Further Pure | Topic 5: Differentiation and Integration

Curriculum Objectives:

• obtain an expression for $\frac{{{d^2}y}}{{d{x^2}}}$ in cases where the relation between y and x is defined parametrically or implicitly
• derive and use reduction formulae for the evaluation of definite integral in simple cases
• use of integration to find: mean values and centroids of two or three dimensional figures ( where equations are expressed in Cartesian co-ordinates, including the use of a parameter) using strips, discs and shells as appropriate, arc lengths ( for curves in Cartesian co-ordinates or in polar form)
• surface area of revolution about one of the axes ( for curves in Cartesian co-ordinates but not for those in polar form )

The second derivative of an implicitly defined function

You are reminded that an implicit function is given by:

$f({x_1},{x_2},...,{x_n}) = 0$

However we are concerned only with the case where there are two variables. For example, the function defining a circle, with the origin as the centre and c as the radius:

${x^2} + {y^2} = {c^2}$

Using this relatively simple example let me illustrate the method of arriving at the second derivative with respect to either variable. We are going to use x, x', x'', y, y', y'', as opposed to $x,\frac{{dx}}{{dy}},\frac{{{d^2}x}}{{d{y^2}}},y,\frac{{dy}}{{dx}},\frac{{{d^2}y}}{{d{x^2}}}$, on account of the, being easier to work with in this case, as you will see.

Differentiating once with respect to x, as per the implicit differentiation you must have learnt in your A Levels P3.

$2x + 2yy' = 0$
$y' =  - \frac{x}{y}$

Differentiating again with respect to x, we find that the term 2x is easily differentiated; differentiation of yy' however is notion as yet foreign. For this we recognise that y' must too be a function of x and that the product rule is then applicable here, and that:

$(y')' = y''$ or $\frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{{{d^2}y}}{{d{x^2}}}$

The differentiation of yy' is then:

$(yy')' = yy'' + {(y')^2}$

Then the second differentiation of the curve should yield:

$2 + 2yy'' + 2{(y')^2} = 0$
$1 + yy'' + {(y')^2} = 0$
$yy'' =  - 1 - {(y')^2}$
$yy'' =  - 1 - {\left( { - \frac{x}{y}} \right)^2}$
$y'' =  - \frac{1}{y} - \frac{{{x^2}}}{{{y^3}}}$
$y'' = \frac{{ - ({y^2} + {x^2})}}{{{y^3}}}$

A Further substitution could yield:

$y'' =  - \frac{{{c^2}}}{{{y^3}}}$


Ex. (October/November 2011, Paper 12, Q 5)

The point P (2, 1) lies on the curve with the equation:

${x^3} - 2{y^3} = 3xy$

Find
(I) the value of $\frac{{dy}}{{dx}}$ at P
(II) the value of $\frac{{{d^2}y}}{{d{x^2}}}$ at P

Sol.

(I) We differentiate the equation, to get:

$3{x^2} - 6{y^2}y' = 3(y + xy')$

Putting in the co-ordinates of P

$3{(2)^2} - 6{(1)^2}y' = 3(1 + 2y')$
$12 - 6y' = 3 + 6y'$
$y' = \frac{3}{4}$

(II) Differentiating we  get

$6x - 6({y^2}y'' + 2y{(y')^2}) = 3y' + 3(y' + xy'')$
$6(2) - 6\left( {y'' + \frac{9}{8}} \right) = \frac{9}{4} + \frac{9}{4} + 6y''$
$12y'' = \frac{3}{4}$
$y'' = \frac{1}{{16}}$

The Second Derivative of a parametrically defined function

Here, too, the method will be explained with the aid of examples, although no particular notation is given precedence.

Ex. (Oct/Nov 2013 Paper 13, Q4)

A curve has the parametric equations:

$x = 2\theta  - \sin (2\theta )$  $y = 1 - \cos (2\theta )$ for  $- 3\pi  \le \theta  \le 3\pi$

Show that

$\frac{{dy}}{{dx}} = \cot \theta $

expect for certain values of $\theta $, which should be stated.

Sol.

