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IB Diploma Mathematics: Analysis & Approaches HL · Further Calculus & Differential Equations

Mini-Lesson

Further Calculus & Differential Equations

5 missions. Guess, learn, then check. Your place is saved.

Mission 1What is $\frac{s i n x}{x}$ (sin x)/x at $x = 0$ x = 0? Your calculator says error.
Mission 2A sliding ladder's top falls faster and faster. Why?
Mission 3Spin $y = x$ y = √x about the x-axis from 0 to 4. What volume is swept?
Mission 4A hot drink cools faster when it's hotter. What equation is that?
Mission 5How could a calculator find $sin 0.3$ sin 0.3 using only + and ×?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

What is $\frac{s i n x}{x}$ (sin x)/x at $x = 0$ x = 0? Your calculator says error.

Make a guess: What is $\frac{s i n x}{x}$ (sin x)/x at $x = 0$ x = 0? Your calculator says error.

Calculus · limits and l'Hôpital's rule

Limits and continuity

$lim_{x \to a} f (x) = L$ lim_(x → a) f(x) = L means $f (x)$ f(x) gets as close as you like to $L$ L as $x$ x gets close to $a$ a. The limit can exist even when $f (a)$ f(a) is undefined: $lim_{x \to 1} \frac{x ^{2} - 1}{x - 1} = lim_{x \to 1} (x + 1) = 2$ lim_(x → 1)(x² − 1)/(x − 1) = lim_(x → 1)(x + 1) = 2.
$f$ f is continuous at $a$ a if $lim_{x \to a} f (x) = f (a)$ lim_(x → a) f(x) = f(a): no gap or jump.
$f$ f is differentiable at $a$ a if the curve has a single tangent there. $y = ∣ x ∣$ y = |x| is continuous at 0 but has a corner, so it is not differentiable there.
A sequence or series that approaches a finite limit converges; otherwise it diverges.

Quick check

?Find

x \to 3 lim \frac{x ^{2} - 9}{x - 3}

Calculus · limits and l'Hôpital's rule

The derivative from first principles

The gradient of the tangent is the limit of chord gradients as the gap $h$ h shrinks to 0:

f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}

Method for a polynomial: expand $f (x + h)$ f(x + h), cancel, divide every term by $h$ h, then let $h \to 0$ h → 0.
Higher derivatives: differentiate again and again, written $f^{''} (x)$ f''(x), $f^{'''} (x)$ f'''(x), $f^{(n)} (x)$ f⁽ⁿ⁾(x) or $\frac{d ^{n} y}{d x ^{n}}$ dⁿy/dxⁿ. For $f (x) = x^{4}$ f(x) = x⁴, $f^{(4)} (x) = 24$ f⁽⁴⁾(x) = 24.

Worked example Calculus · limits and l'Hôpital's rule

Differentiating from first principles

Question

f (x) = 3 x^{2} - 2 x

. Use the definition of the derivative to show that

f^{'} (x) = 6 x - 2

1
Expand: $f (x + h) = 3 x^{2} + 6 x h + 3 h^{2} - 2 x - 2 h$ f(x + h) = 3x² + 6xh + 3h² − 2x − 2h
2
Subtract $f (x)$ f(x): $f (x + h) - f (x) = 6 x h + 3 h^{2} - 2 h$ f(x + h) − f(x) = 6xh + 3h² − 2h
3
Divide by $h$ h (as $h \neq = 0$ h ≠ 0): $\frac{f ( x + h ) - f ( x )}{h} = 6 x + 3 h - 2$ (f(x + h) − f(x))/h = 6x + 3h − 2
4
Let $h \to 0$ h → 0: $f^{'} (x) = lim_{h \to 0} (6 x + 3 h - 2) = 6 x - 2$ f'(x) = lim_(h → 0)(6x + 3h − 2) = 6x − 2

Answer

f^{'} (x) = 6 x - 2

, as required

⚠️ Watch out

putting

h = 0

before dividing gives

\frac{0}{0}

. Cancel the

h

first, and write "

lim_{h \to 0}

" on every line until the last.

Calculate

f (x) = 2 x^{3} - x

. Show that

\frac{f ( 1 + h ) - f ( 1 )}{h} = 2 h^{2} + 6 h + 5

, and hence find

f^{'} (1)

from first principles.

Calculus · limits and l'Hôpital's rule

L'Hôpital's rule

$\frac{0}{0}$ 0/0 and $\frac{\infty}{\infty}$ (∞)/(∞) are indeterminate forms: the limit could be anything.
L'Hôpital's rule: if $lim_{x \to a} \frac{f ( x )}{g ( x )}$ lim_(x → a)(f(x))/(g(x)) is of one of these forms, then

x \to a lim \frac{f ( x )}{g ( x )} = x \to a lim \frac{f ^{'} ( x )}{g ^{'} ( x )}

Differentiate the top and the bottom separately; this is not the quotient rule. The same works for $x \to \infty$ x → ∞.
If the new limit is still indeterminate, apply the rule again. Check the form every time.
Example: $lim_{x \to \infty} \frac{l n x}{x} = lim_{x \to \infty} \frac{1/ x}{1} = 0$ lim_(x → ∞)(ln x)/x = lim_(x → ∞)(1/x)/1 = 0.

