Calculus 2: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

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1Formulas and methods

Integrals

Power rule
∫xⁿ dx = xⁿ⁺¹/(n + 1) + C
Definite integral
∫ab f dx = F(b) − F(a)
Common integrals
∫eˣ dx = eˣ, ∫1/x dx = ln|x|, ∫sin x dx = −cos x
Area between curves
∫ab (f − g) dx

Method

  1. Rewrite the integrand as a sum of simple terms where possible.
  2. Apply the power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1), and standard integrals for ex, 1/x, sin and cos.
  3. Use substitution when you see a function and its derivative together.
  4. For a definite integral, evaluate F(b) − F(a).

Sequences & series

Arithmetic sequence
aₙ = a₁ + (n − 1)d, Sₙ = n(a₁ + aₙ)/2
Geometric sequence
aₙ = a₁rⁿ⁻¹, Sₙ = a₁(1 − rⁿ)/(1 − r)
Infinite geometric series
S = a₁/(1 − r), |r| < 1
p-series
Σ1/kᵖ converges only for p > 1
Ratio test
L = lim |ak+1/ak|: L < 1 converges, L > 1 diverges
Radius of convergence
R = lim |ck/ck+1|
Taylor polynomial
Pₙ(x) = Σ f⁽ᵏ⁾(a)(x − a)ᵏ/k!
Maclaurin series
eˣ = Σxᵏ/k!, sin x = x − x³/3! + …, cos x = 1 − x²/2! + …, ln(1 + x) = x − x²/2 + …
Lagrange error bound
|Rₙ| ≤ M·|x − a|ⁿ⁺¹/(n + 1)!

Method

  1. Identify the type: arithmetic (constant difference), geometric (constant ratio), a standard series, a power series or a Taylor polynomial.
  2. For sums use Sₙ = n(a₁ + aₙ)/2, Sₙ = a₁(1 − rⁿ)/(1 − r), or S = a₁/(1 − r) when |r| < 1, with a₁ the first term of the sum.
  3. For convergence: n-th term test first, then compare with Σ1/kᵖ or a geometric series, and use the ratio or root test for factorials and k-th powers.
  4. For Taylor polynomials list f(a), f′(a), f″(a), …, divide the k-th derivative by k!, and bound the error with M·|x − a|ⁿ⁺¹/(n + 1)!.

Multivariable calculus

Partial derivatives
fx = ∂f/∂x (y constant), fxy = fyx
Gradient
∇f = (fx, fy), largest slope |∇f|
Directional derivative
Duf = ∇f·u, u = v/|v|
Tangent plane
z = f(a, b) + fx(x − a) + fy(y − b)
Chain rule
dz/dt = fx·dx/dt + fy·dy/dt
Second-derivative test
D = fxxfyy − fxy²: D > 0, fxx > 0 minimum; D > 0, fxx < 0 maximum; D < 0 saddle
Double integral
∬R f dA = ∫ab∫g(x)h(x) f dy dx
Polar coordinates
x = r cos θ, y = r sin θ, dA = r dr dθ

Method

  1. For a partial derivative, treat the other variable as a constant and use the one-variable rules.
  2. For gradient questions compute ∇f = (fx, fy) at the point; use |∇f| for the steepest slope and ∇f·u with a unit vector u for a directional derivative.
  3. For extrema solve fx = 0 and fy = 0 together, then classify every point with D = fxx·fyy − fxy².
  4. For a double integral sketch the region, choose the order (or polar coordinates for discs) so the inner limits are the bounding curves, and integrate inside first.

Differential equations

Exponential growth/decay
dy/dt = ky → y = y0ekt
Half-life
k = ln 2 / t½
Newton's law of cooling
T(t) = Ta + (T0 − Ta)e−kt
Integrating factor
y′ + p(x)y = q(x): μ = e∫p dx, (μy)′ = μq
Mixing tank
dm/dt = r·c − (r/V)·m

2Practice problems

1.IntegralsMedium
xy26

Evaluate I = ∫26 4/x dx.

f(x) = 4/x[2, 6]
I =
2.IntegralsHard
xy1.5y = 3xy = 2x²

The region between the line y = 3x and the parabola y = 2x² is rotated about the x-axis. Find the volume V of the solid.

y = 3xy = 2x²
V =
3.Sequences & seriesMedium
AΣ 1/kk = 1 … ∞BΣ k³/3ᵏk = 1 … ∞CΣ k/(k + 6)k = 1 … ∞DΣ (5/3)ᵏk = 1 … ∞

Exactly one of the four series converges. Click it.

