Calculus curriculum

With no domain stated, take every real input the formula can actually be evaluated at. Inputs drop out where a denominator would be zero, where an even root would act on a negative number, and where a logarithm would act on zero or a negative number. Check the formula as written, not a simplified copy: x/x equals 1 everywhere except at 0, where it is undefined.

A function assigns each input exactly one output, so a vertical line can meet its graph at most once; two points on the same vertical would mean two outputs for one x. The curve y squared = x fails, since each positive x pairs with two values, plus and minus root x. Its upper half alone, y = root x, is the graph of a function.

The absolute value |x| is two formulas joined at 0: x for positive inputs and -x for negative ones. Rewrite it that way before differentiating, and the slope is 1 on the right and -1 on the left. At the joint itself neither formula settles the question; the one-sided slopes disagree there, so the graph has a corner and no derivative at 0.

Changes inside the function act on the input, so they run backward: replacing x by x + 2 moves the graph 2 units left, since each old output now arrives 2 inputs sooner. Changes outside act on the output and go the way they look: adding 3 lifts the graph 3 units. Replacing x by -x mirrors the graph left to right, which is how e to the -x relates to e to the x.

In f(g(x)), g acts on x first and f acts on the result, so the function written outside is the one applied last. To split an expression for the chain rule, ask what you would compute last: in sin(x squared) you square first and take the sine last, so sine is the outer function. Swapping the order gives (sin x) squared, a different function.

Far from the origin a polynomial behaves like its leading term, so the degree and the sign of the leading coefficient fix where both ends go. A rational function P/Q is undefined wherever Q is zero. If that zero cancels completely against factors of P, the graph has only a hole; if not, the values run off to plus or minus infinity there.

A function has an inverse only if no two inputs share an output, so a horizontal line may cross its graph at most once. Squaring fails, since x and -x have the same square, and y squared = x defines two functions, root x and minus root x. Restrict x squared to x of at least 0, where it only increases, and the inverse is the single function root x.

An exponential model multiplies by the same factor over each equal time step, so the percentage change stays fixed while the amount of change scales with the current size. That is the difference between b to the t, with the variable in the exponent, and t to the b, a power. Since e to the kt has rate of change k times itself, it models any quantity whose growth is proportional to its current size.

Grow 1 by 100 percent over one period, compounding n times, each step adding 1/n of the current amount, and it ends at (1 + 1/n)^n. As n increases this climbs toward 2.71828..., the number e, rather than without limit. Calculus prefers base e because e to the x is its own derivative, so no extra constant appears when you differentiate it.

The logarithm turns multiplication into addition: ln(xy) = ln x + ln y, so ln(1/x) = -ln x and ln(x to the n) = n ln x. That is why logs simplify a product, quotient or power before you differentiate. Any base converts to natural logs, since b to the x equals e to the (x ln b). The laws need positive inputs: ln(x squared) equals 2 ln|x|, not 2 ln x, when x is negative.

To free a variable from an exponent, take logs of both sides: b to the x = c becomes x ln b = ln c, so x = ln c / ln b. To free it from a logarithm, exponentiate instead. Then check each answer in the original equation, because a log only accepts positive inputs; a solution that makes a log's argument zero or negative must be thrown out.

A radian measures an angle by the arc it cuts off divided by the radius, so a full turn is 2 pi. Calculus needs it because sin x / x tends to 1 only when x is in radians, and that limit is what makes the derivative of sin x equal cos x. Measure in degrees and the derivative picks up an extra factor of pi/180.

A few identities do most of the rewriting: sin squared x + cos squared x = 1, and the double-angle form cos 2x = 1 - 2 sin squared x = 2 cos squared x - 1. Solved for the squares, the last gives sin squared x = (1 - cos 2x)/2 and cos squared x = (1 + cos 2x)/2, which trade a square that has no antiderivative in the basic table for a first-power cosine that does.

A limit at a describes the outputs for inputs near a, never the output at a itself. A function equal to 0 everywhere except at x = 0, where it is 1, still has limit 0 there: moving toward 0 from either side the outputs stay 0, and the lone point above the axis changes nothing. A function need not even be defined at a to have a limit there.

A right-hand limit uses only inputs above a, and a left-hand limit only inputs below. The two-sided limit exists exactly when both one-sided limits exist and agree. The greatest-integer function floor(x) at an integer m tends to m from the right but to m - 1 from the left, so it has no limit at m.

If f and g both have limits at a, the sum, product and quotient have limits equal to the sum, product and quotient of those limits, with the quotient needing a nonzero limit below. Each law assumes the pieces' limits exist first. You cannot split x sin(1/x) as lim x times lim sin(1/x), since sin(1/x) has no limit at 0, yet the product still tends to 0.

Plugging a into the formula gives the limit exactly when the function is continuous at a, which holds for every polynomial everywhere and for a rational function away from the zeros of its denominator. When substitution produces 0/0, that is not an answer: the expression is meaningless at a, but the limit may still exist and has to be found another way.

When a numerator and denominator that are polynomials both vanish at x = a, each has x - a as a factor. Cancel it, which is legitimate because the limit never uses x = a itself, then substitute what is left. Not every 0/0 factors: when a square root causes the zero, multiply top and bottom by the conjugate, the same expression with the sign between its terms flipped.

If g(x) is at most f(x), which is at most h(x), for x near a, and g and h both tend to L, then f is forced to L as well, whatever it does in between. The usual target is a bounded factor times something shrinking: sin(1/x) stays between -1 and 1, so x sin(1/x) is pinned between -|x| and |x| and tends to 0, although sin(1/x) itself has no limit at 0.

Both results come from comparing a small arc with lengths beside it: for x in radians between 0 and pi/2, sin x < x < tan x, which traps sin x / x between cos x and 1. The cosine version follows by multiplying 1 - cos x by 1 + cos x, which turns it into sin squared x. Before using either, make the angle match the denominator: sin(ax)/x is a times sin(ax)/(ax), so its limit is a, not 1.

Saying f(x) tends to infinity as x approaches a records how the limit fails: the values eventually exceed every bound you name. The sign comes from checking each side. 1/x is large and positive just right of 0 and large and negative just left of it, while 1/x squared is positive on both sides; either way the graph runs along the vertical line x = 0.

f is continuous at a when f(a) is defined, the limit as x approaches a exists, and the two are equal. Each clause fails in its own way: a missing value leaves a hole, one-sided limits that disagree leave a jump, and a value plotted off the curve leaves a limit that exists but does not match it.

A removable discontinuity has matching one-sided limits but a missing or wrong value at the point, so redefining that one value repairs it. A jump has one-sided limits that disagree, and no single value can bridge them. At an infinite discontinuity the function becomes unbounded, as 1/x does at 0. Some points fit none of these: sin(1/x) oscillates at 0.