By differentiating with respect to $\theta $ we have that:

$\frac{{dx}}{{d\theta }} = 2 - 2\cos (2\theta )$ and $\frac{{dy}}{{d\theta }} = 2\sin (2\theta )$

We further realise that:

$\frac{{dy}}{{dx}} = \frac{{2\sin (2\theta )}}{{2 - 2\cos (2\theta )}}$

$ = \frac{{\sin (2\theta )}}{{1 - \cos (2\theta )}}$

$ = \frac{{2\sin \theta cos\theta }}{{2{{\sin }^2}\theta }}$

$ = \frac{{\cos \theta }}{{\sin \theta }}$

$ = \cot \theta $

For all the integral multiples of pi will leave us with a zero denominator, n * pi, where n is an integer, are the values of $\theta $ for which this derivative is undefined

Find the value of $\frac{{{d^2}y}}{{d{x^2}}}$ for $\theta $ = (pi)/4

For this we must make the following considerations:

$\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{d}{{dx}}\left( {\cot \theta } \right) = \frac{d}{{d\theta }}(\cot \theta ) \cdot \frac{{d\theta }}{{dx}}$

Recall that the differentiation of $\cot \theta $ is $ - {\csc ^2}\theta $ a that

$\frac{{d\theta }}{{dx}} = \frac{1}{{\frac{{dx}}{{d\theta }}}}$

$\frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - {{\csc }^2}\theta }}{{2 - 2\cos 2\theta }}$

$ = \frac{{ - {{\csc }^2}\theta }}{{4{{\sin }^2}\theta }}$

$ =  - \frac{1}{{4{{\sin }^4}\theta }}$

Putting in the value of $\theta $, we find the value -1

Reduction Formula

In P3, oft times, question involved integrating by parts twice to achieve the desired result. Consider the following example:

$\int {{x^2}{e^x}dx} $

And we let:

$u = {x^2}$ $v' = {e^x}$

$u' = 2x$ $v = {e^x}$

Then we have:

$\int {u\frac{{dv}}{{dx}}dx}  = uv - \int {v\frac{{du}}{{dx}}dx} $

$\int {{x^2}{e^x}dx}  = {x^2}{e^x} - 2\int {{e^x}x} dx$

Now if I were to let $I_2$ and $I_1$ be:

${I_2} = \int {{x^2}{e^x}} dx$

${I_1} = \int {{x}{e^x}} dx$

Which are thusly related:

${I_2} = {x^2}{e^x} - 2{I_1}$

If instead we begin with the general term n.

${I_n} = \int {{x^n}{e^x}dx} $

We have:

$u = {x^n}$  $v' = {e^x}$

$u' = n{x^{n - 1}}$ $v = {e^x}$

$\int {{x^n}{e^x}dx = {x^n}{e^x} - n\int {{x^{n - 1}}{e^x}dx} } $

${I_n} = {x^n}{e^x} - n{I_{n - 1}}$

This is a reduction formula, which in essence reduces a rather complicated integral to a simpler one.

Say, now that we wanted to find $I_3$. We have:

${I_3} = {x^3}{e^x} - 3{I_2} = {x^3}{e^x} - 3({x^2}{e^x} - 2{I_1}) = {x^3}{e^x} - 3({x^2}{e^x} - 2(x{e^x} - {I_0}))$

${I_0} = \int {{x^0}{e^x}dx}  = \int {{e^x}dx}  = {e^x}$

Then you input the value of $I_0$ to get:

${I_3} = {x^3}{e^x} - 3{x^2}{e^x} + 6x{e^x} - 6{e^x} + k$

It is customary not to introduce the constant of integration, until the final integration has been completed.

This topic will require relentless practice to master, as is true for most of calculus.

While the integrating by parts methods has been used to introduce the topic and you will find it insinuated in almost all the question on the topics, other methods can be used to derive reduction formulae. Examples in this regard could be of differentiation and identities.

Ex.

Find the reduction formula for:

${I_n} = \int_0^{\frac{\pi }{4}} {{{\tan }^n}x} dx$

We rewrite and then use a well known identity to establish the result.

