, 0)\).

2. For \(x = t^3 + t\), \(y = t^2\), find \(\dfrac{d^2y}{dx^2}\) at \(t = 1\) and state the concavity of the curve there.

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   \(\dfrac{dy}{dx} = \dfrac{2t}{3t^2 + 1}\). Then \(\dfrac{d}{dt}\left(\dfrac{2t}{3t^2 + 1}\right) = \dfrac{2(3t^2 + 1) - 2t(6t)}{(3t^2 + 1)^2} = \dfrac{2 - 6t^2}{(3t^2 + 1)^2}\), and dividing by \(\dfrac{dx}{dt} = 3t^2 + 1\) gives \(\dfrac{d^2y}{dx^2} = \dfrac{2 - 6t^2}{(3t^2 + 1)^3}\). At \(t = 1\): \(\dfrac{-4}{64} = -\dfrac{1}{16} \lt 0\), so the curve is concave down there.

3. Find the exact length of the curve \(x = 3t^2\), \(y = 2t^3\) for \(0 \le t \le 1\).

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   \(\sqrt{(6t)^2 + (6t^2)^2} = 6t\sqrt{1 + t^2}\) for \(t \ge 0\). With \(u = 1 + t^2\), \(du = 2t\,dt\), limits \(1 \to 2\): \(L = 3\displaystyle\int_1^2 u^{1/2}\,du = 2\Big[u^{3/2}\Big]_1^2 = 2\left(2\sqrt{2} - 1\right) = 4\sqrt{2} - 2 \approx 3.657\).

4. A point on the rim of a rolling wheel follows \(x = t - \sin t\), \(y = 1 - \cos t\) for \(0 \le t \le 2\pi\). Set up an integral for the distance it travels and evaluate it.

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   \(L = \displaystyle\int_0^{2\pi}\sqrt{(1 - \cos t)^2 + \sin^2 t}\,dt = \int_0^{2\pi}\sqrt{2 - 2\cos t}\,dt\). A calculator gives \(8.000\). Exactly: \(2 - 2\cos t = 4\sin^2\tfrac{t}{2}\), so the integrand is \(2\sin\tfrac{t}{2}\) (nonnegative on \([0, 2\pi]\)) and \(L = \Big[-4\cos\tfrac{t}{2}\Big]_0^{2\pi} = 4 + 4 = 8\).

5. A particle's position is \(\vec r(t) = \langle t^3 - 12t,\ t^2 - 4t\rangle\). (a) Find \(\vec v(t)\) and \(\vec a(t)\). (b) At what time(s) is the particle at rest? (c) Find the speed at \(t = 3\).

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   (a) \(\vec v(t) = \langle 3t^2 - 12,\ 2t - 4\rangle\), \(\vec a(t) = \langle 6t,\ 2\rangle\). (b) Both components must be zero: \(3t^2 - 12 = 0\) at \(t = \pm 2\) and \(2t - 4 = 0\) at \(t = 2\), so the particle is at rest only at \(t = 2\). (c) \(\vec v(3) = \langle 15, 2\rangle\), speed \(= \sqrt{225 + 4} = \sqrt{229} \approx 15.13\).

6. A particle moves with velocity \(\vec v(t) = \langle 2t,\ 3\sqrt{t}\rangle\) and is at \((0, 2)\) when \(t = 1\). (a) Find its speed at \(t = 4\). (b) Find its position at \(t = 4\). (c) Find the total distance it travels for \(1 \le t \le 4\).

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   (a) \(\vec v(4) = \langle 8, 6\rangle\), speed \(= \sqrt{64 + 36} = 10\). (b) \(x(4) = 0 + \displaystyle\int_1^4 2t\,dt = 16 - 1 = 15\); \(y(4) = 2 + \displaystyle\int_1^4 3\sqrt{t}\,dt = 2 + 2\Big[t^{3/2}\Big]_1^4 = 2 + 2(8 - 1) = 16\). Position \((15, 16)\). (c) Distance \(= \displaystyle\int_1^4\sqrt{4t^2 + 9t}\,dt \approx 20.596\) (calculator).

7. (a) Convert \(r = 6\sin\theta\) to rectangular form and identify the curve. (b) For \(r = 2 + 2\sin\theta\), find \(\dfrac{dy}{dx}\) at \(\theta = 0\) and write the tangent line there. (c) At \(\theta = 0\), is that curve moving toward or away from the origin? Justify.