On an open interval, continuity means continuity at every point. At the ends of a closed interval [a, b] only one side counts: the right-hand limit at a must equal f(a) and the left-hand limit at b must equal f(b). Root x is continuous at 0 in this sense though it is undefined to the left. The closed interval matters for the big theorems: 1/x is continuous on (0, 1] yet has no largest value there.

Each piece of a piecewise function is usually continuous on its own, since polynomials are continuous and sums, products and compositions of continuous functions stay continuous. So the only point in doubt is the breakpoint. Compute the limit from each side with that side's formula, set both equal to the function's value there, and solve for the unknown constant.

If f is continuous on [a, b] and N lies between f(a) and f(b), then f takes the value N at least once between a and b. Used with N = 0 on a sign change, it proves a root exists but gives no location. Continuity is essential, since a jump can skip values. The converse fails: a discontinuous function can still hit every value in between.

A function continuous on a closed, bounded interval [a, b] reaches a largest and a smallest value somewhere in it. Drop either hypothesis and the guarantee can fail: x - floor(x) on [0, 1] comes arbitrarily close to 1 yet never equals it, because it jumps at 1, and 1/x on (0, 1] has no largest value at all.

The definition is a challenge and response: lim f(x) = L as x approaches a means that for every tolerance epsilon > 0, however small, some delta > 0 exists so that every x with 0 < |x - a| < delta gives |f(x) - L| < epsilon. A proof is a recipe that produces delta from epsilon. The condition 0 < |x - a| leaves out x = a itself on purpose.

f'(a) is the limit of the difference quotient (f(a + h) - f(a))/h as h approaches 0. Writing x = a + h gives the equivalent form, the limit of (f(x) - f(a))/(x - a) as x approaches a. Substituting h = 0 directly always gives 0/0, so simplify first. If the limit does not exist, f has no derivative at a.

The tangent at P is the limiting position of the chord PQ as Q slides along the curve toward P. Each chord's slope is a difference quotient, so the tangent's slope is their limit, f'(a). If chords from the two sides settle on different slopes, the curve has a corner and no tangent there; a vertical tangent has no finite slope at all.

An average rate is the total change divided by the length of the interval, the slope of the chord between two points. The instantaneous rate is the limit of those averages as the interval shrinks to nothing, the slope of the tangent. A question that names an interval wants the first; one that names a single moment wants the second.

If f'(a) exists, f is continuous at a: the difference quotient can only settle to a finite value if f(a + h) - f(a) shrinks to 0. The converse fails. |x| is continuous everywhere, but at 0 its slope is -1 from the left and 1 from the right, so it has a corner and no derivative there.

A derivative fails at any discontinuity, at a corner where the left and right slopes differ, and where the tangent turns vertical, as for the cube root of x at 0. Those are not the only cases: x sin(1/x), given the value 0 at x = 0, is continuous there, yet its difference quotient is sin(1/h), which oscillates and never settles.

Computing f'(x) at every x where the limit exists defines a new function with its own domain, which can be smaller than f's: the cube root of x is defined everywhere but has no derivative at 0. Being a function, f' can be graphed and differentiated again to give f''. It need not even be continuous: x squared sin(1/x), set to 0 at 0, has a derivative everywhere that oscillates near 0.

A derivative carries output units per input unit. If V(t) is the volume of water in a tank in liters and t is time in minutes, V'(t) is in liters per minute and says how fast the volume is changing at that instant. A sentence interpreting f'(a) should name the quantity, the moment, the rate with its units, and whether the quantity is rising or falling.

The derivative of x to the n is n x to the (n - 1) for any real exponent, not only positive whole numbers. To use it, first write roots and reciprocals as powers: root x is x to the 1/2, with derivative (1/2)x to the -1/2, and 1/x is x to the -1, with derivative -x to the -2. The variable must be in the base; b to the x needs the exponential rule.

The derivative of a product is (fg)' = f'g + fg': each factor takes a turn being differentiated while the other is held as it is. Differentiating both factors and multiplying, f'g', is the classic error; for x times x it would give 1 instead of 2x. Linearity covers sums and constant multiples, not products.

The derivative of f/g is (f'g - fg')/g squared, defined where g is not 0. The order in the numerator matters, since reversing it flips the sign, and the denominator is squared, not differentiated. When the denominator is a single power such as x squared, rewriting the quotient as a product with a negative exponent is often quicker.

Many expressions need no product or quotient rule once rewritten. Expand a product of polynomials, split a fraction whose denominator is a single term into separate powers, and turn roots into fractional exponents; then the power rule and the sum rule finish the job. For example, (x squared + 1)/x becomes x + x to the -1, with derivative 1 - x to the -2.

Differentiating f' gives the second derivative f'', written d2y/dx2, and repeating gives the nth derivative. For position s(t), the first derivative is velocity and the second is acceleration. Differentiating x to the m n times gives m(m - 1)...(m - n + 1) x to the (m - n), so every derivative of a polynomial beyond its degree is 0.

The tangent at (x0, y0) passes through that point with slope f'(x0), so its equation is y - y0 = f'(x0)(x - x0). The normal is the perpendicular through the same point, with slope -1/f'(x0) when f'(x0) is not 0. If f'(x0) = 0 the tangent is horizontal and the normal is the vertical line x = x0.

Before differentiating, name the inner and outer functions: the outer is the last operation you would perform when evaluating. In (ax + b) to the m, the inner is ax + b and the outer is the mth power, so the derivative is m a (ax + b) to the (m - 1). Giving the inner expression its own letter, as in u = ax + b, makes the layers visible.

If y = F(f(x)), then y' = F'(f(x)) times f'(x): differentiate the outer function, leaving the inner one untouched inside it, then multiply by the derivative of the inner function. Forgetting that last factor gives an answer that looks right and is wrong. A quick check: the derivative of F(ax + b) must carry a factor a.

With three or more layers the rule repeats: the derivative of F(f(g(x))) is F'(f(g(x))) times f'(g(x)) times g'(x), one factor per layer, working from the outside in. When a product or quotient is also present, decide which operation is outermost. In (x sin x) cubed the cube is outer, so the chain rule comes first and the product rule handles the inner derivative.

When y is not solved for, differentiate both sides of the equation with respect to x, treating y as a function of x, so every term containing y picks up a factor dy/dx by the chain rule. Then gather the dy/dx terms and solve. For x cubed + y cubed = 3axy this gives dy/dx = -(x squared - ay)/(y squared - ax), with no need to solve for y first.

To get d2y/dx2 from an implicit equation, differentiate your expression for dy/dx again with respect to x, remembering that y inside it still depends on x and so produces another dy/dx. Then substitute the dy/dx you already found. For y squared = x, 2y y' = 1 gives y' = 1/(2y), and differentiating again gives y'' = -y'/(2y squared) = -1/(4y cubed).

If g is the inverse of f and f(a) = b, the slopes at corresponding points are reciprocals: g'(b) = 1/f'(a), provided f'(a) is not 0. You do not need a formula for g, only the input a that f sends to b and the slope there. Where f'(a) = 0 the inverse has a vertical tangent and no derivative, as the cube root does at 0.

