Further Pure — Polar coordinates and hyperbolic functions: A-level Further Maths OCR revision notes

4.09

Polar coordinates and curves

Polar coordinates 4.09

Definitions

Pole: the fixed point O. Initial line: the half-line from O corresponding to the positive $x$ -axis.
Polar coordinates $(r, θ)$ : $r$ is the distance from O and $θ$ the angle from the initial line, measured anticlockwise. In this course $r \geq 0$ .

Key results

$x = r cos θ$ , $y = r sin θ$ ; $r^{2} = x^{2} + y^{2}$ ; $tan θ = \frac{y}{x}$ (with the quadrant chosen from a sketch).
The same point has many polar forms: $(r, θ)$ , $(r, θ + 2 π)$ , … The angle is usually given in $0 \leq θ < 2 π$ or $- π < θ \leq π$ .

Method

Polar to Cartesian equation: multiply through by $r$ where it helps, then replace $r cos θ$ by $x$ , $r sin θ$ by $y$ and $r^{2}$ by $x^{2} + y^{2}$ . For example $r = 4 cos θ$ gives $r^{2} = 4 r cos θ$ , so $x^{2} + y^{2} = 4 x$ : a circle with centre $(2, 0)$ and radius 2.
Cartesian to polar: substitute $x = r cos θ$ , $y = r sin θ$ and simplify. The line $y = x$ is $θ = \frac{π}{4}$ .

Notes

$r = k$ is a circle centred at O; $θ = α$ is a half-line from O.
As with complex numbers, the arctan of $\frac{y}{x}$ alone does not fix the angle; the quadrant does.

Sketching polar curves 4.09

Key results

Cardioid $r = a (1 + cos θ)$ : heart-shaped, meeting the pole at $θ = π$ .
Limaçon $r = a + b cos θ$ with $a > b > 0$ : no cusp; a dimple if $a < 2 b$ , convex if $a \geq 2 b$ .
Rose $r = a sin n θ$ : with $r \geq 0$ , it has $n$ petals (the angles where $r < 0$ are not drawn).
Spiral $r = a θ$ : the distance from O grows steadily with the angle.

Method

Tabulate $r$ for convenient angles ( $0, \frac{π}{6}, \frac{π}{4}, \dots$ ), skipping angles where the formula gives $r < 0$ , and join the points smoothly.
Look for symmetry: if $r (θ) = r (- θ)$ (for instance a function of $cos θ$ ), the curve is symmetric in the initial line.
Find where $r = 0$ : the curve passes through the pole there, and the half-line at that angle is a tangent at the pole.
Find the greatest and least values of $r$ , and where they occur.

Notes

Tangents parallel to the initial line occur where $\frac{d}{d θ} (r sin θ) = 0$ , and tangents perpendicular to it where $\frac{d}{d θ} (r cos θ) = 0$ .

Converting between polar and Cartesian equations 4.09

Key results

Use $x = r cos θ$ , $y = r sin θ$ , $r^{2} = x^{2} + y^{2}$ and $tan θ = \frac{y}{x}$ .
$r = a$ is a circle centred on the pole; $θ = α$ is a half-line from the pole; $r = 2 a cos θ$ is the circle $(x - a)^{2} + y^{2} = a^{2}$ through the pole.
$r = a sec θ$ , that is $r cos θ = a$ , is the vertical line $x = a$ .

Method

Polar to Cartesian: multiply through by $r$ if that creates $r cos θ$ , $r sin θ$ or $r^{2}$ , then substitute.
Cartesian to polar: substitute $x = r cos θ$ and $y = r sin θ$ , then make $r$ the subject where possible.

Notes

Multiplying by $r$ can add the pole as an extra point; check whether the original curve passes through it.
State the range of $θ$ that traces the curve once: $r = 4 cos θ$ needs only $- \frac{π}{2} \leq θ \leq \frac{π}{2}$ , since $r \geq 0$ .

4.09

Area enclosed by a polar curve

Area enclosed by a polar curve 4.09

Key results

The area bounded by the curve $r = f (θ)$ and the half-lines $θ = α$ , $θ = β$ is $A = \frac{1}{2} \int_{α}^{β} r^{2} d θ$ .

Method

Square $r$ , then use double-angle identities: $cos^{2} θ = \frac{1}{2} (1 + cos 2 θ)$ and $sin^{2} θ = \frac{1}{2} (1 - cos 2 θ)$ .
Choose the limits from the sketch: for one loop of a rose, the limits are consecutive angles where $r = 0$ .
Use symmetry: the area of the whole cardioid is twice the area for $0 \leq θ \leq π$ .

Notes

Why $\frac{1}{2} r^{2}$ : a thin sector of radius $r$ and angle $δ θ$ has area $\frac{1}{2} r^{2} δ θ$ , and adding sectors gives the integral.
For the region between two curves, subtract the areas, taking care over where each curve is the outer one.