Sort it

?Substitute the limit into each expression. Sort by the form you get.

Tap a statement, then tap the group it belongs in. Then press Check.

Worked example Calculus · limits and l'Hôpital's rule

Using l'Hôpital's rule twice

Question

Find

x \to 0 lim \frac{1 - cos x}{x ^{2}}

1
Substitute: $\frac{1 - 1}{0} = \frac{0}{0}$ (1 − 1)/0 = 0/0, indeterminate
2
Differentiate top and bottom: $lim_{x \to 0} \frac{s i n x}{2 x}$ lim_(x → 0)(sin x)/2x, which is still $\frac{0}{0}$ 0/0
3
Apply the rule again: $lim_{x \to 0} \frac{c o s x}{2}$ lim_(x → 0)(cos x)/2
4
Substitute: $\frac{c o s 0}{2} = \frac{1}{2}$ (cos 0)/2 = 1/2

Answer

x \to 0 lim \frac{1 - cos x}{x ^{2}} = \frac{1}{2}

⚠️ Watch out

using the quotient rule on

\frac{1 - c o s x}{x ^{2}}

is a different calculation. L'Hôpital differentiates numerator and denominator on their own.

Calculate

?Find

x \to 0 lim \frac{e ^{2 x} - 1 - 2 x}{x ^{2}}

Mission check

?What is

x \to 0 lim \frac{sin x}{x}

, with

x

in radians?

Mission 2

A sliding ladder's top falls faster and faster. Why?

Make a guess: A sliding ladder's top falls faster and faster. Why?

Calculus · implicit differentiation and related rates

Implicit differentiation

Some curves are given by an equation in $x$ x and $y$ y that is not solved for $y$ y, such as the circle $x^{2} + y^{2} = 25$ x² + y² = 25.
Differentiate both sides with respect to $x$ x, treating $y$ y as a function of $x$ x. By the chain rule, $\frac{d}{d x} (y^{2}) = 2 y \frac{d y}{d x}$ d/dx(y²) = 2ydy/dx. By the product rule, $\frac{d}{d x} (x y) = y + x \frac{d y}{d x}$ d/dx(xy) = y + xdy/dx.
Collect the $\frac{d y}{d x}$ dy/dx terms on one side and factorise.
For the circle: $2 x + 2 y \frac{d y}{d x} = 0$ 2x + 2ydy/dx = 0, so $\frac{d y}{d x} = - \frac{x}{y}$ dy/dx = −x/y.

Worked example Calculus · implicit differentiation and related rates

Gradient of an implicit curve

Question

Find the gradient of the curve

x^{2} + x y + y^{2} = 7

at the point

(1, 2)

1
Check the point lies on the curve: $1 + 2 + 4 = 7$ 1 + 2 + 4 = 7
2
Differentiate each term with respect to $x$ x: $2 x + y + x \frac{d y}{d x} + 2 y \frac{d y}{d x} = 0$ 2x + y + xdy/dx + 2ydy/dx = 0
3
Collect the $\frac{d y}{d x}$ dy/dx terms: $(x + 2 y) \frac{d y}{d x} = - (2 x + y)$ (x + 2y)dy/dx = −(2x + y)
4
Substitute $x = 1$ x = 1, $y = 2$ y = 2: $5 \frac{d y}{d x} = - 4$ 5dy/dx = −4

Answer

the gradient is

- \frac{4}{5}

⚠️ Watch out

x y

needs the product rule. Writing its derivative as just

\frac{d y}{d x}

, or as

y

, loses a term.

Calculus · implicit differentiation and related rates

Related rates and optimisation

When two linked quantities both change with time, differentiate the equation that links them with respect to $t$ t: $\frac{d}{d t} (y^{2}) = 2 y \frac{d y}{d t}$ d/dt(y²) = 2ydy/dt.
Method: sketch and label the variables; write the equation linking them; differentiate with respect to $t$ t; substitute the values after differentiating.
The chain rule links rates directly: $\frac{d V}{d t} = \frac{d V}{d r} \times \frac{d r}{d t}$ dV/dt = dV/dr × dr/dt.
In optimisation on an interval, compare the stationary values with the values at the end points. The optimum can be at an end point, where $f^{'} (x) \neq = 0$ f'(x) ≠ 0.

Worked example Calculus · implicit differentiation and related rates

The sliding ladder

Question

A 5 m ladder leans against a vertical wall. Its foot slides away from the wall at 0.4 m/s. Find the rate at which the top slides down the wall when the foot is 3 m from the wall.