For a power whose base and exponent both vary, like x to the x, take logs first: ln y = x ln x. Differentiate, writing y'/y on the left, then multiply through by y to get y' = x to the x times (ln x + 1). The same move turns a long product or quotient into a sum of logs. The log of y needs y > 0, or use ln|y|, whose derivative is still y'/y.

With x in radians, the derivative of sin x is cos x and the derivative of cos x is -sin x; the minus sign belongs to cosine. Both come from the sine difference formula and the limit sin h / h tends to 1. A check at 0: sine is rising through 0 with slope cos 0 = 1, while cosine is at its peak with slope -sin 0 = 0.

Write each in sines and cosines and apply the quotient rule: tan x gives sec squared x, sec x gives sec x tan x, cot x gives -csc squared x, and csc x gives -csc x cot x. The co-functions take the minus sign, as cosine does. Each formula holds only where the function itself is defined, so tan and sec fail where cos x = 0.

The derivative of e to the x is e to the x: at every point the slope equals the height. That makes it the natural model for anything whose growth rate is proportional to its current size, since dy/dx = ay is solved by y = K e to the ax. It also means e to the x is positive and increasing everywhere.

Since b to the x equals e to the (x ln b), the chain rule gives its derivative as b to the x times ln b. That factor ln b equals 1 only when b = e; it is below 1 for bases between 1 and e and above 1 for bases past e. Do not confuse b to the x with x to the b: the power rule applies only when the variable sits in the base.

The derivative of ln x is 1/x, on its domain x > 0. With the chain rule, ln(u) has derivative u'/u, the form behind logarithmic differentiation. The slope 1/x is steep near 0 and gentle for large x, which is why ln x grows without bound yet more slowly than any positive power of x.

Each inverse trig derivative comes from the inverse-function rule, and the result is algebraic. Arctan x has derivative 1/(1 + x squared) for every x, and arcsin x has derivative 1/root(1 - x squared) on -1 < x < 1, taking arcsin between -pi/2 and pi/2 where cosine is positive. Arccos and arccot carry the opposite sign.

A related-rates problem asks how one changing quantity's rate depends on another's, which only makes sense once the quantities are tied together. Write that link first, valid at every moment: for a ladder of fixed length l sliding down a wall, the distance x of its foot and the height y it reaches satisfy x squared + y squared = l squared. Differentiating comes second.

Differentiate the relating equation with respect to t, so every changing variable contributes its own rate by the chain rule: x squared + y squared = l squared becomes 2x dx/dt + 2y dy/dt = 0, and the fixed length drops out. A shrinking quantity has a negative rate; as the ladder's foot moves out, dy/dt comes out negative. Treating a changing variable as constant silently deletes a term.

Put in the given instant's values only after differentiating. If you substitute the ladder's current base distance first, the equation treats a changing quantity as a constant, its derivative becomes 0, and the rate you needed disappears. Quantities that truly stay fixed, like the ladder's length, may go in early.

Most setups reuse a short list of relations: Pythagoras for a ladder against a wall, and volume formulas for a tank, cone or balloon. A sphere's V = (4/3) pi r cubed gives dV/dt = 4 pi r squared dr/dt, so at a steady volume rate the radius grows more slowly as the balloon gets bigger. For a cone of fixed shape, use the fixed radius-to-height ratio to remove one variable first.

Near x = a the tangent line L(x) = f(a) + f'(a)(x - a) hugs the graph, so it can stand in for f. Taylor's theorem puts the gap at (1/2) f''(c)(x - a) squared for some c between, so the error shrinks like the square of the distance from a, and it grows quickly as you move away, fastest where the curve bends sharply.

Writing dy = f'(x) dx turns a small input error dx into an estimated output error dy. For a sphere, V = (4/3) pi r cubed gives dV = 4 pi r squared dr, and dividing by V gives dV/V = 3 dr/r: a small percentage error in the radius roughly triples in the volume. Relative error, dy/y, is often the more useful number.

Taylor's theorem of the second order gives f(x) = f(a) + f'(a)(x - a) + (1/2) f''(t)(x - a) squared for some t between a and x. That last term is how far the curve sits from the tangent line, and its sign is the sign of f''. Where f'' is positive (concave up) the tangent estimate runs low; where f'' is negative it runs high. Since t is unknown, check f'' over the whole stretch from a to x, not only at a.

A local maximum is a value that nothing nearby exceeds, within some small neighborhood on both sides; an absolute maximum is one that nothing in the whole domain exceeds. So a function can have a local maximum lower than a local minimum elsewhere. On a closed interval an absolute extreme can sit at an endpoint, where there is no neighborhood on both sides, so endpoints need checking separately.

A critical point is an interior point where f' is 0 or does not exist. Every local extremum of f happens at one, but not every critical point is an extremum: x cubed has derivative 0 at x = 0 and keeps rising through it. The peak of 1 - |x| at 0 is a maximum with no derivative at all, which is why points where f' fails to exist belong on the list.

If f has a local maximum or minimum at an interior point c and f'(c) exists, then f'(c) = 0. A positive slope at c would put larger values just to the right, and a negative slope larger values just to the left. That cuts the search to the points where f' is zero or undefined, plus any endpoints.

To find the absolute extremes of a continuous f on [a, b], list the critical points inside the interval, evaluate f there and at both endpoints, and compare: the largest value is the maximum and the smallest the minimum. No derivative test is needed. Skipping the endpoints is the classic omission, since an extreme can sit there with f' not 0, as for f(x) = x on [0, 1].

Suppose f is continuous on the closed interval from a to b, differentiable inside it, and takes equal values at the two ends; then the derivative is 0 at one or more points strictly between. Every hypothesis matters: 1 - |x| on [-1, 1] has equal end values but a corner at 0, and its slope is never 0. A common use is counting roots, since between two roots of f lies a root of f'.

When f has no breaks on the closed interval [a, b] and a derivative at every interior point, there is some c inside with f'(c) = (f(b) - f(a))/(b - a): at some instant the rate equals the average. Geometrically, somewhere a tangent runs parallel to the chord. The theorem promises that c exists, not where it is, and a single corner can break it.

If f'(x) = 0 at every point of an interval, f is constant there, because the Mean Value Theorem makes f(b) - f(a) = (b - a) times 0 for any two points. Two functions with the same derivative on an interval therefore differ by a constant. The interval matters: floor(x) has derivative 0 wherever it has one, yet it jumps at each integer.

If f' > 0 throughout an interval, f is increasing there; if f' < 0, decreasing. To use it, mark every point where f' is 0 or undefined, then test the sign of f' in each piece between those marks. A positive derivative at a single point is not enough to make f increasing on any interval around it.

At a critical point c, read the sign of f' just to the left and just to the right. A change from positive to negative marks a local maximum, negative to positive a local minimum, and no change marks neither. The test still works where f' does not exist at c, provided f is continuous there.