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   (a) Multiply by \(r\): \(r^2 = 6r\sin\theta\), so \(x^2 + y^2 = 6y\), i.e. \(x^2 + (y - 3)^2 = 9\): a circle of radius 3 centered at \((0, 3)\). (b) At \(\theta = 0\): \(r = 2\), \(\dfrac{dr}{d\theta} = 2\cos\theta = 2\). \(\dfrac{dy}{d\theta} = r'\sin\theta + r\cos\theta = 0 + 2 = 2\) and \(\dfrac{dx}{d\theta} = r'\cos\theta - r\sin\theta = 2 - 0 = 2\), so \(\dfrac{dy}{dx} = 1\). The point is \((2, 0)\); tangent line \(y = x - 2\). (c) \(\dfrac{dr}{d\theta} = 2 \gt 0\), so \(r\) is increasing: the curve is moving away from the origin.

8. Find the area of one petal of the rose \(r = 3\sin 2\theta\).

   Show answer

   \(r = 0\) at \(\theta = 0\) and \(\theta = \tfrac{\pi}{2}\), so one petal is traced for \(0 \le \theta \le \tfrac{\pi}{2}\). \(A = \tfrac{1}{2}\displaystyle\int_0^{\pi/2} 9\sin^2 2\theta\,d\theta = \tfrac{9}{2}\int_0^{\pi/2}\frac{1 - \cos 4\theta}{2}\,d\theta = \tfrac{9}{4}\Big[\theta - \tfrac{1}{4}\sin 4\theta\Big]_0^{\pi/2} = \tfrac{9}{4}\cdot\tfrac{\pi}{2} = \dfrac{9\pi}{8} \approx 3.534\).

9. Find the area of the region inside the cardioid \(r = 2 + 2\cos\theta\) and outside the circle \(r = 3\).

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   Intersections: \(2 + 2\cos\theta = 3 \Rightarrow \cos\theta = \tfrac{1}{2}\), \(\theta = \pm\tfrac{\pi}{3}\); between them the cardioid is outside (at \(\theta = 0\), \(4 \gt 3\)). \(A = \tfrac{1}{2}\displaystyle\int_{-\pi/3}^{\pi/3}\Big[(2 + 2\cos\theta)^2 - 9\Big]d\theta = \tfrac{1}{2}\int_{-\pi/3}^{\pi/3}\big(2\cos 2\theta + 8\cos\theta - 3\big)\,d\theta\) (using \(4\cos^2\theta = 2 + 2\cos 2\theta\)). An antiderivative is \(\sin 2\theta + 8\sin\theta - 3\theta\); by symmetry, \(A = \left(\tfrac{\sqrt{3}}{2} + 4\sqrt{3} - \pi\right) = \dfrac{9\sqrt{3}}{2} - \pi \approx 4.653\).

10. The area of the region enclosed by \(r = 4\cos\theta\) is

    \(4\pi\) \(8\pi\) \(16\pi\) \(2\pi\)

    The curve is the circle \((x - 2)^2 + y^2 = 4\), radius 2, traced once for \(0 \le \theta \le \pi\): \(\tfrac{1}{2}\displaystyle\int_0^{\pi}16\cos^2\theta\,d\theta = 4\int_0^{\pi}(1 + \cos 2\theta)\,d\theta = 4\pi\), which agrees with \(\pi(2)^2\).

    \(8\pi\) is what you get by integrating from \(0\) to \(2\pi\): over that interval the circle is traced twice (once with \(r \gt 0\), once with \(r \lt 0\)), so the area is double-counted. Always check how much of \(\theta\) traces the curve exactly once.

    \(16\pi\) is the area of a circle of radius 4, which comes from reading the coefficient 4 as the radius. For \(r = a\cos\theta\) the number \(a\) is the diameter, so the radius is 2.

    \(2\pi\) is half the correct area: the result of thinking \(0 \le \theta \le \pi\) covers only half the circle and halving again, or of dropping a factor of 2 in \(\cos^2\theta = \tfrac{1}{2}(1 + \cos 2\theta)\). The interval \(0\) to \(\pi\) traces the whole circle exactly once.

Lesson 10.1 · Unit 10 · CED topic 10.1

Defining convergent and divergent infinite series

Adding infinitely many numbers sounds impossible, and sometimes it is. The trick is to add finitely many, watch the running totals, and ask whether they settle down. That question: do the partial sums have a limit?: is the definition everything in this unit rests on.