 ${I_n} = \int_0^{\frac{\pi }{4}} {{{\tan }^n}xdx} $
${I_n} = \int_0^{\frac{\pi }{4}} {{{\tan }^{n - 2}}x \cdot {{\tan }^2}x} dx$
${I_n} = \int_0^{\frac{\pi }{4}} {{{\tan }^{n - 2}}x \cdot } ({\sec ^2}x - 1)dx$
${I_n} = \int_0^{\frac{\pi }{4}} {{{\tan }^{n - 2}}x} {\sec ^2}xdx - \int_0^{\frac{\pi }{4}} {{{\tan }^{n - 2}}x} dx$
${I_n} = \left[ {\frac{{{{\tan }^{n - 1}}x}}{{n - 1}}} \right]_0^{\frac{\pi }{4}} - {I_{n - 2}}$
${I_n} = \frac{1}{{n - 1}} - {I_{n - 2}}$

Provided of course that $n \ge 2$

Let us attempt a couple of past paper questions before moving on.

Ex. (October/November 2012 Paper 13 Question 11)

Show that ${\int {x\left( {1 - {x^2}} \right)} ^{\frac{1}{2}}}dx =  - \frac{1}{3}{\left( {1 - {x^2}} \right)^{\frac{3}{2}}} + c$

Given that ${I_n} = \int_0^1 {{x^n}{{(1 - {x^2})}^{\frac{1}{2}}}dx} $ prove that for $n \ge 2$

$\left( {n + 2} \right){I_n} = \left( {n - 1} \right){I_{n - 2}}$

Use the substitution x = sin u to show that:

${\int_0^1 {\left( {1 - {x^2}} \right)} ^{\frac{1}{2}}}dx = \frac{ \pi }{4}$

Find ${I_4}$
Sol.

For the first part we let:

$u = 1 - {x^2}$
$\frac{{du}}{{dx}} =  - 2x$

We rewrite the given integral thusly:

$ - \frac{1}{2}\int { - 2x{{(1 - {x^2})}^{\frac{1}{2}}}dx} $

Then substituting u and u' into the integral

$ - \frac{1}{2}\int {\frac{{du}}{{dx}}{{(u)}^{\frac{1}{2}}}dx}  =  - \frac{1}{2}\int {{u^{\frac{1}{2}}}} du =  - \frac{1}{2}\left[ {\frac{2}{3}{u^{\frac{3}{2}}}} \right] =  - \frac{1}{3}{\left( {1 - {x^2}} \right)^{\frac{3}{2}}} + c$

Obviously, in having us evaluate a rather simple integral, the examiner wishes us to use it in the next part, but disregarding that at this point, if you consider the following:

$u = {x^n}$      $v' = {\left( {1 - {x^2}} \right)^{\frac{1}{2}}}$

The differentiation of u is easy enough, the integration of v' however is slightly difficult and will not help prove the relationship that is required. Yet the integral is evaluated so to impress upon you the fact, that seldom, does the most apparent method is correct method in a further mathematics paper.

We begin with the substitution:

$x = \sin u$
$\frac{{dx}}{{du}} = \cos u$
$dx = \cos u  du$

Also we have:

$u = {\sin ^{ - 1}}x$

By making the substitutions and using the identity linking the squares of sine and cosine, and unity, we find that:

$\sqrt {1 - {{\sin }^2}u}  = \sqrt {{{\cos }^2}x}  = \cos x$ 

The integral becomes, therefore:

$\int {\sqrt {1 - {x^2}} dx}  = \int {\cos u} dx = \int {\cos u \cdot \cos udu = \int {{{\cos }^2}udu} } $

We know that:

${\cos ^2}u = \frac{{\cos 2u}}{2} + \frac{1}{2}$

Therefore the integral becomes

$\int {\left( {\frac{{\cos 2u}}{2} + \frac{1}{2}} \right)} du$

Integrating it using the rules you must have studied in P3, we get:

$\int {\left( {\frac{{\cos 2u}}{2} + \frac{1}{2}} \right)} du = \frac{{\sin (2u)}}{4} + \frac{u}{2}$

The constant of integration has been disregarded at this point. x is now substituted back. We get:

$\frac{1}{4}\sin (2u) + \frac{1}{2}u = \frac{1}{4}\sin (2{\sin ^{ - 1}}x) + \frac{1}{2}{\sin ^{ - 1}}x$

Using the following identity we may clean the integral up somewhat.