Concave up means the slope f' is increasing, so the graph bends upward and lies on or above its tangent lines; f'' at least 0 throughout signals it. Concave down means f' is decreasing and the graph lies below its tangents. Concavity is about how the slope changes, not whether f rises: a graph can be increasing and concave down at once.

Where f'' > 0 the slope f' is increasing, and where f'' < 0 it is decreasing, so the bend can switch only where f'' changes sign. A zero of f'' just marks a place to look: x to the 4th has f''(0) = 0, yet f'' = 12x squared is positive on both sides, so the slope keeps increasing and the bend never switches. Some older books define the term through f'' = 0 and the tangent's contact instead, so check which sense a text uses.

At a critical point where f'(c) = 0, f''(c) < 0 means a local maximum and f''(c) > 0 a local minimum. If f''(c) = 0 the test says nothing: x to the 4th, -x to the 4th and x cubed all have first and second derivatives 0 at 0, and they give a minimum, a maximum and neither. Then fall back on the sign of f' around c.

On a graph of f', stretches above the axis mark where f increases and stretches below where it decreases. Where f' crosses the axis going down, f has a local maximum; going up, a local minimum. Where f' is rising, f is concave up, and where f' peaks or bottoms out, f has an inflection point.

Divide numerator and denominator by the highest power in the denominator. Every leftover term like 1/x goes to 0, leaving 0 if the numerator's degree is lower, the ratio of leading coefficients if the degrees match, and divergence if the numerator's degree is higher. With a square root, root(x squared) is |x|, so as x heads to minus infinity it equals -x and the sign flips.

A horizontal asymptote y = L says f(x) approaches L as x heads to infinity or minus infinity; it describes only the far ends of the graph. Nothing stops the graph from crossing it, even repeatedly: e to the -x times sin x crosses y = 0 at every multiple of pi, yet tends to 0. The two ends can differ, as arctan x tends to pi/2 on the right and -pi/2 on the left.

Sketching runs on the calculus you already have. Find the domain and where the function is undefined, the behavior near those points and far out, then the sign of f' for rising and falling stretches with their maxima and minima, and the sign of f'' for concavity. Plot those landmarks and join them; each fact checks the others.

If f(x)/g(x) takes the form 0/0 or infinity/infinity as x approaches a, its limit equals the limit of f'(x)/g'(x), provided that limit exists and g' is not 0 near a. Differentiate top and bottom separately, not as a quotient, and check the form before every use: on (x + 1)/(x + 2) at 0, which is not indeterminate, the rule would wrongly give 1 instead of 1/2.

The rule handles only quotients, so the other forms get rewritten first. A product 0 times infinity becomes f divided by 1/g; infinity minus infinity is combined into one fraction; and 1 to the infinity, 0 to the 0 and infinity to the 0 are handled with logs, since ln(f to the g) = g ln f, a 0 times infinity product. Find that limit, then exponentiate.

As x grows, ln x tends to infinity more slowly than any positive power of x, even a tiny one; every power of x is outgrown by e to the x; and for whole numbers n, n! eventually exceeds any fixed b to the n. So in a ratio like (ln x) to the 5th over x, the bottom wins and the limit is 0.

Optimization starts in words: name the quantity to make largest or smallest and write it as a formula. Then write the condition tying its variables together, such as a fixed sum or a fixed amount of material. Two variables and one constraint leave one free variable, which is what single-variable calculus can handle.

Solve the constraint for one variable and substitute it into the objective, so the objective depends on a single variable; then differentiate and set the derivative to 0. If solving is awkward, differentiate the constraint implicitly instead and combine the two dy/dx expressions. Either way, the answer must satisfy both the objective's condition and the constraint.

A model's variable only makes sense on some interval: a length cannot be negative, and a cut cannot be bigger than the sheet. That interval decides the method. On a closed interval, compare critical values with the endpoint values; on an open one there are no endpoints to compare, so you must argue that the critical point really gives the extreme.

Finding f'(c) = 0 does not show that c gives the maximum, since a critical point can be a minimum or neither. Justify it: compare values by the closed interval method; or show f' is positive before c and negative after across the whole domain; or show f'' < 0 everywhere, since then f(x) = f(c) + (1/2) f''(t)(x - c) squared lies below f(c) for every x other than c.

To approximate a root of f(x) = 0, start from a guess and follow the tangent line to the axis: the next guess is x - f(x)/f'(x), and you repeat. Near a simple root each new error is on the order of the previous error squared, so correct digits roughly double. For x squared = 2 from 3/2, one step gives 17/12 and the next 577/408, already right to five decimal places.

The update divides by f'(x), so a guess where the tangent is nearly flat throws the next estimate far away, and f'(x) = 0 stops the method outright. It can also cycle: for x cubed - 2x + 2 started at 0, the guesses bounce between 0 and 1 forever. The fast convergence is guaranteed only close to a simple root, so start near one.

F is an antiderivative of f when F' = f. If F is one, F + C is another for every constant C, and on an interval these are all of them, since two functions with the same derivative there differ by a constant. Leaving off + C names one antiderivative where the question asked for the whole family.

Reverse the power rule: raise the exponent by one and divide by the new exponent, so x to the n integrates to x to the (n + 1) over (n + 1), plus C. That fails exactly at n = -1, where it would divide by 0; there the antiderivative of 1/x is ln|x| + C, valid on either side of 0. Rewrite roots and reciprocals as powers first.

Read the derivative table backward: e to the x integrates to e to the x, cos x to sin x, sin x to -cos x (watch that sign), sec squared x to tan x, 1/(1 + x squared) to arctan x, and 1/root(1 - x squared) to arcsin x, each plus C. With a linear inside, ax + b, divide the result by a.

A condition such as F(0) = 3 picks one member out of the family F(x) + C. Find the general antiderivative, substitute the given point, solve for C, and write out the single function. Then check it twice: differentiate to recover f, and evaluate at the given point to recover the given value.

Velocity is an antiderivative of acceleration and position an antiderivative of velocity, so integrating twice brings in two constants, fixed by the starting velocity and starting position. For constant acceleration a, v(t) = at + v0 and s(t) = (1/2)at squared + v0 t + s0. In free fall a is the constant -g when up is taken as positive.

Split [a, b] into n pieces of width (b - a)/n and add up rectangle areas, taking each height from f at the left end, the right end or the middle of its piece. From a table you can only use the listed x-values, which decides which sum you can build. For a continuous f all three sums approach the same integral as n grows.

For an increasing function each left rectangle sits under the curve and each right rectangle over it, so the left sum underestimates and the right sum overestimates; a decreasing function swaps them. Concavity governs the trapezoid estimate: on a concave-up graph every chord lies on or above the curve, so trapezoids overestimate.

The notation sum of a_k for k = 1 to n means a_1 + a_2 + ... + a_n. Constants factor out and a sum of terms splits into separate sums. The closed form 1 + 2 + ... + n = n(n + 1)/2 turns a Riemann sum for x into a limit you can evaluate, and the sum of the first n squares grows like n cubed over 3, which is where the integral of x squared gets its one third.