Lesson 10.2 · Unit 10 · CED topic 10.2

Working with geometric series

A geometric series multiplies by the same ratio at every step. It is the one family of series whose sum you can always compute exactly, and it is the model that every power series in Lessons 10.9–10.11 is compared against.

Exam tip: a series like \(\displaystyle\sum 2^n x^n\) is geometric with \(r = 2x\); it converges exactly when \(|2x| \lt 1\). That one observation is the seed of the interval of convergence.

Lesson 10.3 · Unit 10 · CED topics 10.3–10.4

The nth-term test for divergence and the integral test

Two tests, opposite in spirit. The nth-term test is a quick way to rule a series out: if the terms don't shrink to zero, stop. The integral test converts a hard sum into an improper integral (Lesson 6.12), which you already know how to evaluate.

Lesson 10.4 · Unit 10 · CED topic 10.5

Harmonic series and p-series

The p-series are the reference family for the whole unit: you'll compare almost every unfamiliar series to one of them. Their behavior comes straight from the integral test and the p-test for improper integrals, and the boundary case \(p = 1\) is the famous harmonic series.

Exam tip: the justification graders want is one line; "p-series with \(p = \tfrac{5}{3} \gt 1\), so it converges." Name the test, state \(p\), compare to 1.

Lesson 10.5 · Unit 10 · CED topic 10.6

Comparison tests for convergence

Most series aren't geometric or p-series, but they behave like one. Comparison tests make that intuition rigorous: bound the terms by a known series, or show the terms have the same size for large \(n\). Both tests need positive terms.

Exam tip: every comparison argument must name the comparison series and say why it converges or diverges. "Compare to \(\tfrac{1}{n^2}\)" alone earns nothing.

Lesson 10.6 · Unit 10 · CED topics 10.7, 10.10

The alternating series test and the alternating series error bound

When signs alternate, terms partly cancel, and a series can converge even when its terms shrink slowly. Better still, the partial sums bounce back and forth across the true sum, which hands you an error bound for free.

Exam tip: the error-bound justification has three parts; say the series alternates and satisfies the AST, write the inequality with the first omitted term, and compute that term. Then stop; don't "improve" the bound.

Lesson 10.7 · Unit 10 · CED topics 10.8–10.9

The ratio test, and absolute versus conditional convergence

The ratio test asks whether a series is eventually geometric: compare each term to the one before it. It is the workhorse for factorials and exponentials, and it's the test that will define the radius of convergence in Lesson 10.9. Its one blind spot is a ratio of exactly 1.

Lesson 10.8 · Unit 10 · CED topics 10.11–10.12

Taylor polynomials and the Lagrange error bound

The tangent line matches a function's value and slope at one point. A Taylor polynomial keeps going: match the second derivative, the third, and so on, and the polynomial hugs the curve over a wider stretch. The Lagrange bound says how far off it can be.

Lesson 10.9 · Unit 10 · CED topic 10.13

Radius and interval of convergence of power series

A power series is a polynomial that never ends: \(\displaystyle\sum c_n(x - a)^n\). Plug in an \(x\) and you get an ordinary series that may or may not converge. The set of \(x\)-values where it does is always an interval centered at \(a\), and the ratio test finds it: except at the two endpoints, which need their own tests.

Exam tip: a full-credit answer has the ratio limit, the inequality solved for \(x\), and a separate one-line verdict for each endpoint with the test named. Skipping the endpoints costs the point every time.

Lesson 10.10 · Unit 10 · CED topic 10.14

Finding Taylor or Maclaurin series for a function

Let the Taylor polynomial run forever and you get the Taylor series. Where it converges, it equals the function. Four series come up on every BC exam; know them cold, including the general term and the interval of convergence.

Exam tip: "first four nonzero terms and the general term" is the standard phrasing. Write the general term with its index and starting value, e.g. \(\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!}\).

Lesson 10.11 · Unit 10 · CED topic 10.15

Representing functions as power series

You rarely build a series from scratch. Instead you start with one of the four memorized series and transform it: substitute into it, multiply by a power of \(x\), differentiate, or integrate. The payoff is that integrals and limits that resisted every earlier technique become polynomial arithmetic.

Unit 10 practice · 10 problems

Unit 10 practice: Infinite Sequences and Series

Ten problems covering the whole unit: the heaviest BC-only topic on the exam. Work each one on paper before revealing the answer. For every convergence question, name the test and check its conditions in writing. No calculator needed.