1
Let the foot be $x$ x m from the wall and the top $y$ y m high: $x^{2} + y^{2} = 25$ x² + y² = 25
2
Differentiate with respect to $t$ t: $2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0$ 2xdx/dt + 2ydy/dt = 0, so $\frac{d y}{d t} = - \frac{x}{y} \frac{d x}{d t}$ dy/dt = −x/ydx/dt
3
When $x = 3$ x = 3: $y = 25 - 9 = 4$ y = √(25 − 9) = 4
4
Substitute: $\frac{d y}{d t} = - \frac{3}{4} \times 0.4 = - 0.3$ dy/dt = −3/4 × 0.4 = −0.3

Answer

the top slides down at 0.3 m/s (the negative sign means

y

is decreasing)

⚠️ Watch out

substitute

x = 3

only after differentiating. Putting

x = 3

in first makes

x

a constant, and its rate disappears.

Calculate

?A spherical balloon is inflated so that its volume increases at a constant 50 cm³ per second. Find the rate at which its radius is increasing when the radius is 10 cm. Give your answer to 3 significant figures.

cm per second

Quick check

?A firm's weekly profit is

P (x) = x^{3} - 6 x^{2} + 9 x + 2

thousand euros, where

x

hundred items are made and

0 \leq x \leq 5

. What is the maximum weekly profit?

Calculus · implicit differentiation and related rates

Further derivatives and their integrals

$f (x)$ f(x)	$f^{'} (x)$ f'(x)	$f (x)$ f(x)	$f^{'} (x)$ f'(x)
$tan x$ tan x	$sec^{2} x$ sec² x	$a^{x}$ a^x	$a^{x} ln a$ a^x ln a
$sec x$ sec x	$sec x tan x$ sec x tan x	$lo g_{a} x$ log_a x	$\frac{1}{x l n a}$ 1/(x ln a)
$cosec x$ cosec x	$- cosec x cot x$ −cosec x cot x	$arcsin x$ arcsin x	$\frac{1}{1 - x ^{2}}$ 1/(√(1 − x²))
$cot x$ cot x	$- cosec^{2} x$ −cosec² x	$arccos x$ arccos x	$- \frac{1}{1 - x ^{2}}$ −1/(√(1 − x²))
		$arctan x$ arctan x	$\frac{1}{1 + x ^{2}}$ 1/(1 + x²)

Reverse each one to integrate, for example $\int \frac{1}{a ^{2} + x ^{2}} d x = \frac{1}{a} arctan (\frac{x}{a}) + C$ ∫ 1/(a² + x²) dx = 1/aarctan (x/a) + C. Complete the square first if needed.
For a linear composite, divide by the coefficient of $x$ x: $\int sec^{2} (2 x + 5) d x = \frac{1}{2} tan (2 x + 5) + C$ ∫ sec²(2x + 5) dx = 1/2tan (2x + 5) + C.
Partial fractions turn some integrands into logarithms: $\frac{1}{x ^{2} + 3 x + 2} = \frac{1}{x + 1} - \frac{1}{x + 2}$ 1/(x² + 3x + 2) = 1/(x+1) − 1/(x+2), so the integral is $ln \frac{x + 1}{x + 2} + C$ ln |(x+1)/(x+2)| + C.

Quick check

?Find

\int \frac{1}{x ^{2} + 2 x + 5} d x

Calculate

?Find

\int_{01} \frac{x + 5}{( x + 1 ) ( x + 2 )} d x

. Give your answer to 3 significant figures.

Mission check

?A ladder's foot slides out at a steady speed. Why does its top fall faster and faster?

Mission 3

Spin $y = x$ y = √x about the x-axis from 0 to 4. What volume is swept?

Make a guess: Spin $y = x$ y = √x about the x-axis from 0 to 4. What volume is swept?

Calculus · further integration and volumes

Integration by substitution

To find $\int k g^{'} (x) f (g (x)) d x$ ∫ k g'(x) f(g(x)) dx, let $u = g (x)$ u = g(x). Then $d u = g^{'} (x) d x$ du = g'(x) dx and the integral becomes one in $u$ u.
If the integral is not of this form, the exam question gives you the substitution.
Replace every $x$ x, including $d x$ dx, before integrating. For a definite integral, change the limits to values of $u$ u as well, so you never need to switch back.