A: Σ 1/kB: Σ k³/3ᵏC: Σ k/(k + 6)D: Σ (5/3)ᵏ
  1. 1A: Σ 1/k
  2. 2B: Σ k³/3ᵏ
  3. 3C: Σ k/(k + 6)
  4. 4D: Σ (5/3)ᵏ
4.Sequences & seriesHard

Recognise the series and find its sum: S = Σk=0∞ (−1)ᵏ·22k/(2k)!.

ak = (−1)ᵏ·22k/(2k)!
S =
5.Multivariable calculusMedium
xy0.5−1∇fP

For f(x, y) = 2 ln(x² + 2y²), find the direction of steepest ascent at (0.5, −1) as the angle θ from the positive x-axis, in degrees between −180° and 180°.

f(x, y) = 2 ln(x² + 2y²)P = (0.5, −1)
θ = °
6.Multivariable calculusHard
xy−20−2−101AB

The marked points are the critical points of f(x, y) = x³ + y³ + 3xy. Click the local maximum.

f(x, y) = x³ + y³ + 3xy
  1. 1A (−1, −1)
  2. 2B (0, 0)
7.Differential equationsMedium
t (min)T (°C)30Ta​28 °C67 °C

A cup of coffee at 84 °C is left in a room at 24 °C. After 30 min its temperature is 28 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 67 °C?

T0 = 84 °CTa = 24 °CT(30) = 28 °CT = 67 °C
t = min
8.Differential equationsHard
t (min)m (kg)t = ?40 kg

A tank holds 1000 L of pure water. Brine with 0.1 kg of salt per litre flows in at 10 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 40 kg of salt?

V = 1000 Lm0 = 0 kgc = 0.1 kg/Lr = 10 L/min
t = min

3Solutions

1.Integrals

Answer: I = 4.394

Definite integrals: ∫ₐᵇ f(x) dx is the signed area between the graph and the x-axis from a to b. Find an antiderivative F (with F′ = f) and evaluate F(b) − F(a). Power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1). The area between two curves is ∫(upper − lower) dx.

Antiderivative

F(x) = 4·ln x

Evaluate at the limits

I = 4·(ln 6 − ln 2) = 4·ln 3 = 4.394

Answer

I = 4.394
2.Integrals

Answer: V = 12.72

Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.

Intersection points

3x = 2x² → x = 0, x = 1.5

Washer method

V = π∫₀1.5 ((3x)² − (2x²)²) dx

Integrate

V = π(9·1.5³/3 − 4·1.5⁵/5) = π·4.05 = 12.72

Answer

V = 12.72
3.Sequences & series

Answer: B: Σ k³/3ᵏ

Convergence: if the terms do not tend to 0 the series diverges (n-th term test). Σ1/kᵖ converges only for p > 1, and Σrᵏ only for |r| < 1. Compare a series with one of these when its terms behave like them for large k. Ratio test: L = lim |ak+1/ak|; root test: L = lim |ak|1/k. L < 1 converges, L > 1 diverges, L = 1 decides nothing. A power series Σck(x − x₀)ᵏ converges for |x − x₀| < R with R = lim |ck/ck+1|; an alternating series with terms decreasing to 0 converges.

Test each series

A: Σ 1/k → A multiple of the harmonic series (p = 1): diverges.
B: Σ k³/3ᵏ → Ratio test: L = 1/3 < 1, converges.
C: Σ k/(k + 6) → The terms tend to 1, not 0: diverges by the n-th term test.
D: Σ (5/3)ᵏ → Geometric with r = 5/3 > 1: the terms grow, so it diverges.