$\sin (2{\sin ^{ - 1}}x) = 2x\sqrt {1 - {x^2}} $

Finally we write that:

$\int {\sqrt {1 - {x^2}} } dx = \frac{x}{2}\sqrt {1 - {x^2}} + \frac{1}{2}{\sin ^{ - 1}}x + c$

This evidently doesn't help us in the least. We know attempt the second part using the integral we have proven in the first part of the question. We must adjust the $I_n$ so that the consequent integration by parts involves:

$\int {x\sqrt {1 - {x^2}} } dx$

Hence we manipulate the given integral thusly:

$\int_0^1 {{x^{n - 1}} \cdot x\sqrt {1 - {x^2}} dx} $

And let:

$u = {x^{n - 1}}$    $v' = x\sqrt {1 - {x^2}} $
$u' = \left( {n - 1} \right){x^{n - 2}}$   $v =  - \frac{1}{3}{\left( {1 - {x^2}} \right)^{\frac{3}{2}}} + c$

The formula for integration by parts then states:

${I_n} = \left[ {\left( {{x^{n - 1}}} \right)\left( { - \frac{1}{3}{{\left( {1 - {x^2}} \right)}^{\frac{3}{2}}}} \right)} \right]_0^1 - \int_0^1 {(n - 1){x^{n - 2}} \cdot \left( { - \frac{1}{3}{{\left( {1 - {x^2}} \right)}^{\frac{3}{2}}}} \right)} dx$

${I_n} = \left[ {\left( {{x^{n - 1}}} \right)\left( { - \frac{1}{3}{{\left( {1 - {x^2}} \right)}^{\frac{3}{2}}}} \right)} \right]_0^1 + \frac{1}{3}(n - 1)\int_0^1 {{x^{n - 2}} \cdot {{\left( {1 - {x^2}} \right)}^{\frac{3}{2}}}} dx$

${I_n} = \frac{1}{3}(n - 1)\int_0^1 {{x^{n - 2}} \cdot {{\left( {1 - {x^2}} \right)}^{\frac{3}{2}}}} dx$

$3{I_n} = (n - 1)\int_0^1 {{x^{n - 2}} \cdot {{\left( {1 - {x^2}} \right)}^{1 + \frac{1}{2}}}} dx$

$3{I_n} = (n - 1)\int_0^1 {{x^{n - 2}} \cdot \left( {1 - {x^2}} \right){{\left( {1 - {x^2}} \right)}^{\frac{1}{2}}}} dx$

$3{I_n} = (n - 1)\int_0^1 {{x^{n - 2}} \cdot {{\left( {1 - {x^2}} \right)}^{\frac{1}{2}}}} dx - (n - 1)\int_0^1 {{x^n} \cdot {{\left( {1 - {x^2}} \right)}^{\frac{1}{2}}}} dx$

$3{I_n} = (n - 1)\int_0^1 {{x^{n - 2}} \cdot {{\left( {1 - {x^2}} \right)}^{\frac{1}{2}}}} dx - (n - 1){I_n}$

$(n + 2){I_n} = (n - 1)\int_0^1 {{x^{n - 2}} \cdot {{\left( {1 - {x^2}} \right)}^{\frac{1}{2}}}} dx = (n - 1){I_{n - 2}}$

The indefinite integral has already been evaluated, you need only complete the calculation with the bounds given, and it's easily proven.