Worked example Calculus · further integration and volumes

A given substitution

Question

Use the substitution

u = x - 1

to find

\int_{12} x x - 1 d x

1
Write everything in $u$ u: $x = u + 1$ x = u + 1 and $d x = d u$ dx = du
2
Change the limits: $x = 1$ x = 1 gives $u = 0$ u = 0; $x = 2$ x = 2 gives $u = 1$ u = 1
3
Rewrite: $\int_{01} (u + 1) u d u = \int_{01} (u^{3/2} + u^{1/2}) d u$ ∫₀¹ (u + 1)√u du = ∫₀¹ (u^(3/2) + u^(1/2)) du
4
Integrate: $[\frac{2}{5} u^{5/2} + \frac{2}{3} u^{3/2}]_{01} = \frac{2}{5} + \frac{2}{3}$ [2/5u^(5/2) + 2/3u^(3/2)]₀¹ = 2/5 + 2/3

Answer

\frac{16}{15}

⚠️ Watch out

keeping the old limits 1 and 2 with the new variable

u

gives a wrong value.

Calculate

?Find

\int_{01} x^{2} (x^{3} + 1)^{4} d x

. Give your answer to 3 significant figures.

Calculus · further integration and volumes

Integration by parts

For a product, use integration by parts:

\int u \frac{d v}{d x} d x = uv - \int v \frac{d u}{d x} d x

Choose $u$ u to be the factor that gets simpler when differentiated (a power of $x$ x, or $ln x$ ln x). For $\int ln x d x$ ∫ ln x dx, take $u = ln x$ u = ln x and $\frac{d v}{d x} = 1$ dv/dx = 1 to get $x ln x - x + C$ xln x − x + C.
Repeated parts: for $\int x^{2} e^{x} d x$ ∫ x² e^x dx apply the rule twice. For $\int e^{x} sin x d x$ ∫ e^x sin x dx, two applications bring back the original integral; collect it on one side and solve for it.

Worked example Calculus · further integration and volumes

Repeated integration by parts

Question

Find

\int x^{2} e^{x} d x

1
Take $u = x^{2}$ u = x² and $\frac{d v}{d x} = e^{x}$ dv/dx = e^x, so $\frac{d u}{d x} = 2 x$ du/dx = 2x and $v = e^{x}$ v = e^x
2
Apply the rule: $\int x^{2} e^{x} d x = x^{2} e^{x} - \int 2 x e^{x} d x$ ∫ x² e^x dx = x² e^x − ∫ 2x e^x dx
3
Parts again with $u = 2 x$ u = 2x: $\int 2 x e^{x} d x = 2 x e^{x} - \int 2 e^{x} d x = 2 x e^{x} - 2 e^{x}$ ∫ 2x e^x dx = 2x e^x − ∫ 2e^x dx = 2x e^x − 2e^x
4
Combine, watching the signs: $x^{2} e^{x} - 2 x e^{x} + 2 e^{x} + C$ x² e^x − 2x e^x + 2e^x + C

Answer

e^{x} (x^{2} - 2 x + 2) + C

⚠️ Watch out

the minus sign in front of the second integral changes the sign of every term inside it.

Quick check

?Find

\int x cos x d x

Calculus · further integration and volumes

Areas with the y-axis and volumes of revolution

Area between a curve and the $y$ y-axis from $y = a$ y = a to $y = b$ y = b: write $x$ x in terms of $y$ y, then $A = \int_{a b} x d y$ A = ∫_a^b x dy (use $∣ x ∣$ |x| if part of the curve is left of the axis).
Spin a region about the $x$ x-axis and each thin slice becomes a disc of radius $y$ y and volume $π y^{2} δ x$ π y² δ x. Adding the discs:

V = \int_{a b} π y^{2} d x

About the $y$ y-axis, the discs have radius $x$ x: $V = \int_{c d} π x^{2} d y$ V = ∫_c^d π x² dy, with limits in $y$ y.

Worked example Calculus · further integration and volumes

Volume about the y-axis

Question

The region between the curve

y = x^{3}

, the

y

-axis and the lines

y = 1

and

y = 8

is rotated through

2 π

about the

y

-axis. Find the exact volume.

1
Write $x$ x in terms of $y$ y: $x = y^{1/3}$ x = y^(1/3), so $x^{2} = y^{2/3}$ x² = y^(2/3)
2
Set up with limits in $y$ y: $V = π \int_{18} y^{2/3} d y$ V = π∫₁⁸ y^(2/3) dy
3
Integrate: $V = π [\frac{3}{5} y^{5/3}]_{18} = \frac{3 π}{5} (32 - 1)$ V = π[3/5y^(5/3)]₁⁸ = 3π/5(32 − 1)

Answer

V = \frac{93 π}{5}

(about 58.4 cubic units)

⚠️ Watch out

about the

y

-axis you need

x^{2}

and

d y

. Using

\int π y^{2} d x

here finds a different solid.

Calculate

?Find the area of the region enclosed by the curve

y = ln x

, the

y

-axis, the

x

-axis and the line

y = 2

. Give your answer to 3 significant figures.

Mission check

?Spin

y = x

about the

x

-axis from

x = 0

x = 4

. What volume is swept out?

