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
xy2y = 3x

The region under y = 3x from x = 0 to x = 2 is rotated about the x-axis. Find the volume V of the solid (a cone).

y = 3x0 ≤ x ≤ 2
V =
2.IntegralsHard
xy0.5y = x²eˣ

Evaluate I = ∫₀0.5 x²·eˣ dx.

∫₀0.5 x²eˣ dx
I =
3.Sequences & seriesMedium
AΣ k²/2ᵏk = 1 … ∞BΣ (5/3)ᵏk = 1 … ∞CΣ 1/k²k = 1 … ∞DΣ (3/4)ᵏk = 1 … ∞

Exactly one of the four series diverges. Click it.

A: Σ k²/2ᵏB: Σ (5/3)ᵏC: Σ 1/k²D: Σ (3/4)ᵏ
  1. 1A: Σ k²/2ᵏ
  2. 2B: Σ (5/3)ᵏ
  3. 3C: Σ 1/k²
  4. 4D: Σ (3/4)ᵏ
4.Sequences & seriesHard

f(x) = sin x is approximated at x = 0.5 by its Maclaurin polynomial of degree n. With the Lagrange bound M·|x|ⁿ⁺¹/(n + 1)! and M = 1, find the smallest degree n that guarantees an error below 0.001.

f(x) = sin xx = 0.5M = 1ε = 0.001
n =
5.Multivariable calculusMedium
xy−11∇fP

For f(x, y) = ln(x² + 4y²), find the maximum rate of change of f at the point (−1, 1), that is |∇f|.

f(x, y) = ln(x² + 4y²)P = (−1, 1)
|∇f| =
6.Multivariable calculusHard
xyy = 1.5y = x1.51.5

The integral I = ∫01.5 ∫x1.5 1/(1 + y²)² dy dx cannot be done in this order. Reverse the order of integration and evaluate it.

f = 1/(1 + y²)²0 ≤ x ≤ y ≤ 1.5
I =
7.Differential equationsMedium
t (min)T (°C)25Ta​18 °C21 °C

A can of soda at 3 °C is taken out of the fridge into a room at 24 °C. After 25 min its temperature is 18 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 21 °C?

T0 = 3 °CTa = 24 °CT(25) = 18 °CT = 21 °C
t = min
8.Differential equationsHard
tyK = 2ymax​ = ?

A system obeys y″ + y′ + 16.25y = 32.5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.

y″ + y′ + 16.25y = 32.5y(0) = 0y′(0) = 0
ymax =

3Solutions

1.Integrals

Answer: V = 75.4

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.

Disc method

V = π∫₀2 9x² dx

Integrate

V = π·9·2³/3 = 75.4

Answer

V = 75.4

Check: a cone of radius 6 and height 2 has (1/3)πr²h = 75.4.

2.Integrals

Answer: I = 0.0609

Integration by parts: ∫u·dv = u·v − ∫v·du, from the product rule. Choose u as the factor that gets simpler when you differentiate it (a power of x, or ln x), and dv as the factor you can integrate (eˣ, sin x, cos x, or x when u = ln x). For a definite integral, evaluate u·v at both limits: ∫ₐᵇ u dv = [u·v]ₐᵇ − ∫ₐᵇ v du. A factor x² needs two rounds.

First round

∫x²eˣ dx = x²eˣ − ∫2x eˣ dx

Second round

∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)

Limits

1.649·(1.25) − 1·2 = 0.0609

Answer

I = 0.0609
3.Sequences & series

Answer: B: Σ (5/3)ᵏ

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 k²/2ᵏ → Ratio test: L = 1/2 < 1, converges.
B: Σ (5/3)ᵏ → Geometric with r = 5/3 > 1: the terms grow, so it diverges.
C: Σ 1/k² → p-series with p = 2 > 1: converges.
D: Σ (3/4)ᵏ → Geometric with r = 3/4, |r| < 1: converges.

The answer

B: Σ (5/3)ᵏ

Answer

B: Σ (5/3)ᵏ
4.Sequences & series

Answer: n = 4

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.

n(n + 1)!bound010.5120.125260.020833240.00260441200.0002604

Bounds for increasing n

n = 3: 0.002604 ≥ 0.001
n = 4: 0.0002604 < 0.001

Smallest degree

n = 4

Answer

n = 4
5.Multivariable calculus

Answer: |∇f| = 1.649

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 = 2x/(x² + 4y²) → fx(−1, 1) = −0.4
fy = 8y/(x² + 4y²) → fy(−1, 1) = 1.6

Length of the gradient

|∇f| = √((−0.4)² + 1.6²) = 1.649

Answer

|∇f| = 1.649
6.Multivariable calculus

Answer: I = 0.3462

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).

Region

0 ≤ x ≤ y, 0 ≤ y ≤ 1.5

Reversed order

I = ∫01.5 ∫0y 1/(1 + y²)² dx dy = ∫01.5 y·1/(1 + y²)² dy

Substitution u = y²

I = [−1/(2(1 + y²))]01.5 = 0.3462

Answer

I = 0.3462
7.Differential equations

Answer: t = 38.83 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(-21/-6)/25 = 0.05011 min⁻¹

Time to reach the target

t = ln(-21/(−3))/0.05011 = 38.83 min

Answer

t = 38.83 min
8.Differential equations

Answer: ymax = 3.35

Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.

Steady value and roots

yss = 2, r = −0.5 ± 4j

Constants from the initial values

y(0) = 2 + A = 0 → A = −2
y′(0) = −0.5A + 4B = 0 → B = −0.25

Peak

t = π/4 = 0.7854
ymax = 2·(1 + e−0.5·π/4) = 3.35

Answer

ymax = 3.35

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