$6{I_4} = 3{I_2} \Rightarrow {I_4} = \frac{1}{2}{I_2}$

$4{I_2} = {I_0} \Rightarrow {I_2} = \frac{1}{4}{I_0}$

${I_4} = \frac{1}{2}\left( {\frac{1}{4}{I_0}} \right) = \frac{1}{8}{I_0} = \frac{\pi }{{32}}$

Ex. (October/November 2009; Paper 01; Q6)

Show that

$\frac{d}{{dx}}\left[ {{x^{n - 1}}\sqrt {4 - {x^2}} } \right] = \frac{{4(n - 1){x^{n - 2}}}}{{\sqrt {4 - {x^2}} }} - \frac{{n{x^n}}}{{\sqrt {4 - {x^2}} }}$

Let

${I_n} = \int_0^1 {\frac{{{x^n}}}{{\sqrt {4 - {x^2}} }}} dx$

where $n \ge 0$. Prove that

$n{I_n} = 4(n - 1){I_{n - 2}} - \sqrt 3 $

for $n \ge 2$

Given that ${I_0} = \frac{1}{6}\pi $, find ${I_4}$, leaving your answer in exact form.

Sol.

$\frac{d}{{dx}}\left[ {{x^{n - 1}}\sqrt {4 - {x^2}} } \right] = {x^{n - 1}}\frac{d}{{dx}}\left[ {\sqrt {4 - {x^2}} } \right] + \sqrt {4 - {x^2}} \frac{d}{{dx}}\left[ {{x^{n - 1}}} \right]$

$ = {x^{n - 1}}\left[ {\frac{1}{2} \cdot \frac{1}{{\sqrt {4 - {x^2}} }} \cdot \left( { - 2x} \right)} \right] + \sqrt {4 - {x^2}} \left[ {\left( {n - 1} \right){x^{n - 2}}} \right]$

$ =  - \frac{{{x^n}}}{{\sqrt {4 - {x^2}} }} + \sqrt {4 - {x^2}} \left[ {\left( {n - 1} \right){x^{n - 2}}} \right]$

$ =  - \frac{{{x^n}\sqrt {4 - {x^2}} }}{{4 - {x^2}}} + \sqrt {4 - {x^2}} (n - 1){x^{n - 2}}$

$ = \sqrt {4 - {x^2}} \left[ { - \frac{{{x^n}}}{{4 - {x^2}}} + (n - 1){x^{n - 2}}} \right]$

$ = \sqrt {4 - {x^2}} \left[ {\frac{{ - {x^n} + (4 - {x^2})(n - 1){x^{n - 2}}}}{{4 - {x^2}}}} \right]$

$ = \sqrt {4 - {x^2}} \left[ {\frac{{ - {x^n} + 4(n - 1){x^{n - 2}} - n{x^{n - 2 + 2}} + {x^{n - 2}}}}{{4 - {x^2}}}} \right]$

$ = \sqrt {4 - {x^2}} \left[ {\frac{{4(n - 1){x^{n - 2}} - n{x^n}}}{{4 - {x^2}}}} \right]$

Furthermore, you need only cancel out the term $\sqrt {4 - {x^2}} $ to achieve the required result.

For the next part, we integrate the relationship we have already proven.

$\int_0^1 {\frac{d}{{dx}}\left[ {{x^{n - 1}}\sqrt {4 - {x^2}} } \right]} dx = \int_0^1 {\frac{{4(n - 1){x^{n - 2}}}}{{\sqrt {4 - {x^2}} }}} dx - \int_0^1 {\frac{{n{x^n}}}{{\sqrt {4 - {x^2}} }}} dx$

$\left[ {{x^{n - 1}}\sqrt {4 - {x^2}} } \right]_0^1 = 4(n - 1)\int_0^1 {\frac{{{x^{n - 2}}}}{{\sqrt {4 - {x^2}} }}dx - n\int_0^1 {\frac{{{x^n}}}{{\sqrt {4 - {x^2}} }}} } dx$

$\sqrt 3  = 4(n - 1){I_{n - 2}} - n{I_n}$

$n{I_n} = 4(n - 1){I_{n - 2}} - \sqrt 3 $

Centroid

The centroid of an object is such a point, that it is the average ( or arithmetic mean ) of all the points in the object. Say an object consists of n number of points, and $x_i$ is the distance of the point from some reference point in the x-axis, then the distance of the x coordinate of the centroid of the object from that reference point is given by:

${C_x} = \frac{{{x_1} + {x_2} + {x_3}...{x_n}}}{n}$

This can obviously be extended to any number of finite dimensions. If you have completed the M2 unit of the CIE A levels Mathematics Course, you might surmise some similitude centroids have with centres of mass. Assuming uniform density, an object has the same point for its centroid and centre of mass. You will not be questioned about this in the pure paper. For the applied paper we will study this topic in greater detail.