Mission 4

A hot drink cools faster when it's hotter. What equation is that?

Make a guess: A hot drink cools faster when it's hotter. What equation is that?

Calculus · differential equations

Modelling and separating variables

A differential equation links a function and its derivatives. "The rate of change of $T$ T is proportional to $T - 20$ T − 20" becomes $\frac{d T}{d t} = - k (T - 20)$ dT/dt = −k(T − 20), with $k > 0$ k > 0 for decrease.
If $\frac{d y}{d x} = f (x) g (y)$ dy/dx = f(x) g(y), the variables are separable: $\int \frac{1}{g ( y )} d y = \int f (x) d x$ ∫ 1/(g(y)) dy = ∫ f(x) dx.
Add one constant $C$ C, then use a given value (an initial condition) to find it. That gives the particular solution.
The logistic equation $\frac{d n}{d t} = kn (a - n)$ dn/dt = kn(a − n) is separable; use partial fractions on $\frac{1}{n ( a - n )}$ 1/(n(a − n)).

Worked example Calculus · differential equations

Newton's law of cooling

Question

A drink cools in a 20 °C room according to

\frac{d T}{d t} = - k (T - 20)

, where

T

°C is its temperature after

t

minutes.

T = 90

when

t = 0

and

T = 62

when

t = 5

. Find

T

when

t = 10

1
Separate and integrate: $\int \frac{1}{T - 20} d T = \int - k d t$ ∫ 1/(T − 20) dT = ∫ −k dt, so $ln ∣ T - 20∣ = - k t + c$ ln |T − 20| = −kt + c
2
Rearrange: $T = 20 + A e^{- k t}$ T = 20 + Ae^(−kt); $t = 0$ t = 0 gives $A = 70$ A = 70
3
Use $t = 5$ t = 5: $62 = 20 + 70 e^{- 5 k}$ 62 = 20 + 70e^(−5k), so $e^{- 5 k} = 0.6$ e^(−5k) = 0.6
4
Then $e^{- 10 k} = (e^{- 5 k})^{2} = 0.36$ e^(−10k) = (e^(−5k))² = 0.36, so $T = 20 + 70 \times 0.36$ T = 20 + 70 × 0.36

Answer

T = 45.2

°C

⚠️ Watch out

the drink does not lose 28 °C every 5 minutes. That would give 34 °C, as if it cooled at a constant rate.

Calculate

?Solve

\frac{d y}{d x} = x y^{2}

given that

y = 1

when

x = 0

. Find the exact value of

y

when

x = 1

Calculus · differential equations

Euler's method

When a DE cannot be solved exactly, step along it using the gradient at the start of each step. With step length $h$ h:

y_{n + 1} = y_{n} + h f (x_{n}, y_{n}), x_{n + 1} = x_{n} + h

Each step follows the tangent, so the estimate drifts from the true curve. If the curve is concave up, Euler's method underestimates.
A smaller step $h$ h gives a better estimate but needs more steps; use a spreadsheet or your GDC for many steps.

Calculate

?Use Euler's method with step length 0.1 to estimate

y

when

x = 0.3

, given

\frac{d y}{d x} = x + y

and

y = 1

when

x = 0

. Give your answer to 4 significant figures.

Calculus · differential equations

Homogeneous equations and the integrating factor

A homogeneous DE has the form $\frac{d y}{d x} = f (\frac{y}{x})$ dy/dx = f(y/x). Substitute $y = vx$ y = vx, so $\frac{d y}{d x} = v + x \frac{d v}{d x}$ dy/dx = v + xdv/dx. The result, $x \frac{d v}{d x} = f (v) - v$ xdv/dx = f(v) − v, is separable.
A linear DE $\frac{d y}{d x} + P (x) y = Q (x)$ dy/dx + P(x)y = Q(x) is solved with the integrating factor $I (x) = e^{\int P (x) d x}$ I(x) = e^(∫ P(x) dx). Multiplying through turns the left side into $\frac{d}{d x} (I y)$ d/dx(Iy):

\frac{d}{d x} (I (x) y) = I (x) Q (x)

Integrate both sides, then divide by $I (x)$ I(x).