The definite integral is the limit of Riemann sums as the widest piece shrinks to 0. Read backward, a limit such as the sum over r from 0 to n - 1 of (1/n)/(1 + (r/n) squared) is a Riemann sum for 1/(1 + x squared) on [0, 1], so it tends to pi/4. The integration variable is a dummy: integrating f(t) dt or f(x) dx over [a, b] gives the same number.

The definite integral counts area above the x-axis as positive and area below as negative, so it measures net accumulation. The integral of sin x from 0 to 2 pi is 0 because the lobes cancel, although the region between the curve and the axis has area 4. For total area, split where f changes sign and add the absolute values.

Definite integrals are linear, so constants factor out and sums split. Adjacent intervals add: the integral from a to b plus the integral from b to c equals the one from a to c. Swapping the limits flips the sign, and an integral from a to a is 0; together these let the additivity rule work even when c lies outside [a, b].

When the integrand's graph is a line or a circle, read the integral as signed area instead of finding an antiderivative. The integral of root(a squared - x squared) from 0 to a is a quarter of a circle of radius a, so it equals (1/4) pi a squared. Line graphs give triangles and trapezoids, and an odd integrand on [-a, a] cancels to 0.

If F(x) is the integral of f from a fixed start a up to x, F tracks the area piled up so far: F(a) = 0, F rises where f is positive and falls where f is negative. So F's turning points sit where f changes sign, and you read them off the graph of f, not off any formula for F.

For continuous f, the derivative of the integral from a to x of f(t) dt is f(x): area grows at a rate equal to the height of the curve at the moving edge. The fixed lower limit does not affect the derivative, because moving it only adds a constant, and the dummy letter t is simply replaced by x.

When the upper limit is a function g(x), substitute g(x) into the integrand and multiply by g'(x); leaving out that factor is the usual slip. With both limits variable, take the upper result minus the lower one: f(g(x))g'(x) - f(h(x))h'(x).

If F is any antiderivative of a continuous f, the integral from a to b equals F(b) - F(a), and the constant of integration cancels in the subtraction. The continuity requirement is real: across a point where f blows up, F(b) - F(a) can give a finite number for an integral that does not exist.

Integrating a rate over an interval gives the net change in the quantity, because the quantity is an antiderivative of its rate. Units multiply: gallons per hour integrated over hours gives gallons. The result is a change, so add the starting amount when you want a final value.

Integrating velocity gives displacement, where travel backward cancels travel forward. Total distance comes from integrating speed, the absolute value of velocity. Find where velocity changes sign, integrate each piece separately, and add the pieces' absolute values.

Look for an inner function whose derivative also appears as a factor, perhaps off by a constant. In the integral of x cos(x^2), u = x^2 works because x dx is half of du. If the needed factor is missing and no constant can supply it, as with cos(x^2) alone, substitution does not finish the job.

After choosing u, write du = u'(x) dx and rewrite the whole integrand in terms of u, so that no x survives. A leftover x can often be expressed through u: with u = x + 1, a stray x becomes u - 1. An integrand that still mixes x and u is not ready to integrate.

For a definite integral, convert the bounds along with the integrand: if u = g(x), the new limits are g(a) and g(b), and you never return to x. The error is mixing the two, plugging the old x-limits into an antiderivative written in u. Change the limits or back-substitute, not half of each.

The integral of 1/u is ln|u| + C; the absolute value covers negative u, where ln u is undefined. Whenever the numerator is the derivative of the denominator, the answer is the log of the denominator, which is why tan x, sin x over cos x, gives -ln|cos x| + C. The secant integral also ends in a log: ln|sec x + tan x| + C.

Because e^u is its own derivative, its integral is e^u + C, and e^(kx) integrates to e^(kx)/k. For another base b, rewrite b^x as e^(x ln b); the antiderivative is b^x / ln b. Dropping that 1/ln(b) factor, or the 1/k, is the usual mistake.

Three denominators signal inverse trig answers: 1/sqrt(1 - x^2) integrates to arcsin x, 1/(1 + x^2) to arctan x, and, for x > 1, 1/(x sqrt(x^2 - 1)) to arcsec x. Watch the square root: 1/(1 + x^2) has none and gives arctan, not arcsin. With a^2 in place of 1, substitute x = au first.

On an interval from -a to a, an odd integrand, with f(-x) = -f(x), gives 0 because the two sides cancel; an even one, with f(-x) = f(x), gives twice the integral from 0 to a. Test the whole integrand: x^3 is odd but x^3 + 1 is neither. The shortcut also needs the integral to exist; 1/x on [-1, 1] is odd yet diverges.

Area under a curve is the integral of its height, so the area between two curves is the integral of top minus bottom. If the curves cross inside the interval, split there and keep each piece positive; one difference integrated across a crossing lets the pieces cancel and understates the area.

When a region's left and right edges are simpler than its top and bottom, slice horizontally: integrate right minus left with respect to y, using limits on the y-axis. A region whose top changes formula partway across needs two vertical integrals but may need only one horizontal one. Rewrite each curve as x in terms of y first.

If every cross-section perpendicular to an axis has a known area A(x), the volume is the integral of A(x) along that axis: thin slabs of thickness dx, added up. Disks and washers are the special case of circular slices. The slices must be perpendicular to the axis you integrate along, or A(x) measures the wrong thing.

Spinning a region around an axis makes circular slices of area pi r^2, so the volume is the integral of pi r^2. When the region does not touch the axis, each slice is a washer and its area is pi(R^2 - r^2). Squaring the difference, pi(R - r)^2, instead of subtracting the squares is the classic error.

A thin strip parallel to the axis of rotation sweeps a cylindrical shell of volume about 2 pi times radius times height times thickness. Shells are added up in the direction perpendicular to the axis, so revolving a region given as y = f(x) about a vertical line needs no solving for x, which washers would demand.

When the axis is a line such as y = -2 or x = 3, every radius is a distance to that line, not to a coordinate axis. Revolving around y = -2, a point at height y sits y + 2 away, so that is the radius to square. Sketch the axis first and measure each radius from it, outer and inner alike.

Over a short step the curve is nearly a straight segment with sides dx and dy, so arc length is the integral of sqrt(1 + (f'(x))^2) dx. Spinning the curve about the x-axis multiplies each bit of that length by the circumference 2 pi |f(x)| it sweeps. Weighting by dx alone ignores the slant and undercounts.

The average value of f on [a, b] is its integral divided by b - a: the height of a rectangle with the same area. Averaging a handful of sampled values only approximates it, since samples can miss what happens between them. For continuous f, the Mean Value Theorem for Integrals says f actually reaches its average somewhere in the interval.