The formulae relevant to the pure paper, are as follows.

For a curve y = f(x) defined between a <  x <  b, the x and y co-ordinates of the centroid is given by:

${C_x} = \frac{{\int_a^b {xy} {\kern 1pt} dx}}{{\int_a^b y {\kern 1pt} dx}}$ ${C_y} = \frac{1}{2}\frac{{\int_a^b {{y^2}} dx}}{{\int_a^b y {\kern 1pt} dx}}$

Ex. Find the co-ordinates of the centroid of the curve $y = \sqrt {1 - x} $ for $0 \le x \le 1$ 

Sol.

We note:

$\int_0^1 {ydx}  = \int_0^1 {{{(1 - x)}^{\frac{1}{2}}}} dx = \left[ { - \frac{2}{3}{{(1 - x)}^{\frac{3}{2}}}} \right]_0^1 = \frac{2}{3}$

We also note that:

$\int_0^1 {xydx}  = \int_0^1 {x{{(1 - x)}^{\frac{1}{2}}}dx}  =  - \int_0^1 {(1 - u){{(u)}^{\frac{1}{2}}}du} $

u = 1 - x

$\int_0^1 {\left( {{u^{\frac{3}{2}}} - {u^{\frac{1}{2}}}} \right)} du = \frac{4}{{15}}$

${C_x} = \frac{{\frac{4}{{15}}}}{{\frac{2}{3}}} = \frac{2}{5}$

$\int_0^1 {{y^2}} dx = \int_0^1 {\left( {1 - x} \right)dx}  = \frac{1}{2}$

${C_y} = \frac{1}{2} \cdot \frac{{\frac{1}{2}}}{{\frac{2}{3}}} = \frac{3}{8}$

Mean Value of a function

For a function y = f(x), defined on the range a < x < b, the mean value of the function is given by:

$\bar y = \frac{1}{{b - a}}\int_0^1 {y{\kern 1pt} dx} $

Ex. (October/November 2012, Paper 12, Question Number 2)

The curve has, for $0 \le x \le 4$, the equation

$y = 2{x^{{\textstyle{1 \over 2}}}}$

Find

(I) the Mean value of y with respect to x
(II) the y-co-ordinate of the centroid of the region enclosed by the curve, the line x = 4 the x-axis

Sol.

(I) the mean value of function is given by:

$\bar y = \frac{1}{{b - a}}\int_0^4 y \,dx = \frac{1}{4}\int_0^4 {2{x^{\frac{1}{2}}}} {\kern 1pt} dx = \frac{1}{2}\left[ {\frac{2}{3}{x^{\frac{3}{2}}}} \right]_0^4 = \frac{8}{3}$

(II) for the y co-ordinate of the centroid, we note:

$\int_a^b {y{\kern 1pt} dx}  = \int_0^4 {2{x^{\frac{1}{2}}}} {\kern 1pt} dx = \left[ {\frac{4}{3}{x^{\frac{3}{2}}}} \right]_0^4 = \frac{{32}}{3}$

$\int_a^b {{y^2}dx}  = \int_0^4 {4x} {\kern 1pt} dx = 2\left[ {{x^2}} \right]_0^4 = 32$

The y co-ordinate is then simply:

$ = \frac{1}{2}\left[ {\frac{{32 \times 3}}{{32}}} \right] = \frac{3}{2}$

Arc Length

The arc length of a curve defined in rectangular, parametric, or polar equations, is found by the application of the following formulae (provided the function is differentiable over the given range):

1) For the function y = f(x), for the region a < x < b

$S = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx$

2) For the function x = f(y) (which is the inverse of the above function), defined for c < y < d