Worked example Calculus · differential equations

A homogeneous equation

Question

Solve

\frac{d y}{d x} = \frac{x ^{2} + y ^{2}}{x y}

for

x > 0

, given that

y = 2

when

x = 1

. Give

y^{2}

in terms of

x

1
Substitute $y = vx$ y = vx: $v + x \frac{d v}{d x} = \frac{x ^{2} + v ^{2} x ^{2}}{v x ^{2}} = \frac{1}{v} + v$ v + xdv/dx = (x² + v²x²)/vx² = 1/v + v
2
Simplify: $x \frac{d v}{d x} = \frac{1}{v}$ xdv/dx = 1/v
3
Separate and integrate: $\int v d v = \int \frac{1}{x} d x$ ∫ v dv = ∫ 1/x dx, so $\frac{v ^{2}}{2} = ln x + c$ v²/2 = ln x + c
4
Put $v = \frac{y}{x}$ v = y/x and use $x = 1$ x = 1, $y = 2$ y = 2: $\frac{4}{2} = 0 + c$ 4/2 = 0 + c, so $c = 2$ c = 2

Answer

y^{2} = 2 x^{2} (ln x + 2)

⚠️ Watch out

\frac{d y}{d x}

becomes

v + x \frac{d v}{d x}

, not just

\frac{d v}{d x}

(product rule on

vx

Worked example Calculus · differential equations

Using an integrating factor

Question

Solve

\frac{d y}{d x} + \frac{2}{x} y = 4 x

for

x > 0

, given that

y = 3

when

x = 1

1
Find the integrating factor: $e^{\int \frac{2}{x} d x} = e^{2 l n x} = x^{2}$ e^(∫ 2/x dx) = e^(2ln x) = x²
2
Multiply through: $x^{2} \frac{d y}{d x} + 2 x y = 4 x^{3}$ x²dy/dx + 2xy = 4x³, that is $\frac{d}{d x} (x^{2} y) = 4 x^{3}$ d/dx(x² y) = 4x³
3
Integrate: $x^{2} y = x^{4} + C$ x² y = x⁴ + C
4
Use $x = 1$ x = 1, $y = 3$ y = 3: $3 = 1 + C$ 3 = 1 + C, so $C = 2$ C = 2

Answer

y = x^{2} + \frac{2}{x ^{2}}

⚠️ Watch out

divide the constant by

x^{2}

too. Writing

y = x^{2} + 2

is a common slip.

Sort it

?Choose the method to use first for each differential equation.

Tap a statement, then tap the group it belongs in. Then press Check.

Mission check

?A hot drink cools faster when it's hotter. Which equation models that, with

k > 0

Mission 5

How could a calculator find $sin 0.3$ sin 0.3 using only + and ×?

Make a guess: How could a calculator find $sin 0.3$ sin 0.3 using only + and ×?

Calculus · Maclaurin series

Maclaurin series

A Maclaurin series writes a function as a polynomial whose value and derivatives at $x = 0$ x = 0 match the function's:

f (x) = f (0) + x f^{'} (0) + \frac{x ^{2}}{2 !} f^{''} (0) + \frac{x ^{3}}{3 !} f^{'''} (0) + \dots

Find $f (0)$ f(0), $f^{'} (0)$ f'(0), $f^{''} (0)$ f''(0), … and substitute.
Near $x = 0$ x = 0, the first few terms give a good approximation. Each extra term improves it, as the graphs show.

Worked example Calculus · Maclaurin series

The series for $sin x$ sin x

Question

Find the Maclaurin series for

sin x

up to the term in

x^{5}

1
Differentiate repeatedly: $sin x$ sin x, $cos x$ cos x, $- sin x$ −sin x, $- cos x$ −cos x, $sin x$ sin x, $cos x$ cos x
2
Evaluate at $x = 0$ x = 0: $0, 1, 0, - 1, 0, 1$ 0, 1, 0, −1, 0, 1
3
Substitute: $0 + x (1) + \frac{x ^{2}}{2 !} (0) + \frac{x ^{3}}{3 !} (- 1) + \frac{x ^{4}}{4 !} (0) + \frac{x ^{5}}{5 !} (1)$ 0 + x(1) + x²/(2!)(0) + x³/(3!)(−1) + x⁴/(4!)(0) + x⁵/(5!)(1)

Answer

sin x = x - \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} - \dots

⚠️ Watch out

x

must be in radians. The derivative

\frac{d}{d x} sin x = cos x

holds only in radians.

Calculate

?Use the first three non-zero terms of the Maclaurin series

cos x = 1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \dots

to estimate

cos 0.5

. Give your answer to 5 decimal places.

Calculus · Maclaurin series

New series from known ones

Standard series, valid for all $x$ x unless stated:

$e^{x} = 1 + x + \frac{x ^{2}}{2 !} + \dots ln (1 + x) = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \dots (- 1 < x \leq 1)$ e^x = 1 + x + x²/(2!) + … ln (1 + x) = x − x²/2 + x³/3 − … (−1 < x ≤ 1) $(1 + x)^{p} = 1 + p x + \frac{p ( p - 1 )}{2 !} x^{2} + \dots (∣ x ∣ < 1)$ (1 + x)^p = 1 + px + (p(p−1))/(2!)x² + … (|x| < 1)

Substitute: replace $x$ x by $x^{2}$ x², $- x$ −x or $2 x$ 2x throughout a known series.
Multiply two series, keeping only the terms up to the power you need.
Differentiate or integrate a series term by term.