The answer

B: Σ k³/3ᵏ

Answer

B: Σ k³/3ᵏ
4.Sequences & series

Answer: S = −0.4161

Taylor polynomials: near x = a a smooth function is approximated by Pₙ(x) = f(a) + f′(a)(x − a) + f″(a)(x − a)²/2! + … + f⁽ⁿ⁾(a)(x − a)ⁿ/n!; for a = 0 this is the Maclaurin polynomial. Standard series: eˣ = Σxᵏ/k!, sin x = x − x³/3! + x⁵/5! − …, cos x = 1 − x²/2! + x⁴/4! − …, ln(1 + x) = x − x²/2 + x³/3 − …, 1/(1 − x) = Σxᵏ. Lagrange error bound: |f(x) − Pₙ(x)| ≤ M·|x − a|ⁿ⁺¹/(n + 1)!, where M is the largest value of |f⁽ⁿ⁺¹⁾| between a and x.

Standard series

Σ (−1)ᵏx2k/(2k)! = cos x

Substitute

x = 2
S = cos 2 = −0.4161

Answer

S = −0.4161
5.Multivariable calculus

Answer: θ = −75.96 °

Gradient: ∇f = (fx, fy) points in the direction of steepest ascent, and |∇f| is the largest rate of change. The directional derivative along a direction v is Duf = ∇f·u with the unit vector u = v/|v|. The tangent plane at (a, b) is z = f(a, b) + fx(a, b)(x − a) + fy(a, b)(y − b); it gives the linear approximation of f near that point. Chain rule: dz/dt = fx·dx/dt + fy·dy/dt.

Partial derivatives

fx = 4x/(x² + 2y²) → fx(0.5, −1) = 0.8889
fy = 8y/(x² + 2y²) → fy(0.5, −1) = −3.556

Direction

θ = atan2(−3.556, 0.8889) = −75.96°

Answer

θ = −75.96°
6.Multivariable calculus

Answer: A (−1, −1)

Critical points and double integrals: a critical point has fx = fy = 0. Second-derivative test with D = fxxfyy − fxy²: D > 0 and fxx > 0 is a local minimum, D > 0 and fxx < 0 a local maximum, D < 0 a saddle point, and D = 0 decides nothing. For a double integral over a general region, let the inner limits be the curves that bound the region; for discs and sectors use polar coordinates, x = r cos θ, y = r sin θ and dA = r dr dθ (do not forget the factor r).

Critical points

Solve fx = 0 and fy = 0 together.

fx = 3x² + 3y = 0
fy = 3y² + 3x = 0
(0, 0), (−1, −1)

Second-derivative test

(0, 0): fxx = 0, fyy = 0, fxy = 3, D = −9 → saddle point
(−1, −1): fxx = −6, fyy = −6, fxy = 3, D = 27 → local maximum

The point

A (−1, −1): local maximum

Answer

A (−1, −1)
7.Differential equations

Answer: t = 3.691 min

First-order differential equations: dy/dt = k·y has the solution y = y₀·ekt (growth for k > 0, decay for k < 0); a half-life T gives k = ln 2/T. Newton's law of cooling, dT/dt = −k·(T − Tₐ), gives T(t) = Tₐ + (T₀ − Tₐ)·e−kt. A linear equation y′ + p(x)·y = q(x) is solved with the integrating factor μ = e∫p dx: (μy)′ = μq, so μy = ∫μq dx + C.

Cooling constant

k = ln(60/4)/30 = 0.09027 min⁻¹

Time to reach the target

t = ln(60/43)/0.09027 = 3.691 min

Answer

t = 3.691 min
8.Differential equations

Answer: t = 51.08 min

First-order differential equations: dy/dt = k·y has the solution y = y₀·ekt (growth for k > 0, decay for k < 0); a half-life T gives k = ln 2/T. Newton's law of cooling, dT/dt = −k·(T − Tₐ), gives T(t) = Tₐ + (T₀ − Tₐ)·e−kt. A linear equation y′ + p(x)·y = q(x) is solved with the integrating factor μ = e∫p dx: (μy)′ = μq, so μy = ∫μq dx + C.

Result

m(t) = 100 + (−100)·e−t/100

Solve for t

e−t/100 = (100 − 40)/(100 − 0) = 0.6
t = 100·ln(1.667) = 51.08 min

Answer

t = 51.08 min

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