Worked examples

Worked example

Find the area of the region enclosed by the cardioid $r = 2 (1 + cos θ)$ .

Show worked solution

A = \frac{1}{2} \int_{0}^{2 π} 4 (1 + cos θ)^{2} d θ = 2 \int_{0}^{2 π} (1 + 2 cos θ + \frac{1}{2} + \frac{1}{2} cos 2 θ) d θ = 2 [\frac{3}{2} θ + 2 sin θ + \frac{1}{4} sin 2 θ]_{0}^{2 π} = 2 \cdot 3 π = 6 π

Worked example

The curve $r = 2 sin 2 θ$ is drawn for $0 \leq θ \leq \frac{π}{2}$ , forming one petal.

Find the area of the petal.

Show worked solution

$r = 0$ at $θ = 0$ and:

θ = \frac{π}{2}

A = \frac{1}{2} \int_{0}^{π /2} 4 sin^{2} 2 θ d θ = \int_{0}^{π /2} (1 - cos 4 θ) d θ = [θ - \frac{1}{4} sin 4 θ]_{0}^{π /2} = \frac{π}{2}

4.07

Hyperbolic functions and identities

The hyperbolic functions 4.07

Definitions

$cosh x = \frac{e ^{x} + e ^{- x}}{2}$ , $sinh x = \frac{e ^{x} - e ^{- x}}{2}$ , $tanh x = \frac{sinh x}{cosh x} = \frac{e ^{2 x} - 1}{e ^{2 x} + 1}$ .
Reciprocals: $sech x = \frac{1}{cosh x}$ , $cosech x = \frac{1}{sinh x}$ , $coth x = \frac{1}{tanh x}$ .

Key results

$cosh x$ is even with minimum value $cosh 0 = 1$ ; $sinh x$ is odd and takes every real value; $tanh x$ is odd with $- 1 < tanh x < 1$ and asymptotes $y = \pm 1$ .
$cosh x + sinh x = e^{x}$ and $cosh x - sinh x = e^{- x}$ .

Method

To solve an equation such as $a cosh x + b sinh x = c$ : replace each function by its exponential definition, multiply through by $e^{x}$ , and solve the resulting quadratic in $e^{x}$ . Reject any root with $e^{x} \leq 0$ .

Notes

A cable hanging under its own weight takes the shape of $y = a cosh \frac{x}{a}$ , the catenary, not a parabola.
The name comes from the hyperbola: $(cosh t, sinh t)$ lies on $x^{2} - y^{2} = 1$ , as $(cos t, sin t)$ lies on the circle $x^{2} + y^{2} = 1$ .

Hyperbolic identities 4.07

Key results

$cosh^{2} x - sinh^{2} x = 1$ .
$sinh 2 x = 2 sinh x cosh x$ ; $cosh 2 x = cosh^{2} x + sinh^{2} x = 2 cosh^{2} x - 1 = 1 + 2 sinh^{2} x$ .
$sinh (A \pm B) = sinh A cosh B \pm cosh A sinh B$ ; $cosh (A \pm B) = cosh A cosh B \pm sinh A sinh B$ .

Method

Prove an identity from the exponential definitions: for example $cosh^{2} x - sinh^{2} x = \frac{( e ^{x} + e ^{- x} ) ^{2} - ( e ^{x} - e ^{- x} ) ^{2}}{4} = \frac{4}{4} = 1$ .

Notes

Osborn's rule: a trigonometric identity becomes the hyperbolic one on replacing $cos$ by $cosh$ and $sin$ by $sinh$ , and changing the sign of every term containing a product of two sines (including $tan^{2}$ , which is $\frac{sin ^{2}}{cos ^{2}}$ ). So $cos^{2} x + sin^{2} x = 1$ becomes $cosh^{2} x - sinh^{2} x = 1$ .
Osborn's rule is a way to recall identities, not a proof. If asked to prove one, use the exponential definitions.

Calculus of hyperbolic functions 4.07

Key results

$\frac{d}{d x} sinh x = cosh x$ , $\frac{d}{d x} cosh x = sinh x$ (no minus sign), $\frac{d}{d x} tanh x = sech^{2} x$ .
$\int sinh x d x = cosh x + c$ , $\int cosh x d x = sinh x + c$ .

Method

Integrate $sinh^{2} x$ and $cosh^{2} x$ with the double-angle forms: $sinh^{2} x = \frac{1}{2} (cosh 2 x - 1)$ and $cosh^{2} x = \frac{1}{2} (cosh 2 x + 1)$ .
For powers like $cosh^{3} x$ , write $cosh^{3} x = cosh x (1 + sinh^{2} x)$ and integrate by inspection.