Worked example Calculus · Maclaurin series

A product of two series

Question

Find the Maclaurin series for

e^{x} sin x

up to the term in

x^{3}

1
Write each series up to $x^{3}$ x³: $(1 + x + \frac{x ^{2}}{2} + \frac{x ^{3}}{6}) (x - \frac{x ^{3}}{6})$ (1 + x + x²/2 + x³/6)(x − x³/6)
2
Collect the $x$ x and $x^{2}$ x² terms: $x$ x and $x \cdot x = x^{2}$ x · x = x²
3
Collect the $x^{3}$ x³ terms: $\frac{x ^{2}}{2} \cdot x - 1 \cdot \frac{x ^{3}}{6} = \frac{1}{2} x^{3} - \frac{1}{6} x^{3} = \frac{1}{3} x^{3}$ x²/2 · x − 1 · x³/6 = 1/2x³ − 1/6x³ = 1/3x³

Answer

e^{x} sin x = x + x^{2} + \frac{x ^{3}}{3} + \dots

⚠️ Watch out

collect every pair of terms whose powers add to 3. Missing

1 \times (- \frac{x ^{3}}{6})

gives

\frac{x ^{3}}{2}

Quick check

?Find the Maclaurin series for

e^{- x^{2}}

up to the term in

x^{4}

Calculate

?Use your answer to the last question to estimate

\int_{0 0.5} e^{- x^{2}} d x

. Give your answer to 4 significant figures.

Calculus · Maclaurin series

Series from a differential equation

If $\frac{d y}{d x} = f (x, y)$ dy/dx = f(x, y) and $y (0)$ y(0) are given, you can build the Maclaurin series of $y$ y without solving the equation.
Put $x = 0$ x = 0 into the DE to get $y^{'} (0)$ y'(0).
Differentiate the DE (implicitly, with the chain rule on terms in $y$ y) and substitute again to get $y^{''} (0)$ y''(0), $y^{'''} (0)$ y'''(0), ….
Then use $y = y (0) + x y^{'} (0) + \frac{x ^{2}}{2 !} y^{''} (0) + \frac{x ^{3}}{3 !} y^{'''} (0) + \dots$ y = y(0) + x y'(0) + x²/(2!)y''(0) + x³/(3!)y'''(0) + …

Worked example Calculus · Maclaurin series

A series from a DE

Question

\frac{d y}{d x} = x + y^{2}

and

y = 1

when

x = 0

. Find the Maclaurin series for

y

up to the term in

x^{3}

1
At $x = 0$ x = 0: $y^{'} (0) = 0 + 1^{2} = 1$ y'(0) = 0 + 1² = 1
2
Differentiate: $y^{''} = 1 + 2 y y^{'}$ y'' = 1 + 2y y', so $y^{''} (0) = 1 + 2 (1) (1) = 3$ y''(0) = 1 + 2(1)(1) = 3
3
Differentiate again: $y^{'''} = 2 (y^{'})^{2} + 2 y y^{''}$ y''' = 2(y')² + 2y y'', so $y^{'''} (0) = 2 + 6 = 8$ y'''(0) = 2 + 6 = 8
4
Substitute: $y = 1 + x + \frac{3}{2 !} x^{2} + \frac{8}{3 !} x^{3}$ y = 1 + x + 3/(2!)x² + 8/(3!)x³

Answer

y = 1 + x + \frac{3}{2} x^{2} + \frac{4}{3} x^{3} + \dots

⚠️ Watch out

\frac{d}{d x} (y^{2}) = 2 y \frac{d y}{d x}

, and differentiating

2 y y^{'}

needs the product rule.

Mission check

?How could a calculator find

sin 0.3

using only adding and multiplying?