1. Find the sum of \(\displaystyle\sum_{n=2}^{\infty}\frac{1}{n^2 - 1}\).

   Show answer

   Partial fractions: \(\dfrac{1}{(n - 1)(n + 1)} = \dfrac{1}{2}\left(\dfrac{1}{n - 1} - \dfrac{1}{n + 1}\right)\). The partial sum telescopes: \(S_N = \dfrac{1}{2}\left(1 + \dfrac{1}{2} - \dfrac{1}{N} - \dfrac{1}{N + 1}\right) \to \dfrac{1}{2}\cdot\dfrac{3}{2} = \dfrac{3}{4}\).

2. (a) Find \(\displaystyle\sum_{n=1}^{\infty}\frac{2^{n+1}}{3^n}\). (b) Write \(0.454545\ldots\) as a fraction.

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   (a) \(\dfrac{2^{n+1}}{3^n} = 2\left(\dfrac{2}{3}\right)^n\); the first term (\(n = 1\)) is \(\tfrac{4}{3}\) and \(r = \tfrac{2}{3}\), so the sum is \(\dfrac{4/3}{1 - 2/3} = 4\). (b) \(0.45 + 0.0045 + \cdots\) with \(a = 0.45\), \(r = 0.01\): \(\dfrac{0.45}{0.99} = \dfrac{45}{99} = \dfrac{5}{11}\).

3. Decide whether each series converges or diverges, naming the test: (a) \(\displaystyle\sum_{n=1}^{\infty}\frac{n}{3n + 1}\); (b) \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{0.8}}\); (c) \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2 + 5}\).

   Show answer

   (a) \(a_n \to \tfrac{1}{3} \ne 0\): diverges by the nth-term test. (b) p-series with \(p = 0.8 \le 1\): diverges. (c) \(0 \lt \dfrac{1}{n^2 + 5} \lt \dfrac{1}{n^2}\) and \(\displaystyle\sum\frac{1}{n^2}\) converges (p-series, \(p = 2\)): converges by the direct comparison test.

4. Use the integral test to determine whether \(\displaystyle\sum_{n=2}^{\infty}\frac{1}{n(\ln n)^2}\) converges. State the hypotheses you check.

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   \(f(x) = \dfrac{1}{x(\ln x)^2}\) is positive, continuous, and decreasing on \([2, \infty)\) (the denominator increases). With \(u = \ln x\): \(\displaystyle\int_2^{\infty}\frac{dx}{x(\ln x)^2} = \lim_{b\to\infty}\Big[-\frac{1}{\ln x}\Big]_2^b = \frac{1}{\ln 2} \approx 1.443\), finite. The integral converges, so the series converges. (The sum is not \(\tfrac{1}{\ln 2}\); the test decides convergence only.)

5. Which of the following series converges?

   \(\displaystyle\sum_{n=1}^{\infty}\frac{n}{2n + 1}\) \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}\) \(\displaystyle\sum_{n=1}^{\infty}\frac{n + 1}{n^3 + 1}\) \(\displaystyle\sum_{n=2}^{\infty}\frac{1}{n\ln n}\)

   The terms approach \(\tfrac{1}{2}\), not 0, so the series diverges by the nth-term test before any other test is needed. A series cannot converge unless \(a_n \to 0\); terms that settle at a nonzero value always mean divergence.

   This is a p-series with \(p = \tfrac{1}{2} \le 1\), so it diverges; the terms do go to 0, just not fast enough. Students who pick it are usually applying the p-test backwards or confusing it with \(\sum 1/n^2\).

   Limit comparison with \(b_n = \dfrac{1}{n^2}\): \(\displaystyle\lim_{n\to\infty}\frac{(n + 1)/(n^3 + 1)}{1/n^2} = \lim_{n\to\infty}\frac{n^3 + n^2}{n^3 + 1} = 1\), finite and positive, and \(\sum 1/n^2\) converges, so this series converges too.

   Even though \(n\ln n\) grows faster than \(n\), this diverges by the integral test: \(\displaystyle\int_2^{\infty}\frac{dx}{x\ln x} = \lim_{b\to\infty}\Big[\ln(\ln x)\Big]_2^b = \infty\). The logarithm grows too slowly to rescue it.

6. Consider \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n\,2^n}\). (a) Show that it converges. (b) Approximate the sum by \(S_3\) and give a bound on the error.