Notes

The derivatives follow straight from the definitions: $\frac{d}{d x} \frac{e ^{x} - e ^{- x}}{2} = \frac{e ^{x} + e ^{- x}}{2}$ .
Exact answers involving $ln$ often simplify using $cosh (ln a) = \frac{a + a ^{- 1}}{2}$ and $sinh (ln a) = \frac{a - a ^{- 1}}{2}$ .

Worked examples

Worked example

Solve $3 sinh x - 2 cosh x = 2$ , giving the answer exactly.

Show worked solution

Using the definitions,

\frac{3}{2} (e^{x} - e^{- x}) - (e^{x} + e^{- x}) = 2

so:

\frac{1}{2} e^{x} - \frac{5}{2} e^{- x} = 2

Multiply by $2 e^{x}$ :

e^{2 x} - 4 e^{x} - 5 = 0

so:

(e^{x} - 5) (e^{x} + 1) = 0

Since $e^{x} > 0$ , $e^{x} = 5$ and $x = ln 5$ .

Worked example

Find the exact value of:

\int_{0}^{l n 2} sinh^{2} x d x

Show worked solution

sinh^{2} x = \frac{1}{2} (cosh 2 x - 1)

so the integral is:

[\frac{1}{4} sinh 2 x - \frac{1}{2} x]_{0}^{l n 2}

Now:

sinh (2 ln 2) = sinh (ln 4) = \frac{4 - \frac{1}{4}}{2} = \frac{15}{8}

So the value is $\frac{15}{32} - \frac{1}{2} ln 2$ .

4.07

Inverse hyperbolic functions

Inverse hyperbolic functions 4.07

Key results

$arsinh x = ln (x + x^{2} + 1)$ for all real $x$ .
$arcosh x = ln (x + x^{2} - 1)$ for $x \geq 1$ (the principal value, $\geq 0$ ).
$artanh x = \frac{1}{2} ln (\frac{1 + x}{1 - x})$ for $∣ x ∣ < 1$ .
$\frac{d}{d x} arsinh x = \frac{1}{x ^{2} + 1}$ , $\frac{d}{d x} arcosh x = \frac{1}{x ^{2} - 1}$ , $\frac{d}{d x} artanh x = \frac{1}{1 - x ^{2}}$ .

Method

Derive the logarithmic form: let $y = arsinh x$ , so $x = sinh y = \frac{e ^{y} - e ^{- y}}{2}$ . Multiply by $2 e^{y}$ : $e^{2 y} - 2 x e^{y} - 1 = 0$ , so $e^{y} = x \pm x^{2} + 1$ . Since $e^{y} > 0$ , take the positive sign: $y = ln (x + x^{2} + 1)$ .

Notes

$cosh$ is not one-to-one, so it is restricted to $x \geq 0$ before inverting; that is why $arcosh x \geq 0$ and the equation $cosh x = k$ (with $k > 1$ ) has the two solutions $\pm arcosh k$ .
OCR writes these as $arsinh$ , $arcosh$ , $artanh$ ("area" functions); calculators may show $sinh^{- 1}$ .

Integrals giving inverse hyperbolic functions 4.07

Key results

$\displaystyle\int\dfrac{dx}{\sqrt{x^2+a^2}}=\operatorname{arsinh}\left(\dfrac xa\right)+c=\ln\left(x+\sqrt{x^2+a^2}\right)+c'$ .
$\int \frac{d x}{x ^{2} - a ^{2}} = arcosh (\frac{x}{a}) + c$ , for $x > a$ .

Method

Complete the square under the root: $x^{2} + 6 x + 13 = (x + 3)^{2} + 4$ , then integrate to $arsinh (\frac{x + 3}{2})$ .
With a coefficient on $x^{2}$ , take it out first: $4 x^{2} + 9 = 2 x^{2} + \frac{9}{4}$ .

Notes

Compare the family: $\frac{1}{a ^{2} - x ^{2}}$ gives arcsin, $\frac{1}{x ^{2} + a ^{2}}$ gives arsinh, $\frac{1}{x ^{2} - a ^{2}}$ gives arcosh, $\frac{1}{a ^{2} + x ^{2}}$ gives arctan. Recognising which applies is most of the work.
Exam questions usually want the answer as a logarithm: convert using the logarithmic forms above.

Worked examples

Worked example

Solve $cosh x = 2$ , giving your answers as logarithms.

Show worked solution

x = \pm arcosh 2 = \pm ln (2 + 3)

(Equivalently the negative root is $ln (2 - 3)$ , since:

(2 + 3) (2 - 3) = 1

)

Worked example

Show that:

\int_{0}^{4} \frac{d x}{x ^{2} + 9} = ln 3

Show worked solution

\int_{0}^{4} \frac{d x}{x ^{2} + 9} = [arsinh \frac{x}{3}]_{0}^{4} = arsinh \frac{4}{3} = ln (\frac{4}{3} + \frac{16}{9} + 1) = ln (\frac{4}{3} + \frac{5}{3}) = ln 3