Work is force times distance only when the force stays constant; when it varies with position, add up the work over short steps, giving the integral of F(x) dx. A spring's pull grows in proportion to how far it is stretched, so the work to stretch it is the integral of that rising force from rest to the final stretch; multiplying the final force by the distance overstates it.

Slice the liquid into thin horizontal layers. Each layer's weight is density times gravity times its volume, and each must be lifted its own distance to the outlet, so the integral adds weight times distance over all layers. A hanging cable works the same way: lower pieces travel farther than upper ones.

Pressure is force per unit area, and in a liquid it grows steadily with depth. A submerged plate therefore feels more push near its bottom edge. Slice it into horizontal strips, each at nearly one depth, multiply that depth's pressure by the strip's area, and integrate; using the pressure at one edge for the whole plate is the usual error.

When density varies, mass is the integral of density over the object: a rod with density rho(x) per unit length has mass equal to the integral of rho(x) dx along it. With radial symmetry, slice into thin rings of area 2 pi r dr and integrate rho(r) times that area. Multiplying one density by the total size is right only when density is constant.

The center of mass is the moment, the integral of position times mass, divided by the total mass: the balance point. If the object has a line of symmetry and uniform density, the center lies on that line, which can save a full computation. For a uniform triangle it sits one-third of the height above the base.

Integration by parts, the integral of u dv = uv minus the integral of v du, is the product rule read backward. For a definite integral the uv term is evaluated between the limits too, and forgetting it is common. The method trades one integral for another, so it helps only when the new one is easier.

Choose u to be a factor that gets simpler when you differentiate it, such as a power of x or ln x, and dv a factor you can integrate. For x e^x, u = x works at once. The reverse choice, u = e^x with dv = x dx, produces an integral containing x^2, harder than the one you started with.

A factor x^n may need parts n times, each round lowering the power by one. With e^x sin x, two rounds bring back the original integral; call it I, move it to one side and solve, giving e^x(sin x - cos x)/2 + C. Keep the same kind of choice for u both times, or the second round undoes the first.

For powers of sine and cosine, if one power is odd, peel off one factor of that function, convert the rest with sin^2 x + cos^2 x = 1, and substitute. If both are even, lower the degree with sin^2 x = (1 - cos 2x)/2 and cos^2 x = (1 + cos 2x)/2. Tangents and secants pair the same way using sec^2 x = 1 + tan^2 x.

sqrt(a^2 - x^2) suggests x = a sin(theta), sqrt(a^2 + x^2) suggests x = a tan(theta), and sqrt(x^2 - a^2) suggests x = a sec(theta), because each Pythagorean identity turns the radicand into a perfect square. After integrating, draw the right triangle the substitution describes and read the answer back in x.

A proper rational function whose denominator factors into distinct linear pieces splits into A/(x - r) + B/(x - s) + ..., and each term integrates to a logarithm. Find the constants by clearing denominators and substituting each root. If the numerator's degree is not below the denominator's, divide first.

A repeated factor such as (x - r)^3 needs a term for every power: A/(x - r) + B/(x - r)^2 + C/(x - r)^3. An irreducible quadratic factor gets a linear numerator, (Bx + C)/(x^2 + px + q), which integrates to a logarithm plus an arctangent. Leaving out a lower power makes the constants impossible to match.

Try algebra and basic formulas first, then substitution if a derivative is in view. A product of unlike types points to parts, powers of sine and cosine to identities, sqrt(a^2 +/- x^2) to trig substitution, and rational functions to partial fractions. Some integrands, such as e^(-x^2), have no elementary antiderivative at all.

An integral out to infinity is defined as a limit: integrate from a to a finite b, then let b grow. A finite limit means the integral converges; otherwise it diverges. The integral of 1/x^2 from 1 to infinity converges to 1, while 1/x keeps growing like ln b and diverges, though both integrands shrink to 0.

If the integrand blows up at an endpoint, define the integral as a limit as you approach that endpoint. If the trouble point is inside the interval, split there and require each piece to converge on its own. Ignoring it misleads: ln|x| evaluated from -1 to 1 gives 0, yet the integral of 1/x there does not exist.

The integral of 1/x^p from 1 to infinity converges exactly when p > 1, since the tail must thin out fast enough. Near zero the test reverses: the integral from 0 to 1 converges exactly when p < 1. At p = 1 both diverge, because the antiderivative ln x is unbounded at each end.

If 0 <= f(x) <= g(x) and the integral of g converges, so does the integral of f; if f's diverges, so does g's. The inequality must point the right way, since lying below a divergent integral proves nothing. For x >= 1, e^(-x^2) <= e^(-x), so its tail converges even without an antiderivative.

The trapezoidal rule averages the left and right sums, joining neighboring points with straight chords. Where the function is concave up the chords lie above the curve, so the rule overestimates; where it is concave down, it underestimates. Its error shrinks with the square of the step size.

Simpson's rule weights the sample values 1, 4, 2, 4, ..., 4, 1 and multiplies by h/3. Each parabola spans a pair of subintervals, so the number of subintervals must be even. Its error depends on the fourth derivative, which makes it exact for every polynomial of degree three or less.

The trapezoidal error is at most K(b - a)^3/(12n^2), where K bounds |f''|, and Simpson's is at most M(b - a)^5/(180n^4), where M bounds the fourth derivative. Doubling n cuts the first bound by 4 and the second by 16. A large derivative bound means more subintervals are needed.

To check a proposed solution, differentiate it as many times as the equation requires, substitute, and confirm the two sides agree for every x in the interval, not just at one point. How the candidate was found is irrelevant; a solution can look nothing like its equation. For instance, y = 5e^(-3x) satisfies y' = -3y.

Integrating leaves arbitrary constants, so a general solution is a family of curves: dy/dx = -ky gives y = Ce^(-kx) with one constant, while y'' = -n^2 y gives A sin(nx) + B cos(nx) with two. An initial condition picks one member: setting x = 0 in Ce^(-kx) shows C is the starting value. Apply conditions after integrating, never before.

A slope field draws, at many points, a short segment with the slope the equation dy/dx = f(x, y) assigns there; solution curves run tangent to the segments. If f is zero for every x at some fixed y = c, the horizontal line y = c is a solution. It is stable when nearby segments point toward it and unstable when they point away.

Euler's method steps along tangent lines: from (x, y) move to (x + h, y + h f(x, y)), then repeat from the new point. Each step treats the slope as fixed, so smaller steps follow the curve more closely but take more work. Where solutions bend sharply, large steps drift away from the true curve.

If dy/dx is a function of x times a function of y, put every y and dy on one side and every x and dx on the other, then integrate both sides. One constant on one side is enough; use the initial condition to find it. Dividing by h(y) can lose constant solutions where h(y) = 0, so check those separately.

When a quantity changes at a rate proportional to itself, dy/dt = ky, the solution is y = y0 e^(kt): growth for k > 0, decay for k < 0. Doubling time ln 2 / k and half-life ln 2 / |k| do not depend on the starting amount, and that fixed waiting time per doubling or halving is the signature of exponential change.