Recap

Big ideas

A limit is the value approached; it can exist where the function is undefined. First principles: $f^{'} (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$ f'(x) = lim_(h → 0)(f(x + h) − f(x))/h.
For $\frac{0}{0}$ 0/0 or $\frac{\infty}{\infty}$ (∞)/(∞), use l'Hôpital: $lim \frac{f}{g} = lim \frac{f ^{'}}{g ^{'}}$ lim f/g = lim (f')/(g'), again if needed.
Check the form first: if it is not indeterminate, just substitute.
Implicit differentiation: differentiate every term with respect to $x$ x, using $\frac{d}{d x} (y^{n}) = n y^{n - 1} \frac{d y}{d x}$ d/dx(yⁿ) = nyⁿ⁻¹dy/dx and the product rule, then make $\frac{d y}{d x}$ dy/dx the subject.
Related rates: link the quantities, differentiate with respect to $t$ t, then substitute. On an interval, check the end points for the optimum.
Learn the derivatives of $tan x$ tan x, $sec x$ sec x, $cosec x$ cosec x, $cot x$ cot x, $a^{x}$ a^x, $lo g_{a} x$ log_a x and the inverse trig functions; reverse them (with completing the square or partial fractions) to integrate.
Substitution: replace $x$ x and $d x$ dx by $u$ u and $d u$ du, and change the limits. Parts: $\int u \frac{d v}{d x} d x = uv - \int v \frac{d u}{d x} d x$ ∫ udv/dx dx = uv − ∫ vdu/dx dx, repeated if needed.
Area between a curve and the $y$ y-axis: $\int x d y$ ∫ x dy with limits in $y$ y.
Volumes of revolution: $\int π y^{2} d x$ ∫ π y² dx about the $x$ x-axis, $\int π x^{2} d y$ ∫ π x² dy about the $y$ y-axis.
Separable: $\int \frac{1}{g ( y )} d y = \int f (x) d x$ ∫ 1/(g(y)) dy = ∫ f(x) dx, then use the initial condition. Homogeneous: substitute $y = vx$ y = vx to make it separable.
Linear $\frac{d y}{d x} + P (x) y = Q (x)$ dy/dx + P(x)y = Q(x): multiply by $e^{\int P (x) d x}$ e^(∫ P(x) dx) and integrate $\frac{d}{d x} (I y) = I Q$ d/dx(Iy) = IQ.
Euler's method: $y_{n + 1} = y_{n} + h f (x_{n}, y_{n})$ yₙ₊₁ = yₙ + h f(xₙ, yₙ); a smaller $h$ h gives a better estimate.
Maclaurin series: $f (x) = f (0) + x f^{'} (0) + \frac{x ^{2}}{2 !} f^{''} (0) + \dots$ f(x) = f(0) + x f'(0) + x²/(2!)f''(0) + …; a few terms approximate $f$ f near $x = 0$ x = 0.
Build new series from $e^{x}$ e^x, $sin x$ sin x, $cos x$ cos x, $ln (1 + x)$ ln (1+x) and $(1 + x)^{p}$ (1+x)^p by substituting, multiplying, differentiating or integrating.
From a DE: put $x = 0$ x = 0 to get $y^{'} (0)$ y'(0), differentiate the DE for higher derivatives, then substitute into the Maclaurin formula.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Further Calculus & Differential Equations for IB Diploma Mathematics: Analysis & Approaches HL. 🎉

Mission 1What is $\frac{s i n x}{x}$ (sin x)/x at $x = 0$ x = 0? Your calculator says error.
Mission 2A sliding ladder's top falls faster and faster. Why?
Mission 3Spin $y = x$ y = √x about the x-axis from 0 to 4. What volume is swept?
Mission 4A hot drink cools faster when it's hotter. What equation is that?
Mission 5How could a calculator find $sin 0.3$ sin 0.3 using only + and ×?

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Specification points and sources for this lesson are listed at the bottom of the page.

Sources (5)

The specification points and sources behind this lesson.

1 IB Mathematics: analysis and approaches guide (first assessment 2021), AHL 5.12, AHL 5.13
2 IB Mathematics: analysis and approaches guide (first assessment 2021), AHL 5.14 and AHL 5.15 content and guidance
3 IB Mathematics: analysis and approaches guide (first assessment 2021), AHL 5.16 and AHL 5.17 content and guidance
4 IB Mathematics: analysis and approaches guide (first assessment 2021), AHL 5.18 content, guidance and connections
5 IB Mathematics: analysis and approaches guide (first assessment 2021), AHL 5.19 content and guidance

Further Calculus & Differential Equations

What is xsinx​(sin x)/x at x=0x = 0? Your calculator says error.

Limits and continuity

The derivative from first principles

Differentiating from first principles

L'Hôpital's rule

Using l'Hôpital's rule twice

A sliding ladder's top falls faster and faster. Why?

Implicit differentiation

Gradient of an implicit curve

Related rates and optimisation

The sliding ladder

Further derivatives and their integrals

Spin y=x​y = √x about the x-axis from 0 to 4. What volume is swept?

Integration by substitution

A given substitution

Integration by parts

Repeated integration by parts

Areas with the y-axis and volumes of revolution

Volume about the y-axis

A hot drink cools faster when it's hotter. What equation is that?

Modelling and separating variables

Newton's law of cooling

Euler's method

Homogeneous equations and the integrating factor

A homogeneous equation

Using an integrating factor

How could a calculator find sin0.3sin 0.3 using only + and ×?

Maclaurin series

The series for sinxsin x

New series from known ones

A product of two series

Series from a differential equation

A series from a DE

Big ideas

Mini-lesson complete!

📣 Smashed it? Share your score

What is $\frac{s i n x}{x}$ (sin x)/x at $x = 0$ x = 0? Your calculator says error.

Spin $y = x$ y = √x about the x-axis from 0 to 4. What volume is swept?

How could a calculator find $sin 0.3$ sin 0.3 using only + and ×?

The series for $sin x$ sin x