   Show answer

   (a) \(b_n = \dfrac{1}{n\,2^n}\) is positive, decreasing (both \(n\) and \(2^n\) increase), and \(\to 0\), so the series converges by the alternating series test. (b) \(S_3 = \tfrac{1}{2} - \tfrac{1}{8} + \tfrac{1}{24} = \tfrac{5}{12} \approx 0.4167\). By the alternating series error bound, \(|S - S_3| \le b_4 = \dfrac{1}{4\cdot 16} = \dfrac{1}{64} \approx 0.0156\). (The sum is \(\ln\tfrac{3}{2} \approx 0.4055\); the actual error, 0.011, is inside the bound.)

7. (a) Use the ratio test on \(\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{n^2}{2^n}\). (b) Is \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{2n + 1}\) absolutely convergent, conditionally convergent, or divergent?

   Show answer

   (a) \(\left|\dfrac{a_{n+1}}{a_n}\right| = \dfrac{(n + 1)^2}{2^{n+1}}\cdot\dfrac{2^n}{n^2} = \dfrac{1}{2}\left(\dfrac{n + 1}{n}\right)^2 \to \dfrac{1}{2} \lt 1\): converges absolutely. (b) \(\displaystyle\sum\frac{1}{2n + 1}\) diverges by limit comparison with the harmonic series (\(\dfrac{1/(2n+1)}{1/n} \to \tfrac{1}{2}\)), so the convergence is not absolute. But \(b_n = \dfrac{1}{2n + 1}\) is positive, decreasing, and \(\to 0\), so the series converges by the alternating series test. Conditionally convergent.

8. A function \(f\) has \(f(1) = 2\), \(f'(1) = -3\), and \(f''(1) = 8\). (a) Write the second-degree Taylor polynomial \(P_2(x)\) for \(f\) about \(x = 1\). (b) Use it to approximate \(f(1.2)\). (c) Given that \(|f'''(z)| \le 15\) for all \(z\) in \([1, 1.2]\), use the Lagrange error bound to bound the error in (b).

   Show answer

   (a) \(P_2(x) = 2 - 3(x - 1) + \dfrac{8}{2!}(x - 1)^2 = 2 - 3(x - 1) + 4(x - 1)^2\). (b) \(P_2(1.2) = 2 - 0.6 + 4(0.04) = 1.56\). (c) \(|f(1.2) - P_2(1.2)| \le \dfrac{15\,(0.2)^3}{3!} = \dfrac{15(0.008)}{6} = 0.02\).

9. Find the radius and interval of convergence of \(\displaystyle\sum_{n=1}^{\infty}\frac{n\,(x - 4)^n}{3^n}\).

   Show answer

   Ratio test: \(\left|\dfrac{(n + 1)(x - 4)^{n+1}}{3^{n+1}}\cdot\dfrac{3^n}{n(x - 4)^n}\right| = \dfrac{|x - 4|}{3}\cdot\dfrac{n + 1}{n} \to \dfrac{|x - 4|}{3}\). Converges when \(|x - 4| \lt 3\), so \(R = 3\) and \(1 \lt x \lt 7\). Endpoints: at \(x = 7\), \(\displaystyle\sum n\) diverges (nth-term test); at \(x = 1\), \(\displaystyle\sum(-1)^n n\) diverges (terms don't approach 0). Interval of convergence: \((1, 7)\).

10. Let \(f(x) = \dfrac{x}{1 + x^2}\). (a) Find the first four nonzero terms and the general term of the Maclaurin series for \(f\). (b) Find \(f^{(5)}(0)\). (c) Use series to evaluate \(\displaystyle\lim_{x\to 0}\frac{\sin x - x}{x^3}\).

    Show answer

    (a) Substitute \(-x^2\) for \(x\) in \(\dfrac{1}{1 - x} = \sum x^n\): \(\dfrac{1}{1 + x^2} = 1 - x^2 + x^4 - x^6 + \cdots\), then multiply by \(x\): \(f(x) = x - x^3 + x^5 - x^7 + \cdots = \displaystyle\sum_{n=0}^{\infty}(-1)^n x^{2n+1}\), for \(|x| \lt 1\). (b) The coefficient of \(x^5\) is \(\dfrac{f^{(5)}(0)}{5!} = 1\), so \(f^{(5)}(0) = 5! = 120\). (c) \(\sin x - x = -\dfrac{x^3}{6} + \dfrac{x^5}{120} - \cdots\), so \(\dfrac{\sin x - x}{x^3} = -\dfrac{1}{6} + \dfrac{x^2}{120} - \cdots \to -\dfrac{1}{6}\).

Unit recap

Unit recap

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