Newton's law of cooling says an object's temperature changes at a rate proportional to its difference from the surroundings. It is the gap, not the temperature, that decays exponentially: T = Ts + (T0 - Ts)e^(-kt). The temperature approaches the surrounding value Ts without ever crossing it.

Logistic growth follows dP/dt = rP(1 - P/K), where K is the carrying capacity. A small population grows almost exponentially, since 1 - P/K is near 1; growth is fastest at P = K/2, the S-curve's inflection point; and the curve levels off toward K. A population starting above K shrinks toward it.

A sequence converges to L when, for any tolerance however small, all terms past some point stay within it of L. If a_n = f(n) and f(x) has a limit as x grows, the sequence shares that limit, so function tools apply. The converse fails: sin(pi n) is 0 for every n, while sin(pi x) has no limit.

A sequence that never decreases and has an upper bound must converge, and so must one that never increases and has a lower bound. The theorem guarantees a limit exists without saying what it is. Bounded alone is not enough: (-1)^n stays between -1 and 1 and never settles.

For a recursion a_(n+1) = f(a_n) with continuous f, any limit L must satisfy L = f(L), so solving that equation lists the candidates. First show a limit exists, often by proving the sequence is monotone and bounded; otherwise the equation can offer a value for a sequence that never converges.

A series means the limit of its partial sums S_n, the running totals of the first n terms; it converges exactly when those totals settle. Whether the terms themselves approach 0 is a separate question: terms going to 0 is required for convergence but never enough to guarantee it.

A geometric series a + ar + ar^2 + ... converges exactly when |r| < 1, and then its sum is a/(1 - r), where a is the first term actually present. A series that starts at r^3 has first term r^3, not 1. When |r| >= 1 the terms never shrink to 0, so it diverges.

In a telescoping series each term splits into a difference, so neighboring pieces cancel in the partial sums. Since 1/(n(n + 1)) = 1/n - 1/(n + 1), the nth partial sum collapses to 1 - 1/(n + 1), and the series sums to 1. Always write the partial sum and take its limit rather than canceling infinitely many terms by eye.

The harmonic series 1 + 1/2 + 1/3 + ... diverges although its terms go to 0. Group it: 1/3 + 1/4 exceeds 1/2, the next four terms exceed 1/2, the next eight do too, and so on, so the partial sums grow without bound, if very slowly. Shrinking terms are necessary for convergence, never sufficient.

If the terms of a series do not approach 0, the series diverges: the terms n/(n + 1) approach 1, so their sum has no limit. If the terms do approach 0, the test is silent. Use it as a quick first filter, never as evidence of convergence.

If f is positive, continuous and decreasing, the series of f(n) and the integral of f from 1 to infinity converge or diverge together. Applied to 1/x^p, it shows the p-series converges exactly when p > 1. The test decides convergence only; the integral's value is not the series' sum.

For positive terms, a series smaller than a convergent one converges, and a series larger than a divergent one diverges. The other two directions decide nothing: lying below a divergent series or above a convergent one gives no verdict. The inequality only has to hold from some term onward.

If a_n and b_n are positive and a_n/b_n tends to a finite nonzero limit, the two series converge or diverge together. Choose b_n by keeping the dominant powers: (2n + 1)/(n^3 + 5) behaves like 2/n^2, so compare with the convergent 1/n^2. A limit of 0 or infinity gives only a one-way conclusion.

An alternating series converges when the sizes of its terms decrease steadily to 0. Both parts matter: alternating terms that shrink to 0 but not steadily can still give a divergent series. The alternating harmonic series 1 - 1/2 + 1/3 - ... passes, even though the plain harmonic series diverges.

When the alternating series test applies, the sum lies between any two consecutive partial sums, so stopping after n terms leaves an error no larger than the first term you left out. To guarantee an error under 0.001, keep adding terms until the next one is smaller than 0.001.

A series converges absolutely when the series of absolute values converges, and absolute convergence implies convergence. A series that converges only through cancellation, such as the alternating harmonic series, is conditionally convergent, and its terms can be rearranged to reach any sum. Test the absolute values first.

If |a_(n+1)/a_n| tends to L, the series converges absolutely when L < 1 and diverges when L > 1. At L = 1 the test fails, and every p-series lands there. It shines on factorials and powers, whose ratios simplify: for 1/n!, the ratio 1/(n + 1) goes to 0.

The root test takes the nth root of |a_n|: a limit below 1 means convergence, above 1 divergence, and exactly 1 decides nothing. It suits terms raised to the nth power. A sensible order: check that terms go to 0, spot geometric or p-series, compare with one, then reach for the ratio or root test.

A power series centered at c converges for |x - c| < R and diverges for |x - c| > R, where R is the radius; the ratio test usually finds it. Report the interval, not just R, because the endpoints need their own checks. R can also be 0, converging only at c, or infinite, converging everywhere.

At an endpoint the ratio test limit equals 1, so it cannot decide. Substitute the endpoint and test the resulting number series on its own, with the p-series, comparison or alternating series test. The two ends can differ: the series for ln(1 + x) converges at x = 1 but diverges at x = -1.

To expand a function, force it into the form a/(1 - r) and write a + ar + ar^2 + ..., valid when |r| < 1. For instance, 1/(1 + x^2) = 1/(1 - (-x^2)) gives 1 - x^2 + x^4 - ..., valid for |x| < 1. The validity condition comes from the r you built, not from x itself.

Integrating 1 - t + t^2 - ..., the series for 1/(1 + t), term by term from 0 to x gives x - x^2/2 + x^3/3 - ..., the series for ln(1 + x). The geometric series diverges at t = 1, yet the integrated series converges at x = 1, so endpoints can change. Differentiating e^x's series term by term returns the same series.

The degree-n Taylor polynomial at a matches f and its first n derivatives at a, using coefficients f^(k)(a)/k! on powers of (x - a). Raising the degree usually improves the fit near a, since the error shrinks like a high power of x - a, but the approximation can still be poor far from the center.

The Taylor series of f about a has coefficients f^(k)(a)/k! on powers of (x - a); a Maclaurin series is the case a = 0. Centering elsewhere means powers of (x - a), not of x. Continuous derivatives of every order do not by themselves make the series equal f; you must show the remainder tends to 0. The series may also converge on a smaller set than f is defined on: ln(1 + x) exists for all x > -1, but its series converges only for -1 < x <= 1.

e^x is the sum of x^n/n!; sin x and cos x use odd and even powers over factorials with alternating signs; 1/(1 - x) is the sum of x^n; and ln(1 + x) = x - x^2/2 + x^3/3 - .... The first three converge for every x, the last two only for |x| < 1, plus x = 1 for ln(1 + x).

Rather than computing derivatives, substitute into a known series: replacing x with -x^2 in the series for e^x gives e^(-x^2) = 1 - x^2 + x^4/2! - .... You can also multiply by powers of x or add series. The new series is valid wherever the substitution keeps you inside the original interval.

The Lagrange remainder says the error of a degree-n Taylor polynomial equals f^(n+1)(c)/(n + 1)! times (x - a)^(n+1) for some c between a and x. You never learn c, so bound that derivative over the whole interval to certify the accuracy. The factorial in the denominator is why few terms often suffice.

Swapping a function for its series can settle awkward limits: e^x - 1 - x begins x^2/2 + x^3/6, so (e^x - 1 - x)/x^2 tends to 1/2 as x tends to 0. Series also integrate functions with no elementary antiderivative, such as e^(-x^2), term by term.

Parametric equations give x and y as functions of a parameter t, so a curve carries a direction of travel and a timing. The circle x = cos t, y = sin t is traced counterclockwise once as t runs from 0 to 2 pi, but twice over an interval of length 4 pi; the equation x^2 + y^2 = 1 shows neither fact.

Eliminate the parameter by solving one equation for t and substituting, or by an identity such as cos^2 t + sin^2 t = 1. The resulting equation can include points the path never visits: x = t^2, y = t^2 gives y = x, but the path covers only the half with x >= 0.

The chain rule gives dy/dx = (dy/dt)/(dx/dt) wherever dx/dt is not zero. Tangents are horizontal where dy/dt = 0 while dx/dt is not, and vertical where dx/dt = 0 while dy/dt is not. If both vanish at once the formula decides nothing, and the point needs a closer look.

The second derivative is not the ratio of d^2y/dt^2 to d^2x/dt^2. Differentiate the slope dy/dx with respect to t, then divide by dx/dt once more, because each x-derivative is a t-derivative divided by dx/dt. Its sign gives the curve's concavity at that point.

A moving point's speed is sqrt((dx/dt)^2 + (dy/dt)^2), and integrating speed over a time interval gives the distance traveled. If the point retraces part of its path, that part is counted twice, so the result can exceed the curve's length; restrict t to a single pass.

Polar and Cartesian coordinates are linked by x = r cos(theta), y = r sin(theta) and r^2 = x^2 + y^2. Recovering theta from tan(theta) = y/x needs a quadrant check, because y/x is 1 at both (1, 1) and (-1, -1).

r = a cos(theta) is a circle, r = a(1 + cos(theta)) a cardioid, r = a + b cos(theta) a limacon, and r = a cos(n theta) a rose. A rose has n petals for odd n but 2n for even n: for odd n, the angle theta + pi gives r the opposite sign, which plots the same point, so the second half-turn retraces the first.

The area swept by a polar curve between two angles is one-half the integral of r^2 d(theta), built from thin sectors of area r^2 d(theta)/2. Choose limits that cover the region exactly once: for a rose with an odd number of petals, a full turn from 0 to 2 pi traces every petal twice and doubles the area.

A vector has a size and a direction; in components (a, b) its magnitude is sqrt(a^2 + b^2). Dividing a nonzero vector by its magnitude gives a unit vector pointing the same way. Vectors add component by component, which matches placing them head to tail, and multiplying by a negative number reverses the direction.

In space, the distance between two points is the square root of the sum of the squared differences in x, y and z, the Pythagorean theorem applied twice. The points at distance r from a center (a, b, c) form the sphere (x - a)^2 + (y - b)^2 + (z - c)^2 = r^2. Dropping the z term understates any distance between points at different heights.

The dot product multiplies matching components and adds them, giving a number, not a vector. It also equals |u||v| cos(theta), so dividing by both magnitudes gives the cosine of the angle between the vectors. A zero dot product means the vectors are perpendicular, and a negative one means the angle is obtuse.

The dot product u . v equals |v| times the signed length of u's shadow on v, so the projection of u onto v is (u . v / |v|^2) times v, the part of u running along v. Subtracting it from u leaves the part perpendicular to v. A negative dot product means the projection points opposite to v.

The cross product u x v is a vector perpendicular to the plane of u and v, with length |u||v| sin(theta), the area of the parallelogram they span. Order matters: v x u = -(u x v), so swapping the factors flips the direction. Parallel vectors span no area, so their cross product is the zero vector.

A line through the point (x0, y0, z0) with direction (a, b, c) is x = x0 + at, y = y0 + bt, z = z0 + ct. Solving each equation for t and setting the results equal gives symmetric equations, which need every component nonzero. In space, two lines can be neither parallel nor intersecting; such lines are called skew.

A plane through (x0, y0, z0) with normal vector (a, b, c) is a(x - x0) + b(y - y0) + c(z - z0) = 0, since every vector lying in the plane is perpendicular to the normal. Expanding gives the general form ax + by + cz + d = 0, so the coefficients of any plane equation read off a normal. Two different points do not fix a plane; one point and one normal do.

A function of two variables assigns a number to each point of a region in the plane, its domain, and its graph z = f(x, y) is a surface in space. Find the domain by excluding points where the formula breaks, such as negative values under a square root; the answer is a region, not an interval.

A level curve is the set of points where f(x, y) equals a fixed constant; a contour map draws several, and closely spaced contours mean the surface is steep there. A trace is the curve cut from the surface by a plane such as x = constant. Level curves give the view from above, traces the side views.

A limit of f(x, y) at a point must be the same along every path of approach, so two paths with different values prove there is no limit. For xy/(x^2 + y^2) at the origin, both axes give 0 while the line y = x gives 1/2. Each variable alone can behave well even when the pair does not.

The partial derivative with respect to x holds every other variable fixed and differentiates as usual, giving the surface's slope in the x-direction; nothing new is involved beyond that. It relies on the variables being independent: if y depended on x, you could not change x alone. Differentiating the x-partial again with respect to x gives a second partial.

If z depends on x and y and each depends on t, then dz/dt is the partial of z with respect to x times dx/dt, plus the partial with respect to y times dy/dt. A tree diagram lists every route from z down to t; multiply along each branch and add the branches. Missing a route is the usual error.

When the partials are continuous near (a, b), the change in f is close to f_x times the step in x plus f_y times the step in y, so the tangent plane z = f(a, b) + f_x(x - a) + f_y(y - b) estimates f for small steps. Partials alone are not enough: 2xy/(x^2 + y^2), set to 0 at the origin, has both partials there yet is not continuous.

The rate of change of f along a unit vector u = (u1, u2) is f_x u1 + f_y u2, the gradient dotted with u. The direction must be a unit vector; using an unnormalized one scales the answer by its length. Along the x-axis direction it reduces to the ordinary partial f_x.

The gradient collects the partial derivatives into a vector. Since the directional derivative is the gradient dotted with a unit vector, it is largest in the gradient's own direction, with value equal to the gradient's length. Along a level curve f does not change, so the gradient is perpendicular to the level curve through the point.

A double integral adds f(x, y) times small areas over a region, giving the volume under the surface when f is positive. For continuous f it can be computed as an iterated integral in either order when the region allows both descriptions. The inner limits may depend on the outer variable, never the reverse.