$S = \int_c^d {\sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} } dy$

3) For the function defined parametrically, y = f(t); x = f(t), for the region e < t < f

$S = \int_e^f {\sqrt {{{\left( {\frac{{dy}}{{dt}}} \right)}^2} + {{\left( {\frac{{dx}}{{dt}}} \right)}^2}} } dt$

4) For the polar function defined r = f($ \theta $), for g < $ \theta $ < h,

$S = \int_g^h {\sqrt {{{\left( r \right)}^2} + {{\left( {\frac{{dr}}{{d\theta }}} \right)}^2}} d\theta } $

Area of Surface of Revolution

For the function $y = f(x)$, defined for a < x < b, is rotated through $2\pi$, about the x-axis,the area of the surface formed is given by:

$Surface\,Area = \int_a^b {2\pi y} \sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \,dx$

For the function, defined for c < y < d rotated through $2\pi$, about the y-axis, the area of surface generated is given by:

$Surface\,Area = \int_c^d {2\pi x} \sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} \,dy$

For the function defined parametrically ( y = f(t); x = g(t) ), defined for e < t < f, the area of surface generated is given by:

$Surface\,Area = \int_c^d {2\pi y} \sqrt {{{\left( {\frac{{dy}}{{dt}}} \right)}^2} + {{\left( {\frac{{dx}}{{dt}}} \right)}^2}} \,dt$

By replacing y with x in the above formula, you get the formula for the parametric analogue of the function generated by rotating about x-axis.

Ex. (October/November 2012, Paper 12, Question 8)

The curve, C, has the parametric equations:

$x = {\textstyle{1 \over 3}}{t^3} - \ln t$
$y = {\textstyle{4 \over 3}}{t^{\frac{3}{2}}}$

for $1 \le t \le 3$. Find the arc length of C.
Find also the area of the surface generated when C is rotated through $2\pi# radians about the x-axis.

Sol.

We find the derivatives, with respect to t.

$\dot x = {t^2} - \frac{1}{t} = \frac{{{t^3} - 1}}{t}$
$\dot y = 2{t^{\frac{1}{2}}}$

We find the squares:

${\left( {\dot x} \right)^2} = {\left( {\frac{{{t^3} - 1}}{t}} \right)^2} = \frac{{{t^9} - 2{t^3} + 1}}{{{t^2}}}$

${\left( {\dot y} \right)^2} = 4t$

Performing the root extraction of the sum of the two squares of the derivatives:


$\sqrt {{{\left( {\dot x} \right)}^2} + {{\left( {\dot y} \right)}^2}}  = \sqrt {\frac{{{t^9} - 2{t^3} + 1}}{{{t^2}}} + 4t}  = \sqrt {\frac{{{t^9} + 2{t^3} + 1}}{{{t^2}}}} $

$ = \sqrt {\frac{{{{({t^3} + 1)}^2}}}{{{t^2}}}}  = \frac{{{t^3} + 1}}{t} = {t^2} + \frac{1}{t}$

Now we need only integrate this result, from 1 to 3.

For the area of the surface we have:

$\int_a^b {2\pi y\sqrt {{{\left( {\frac{{dy}}{{dt}}} \right)}^2} + {{\left( {\frac{{dx}}{{dt}}} \right)}^2}} {\kern 1pt} dt} $

We have already evaluated the following:

${\sqrt {{{\left( {\frac{{dy}}{{dt}}} \right)}^2} + {{\left( {\frac{{dx}}{{dt}}} \right)}^2}} {\kern 1pt} }$

We have therefore:

$2\pi \int_1^3 {y\left( {{t^2} + \frac{1}{t}} \right)\,dt} $

$2\pi \int_1^3 {\frac{4}{3}{t^{\frac{3}{2}}}\left( {{t^2} + \frac{1}{t}} \right)\,dt}  = \frac{{8\pi }}{3}\int_1^3 {{t^{\frac{7}{2}}}}  + {t^{\frac{1}{2}}}\,dt$

The integral is simple enough, and hence left to the reader.

Further Pure | Topic 4: Mathematical Induction

Curriculum Objectives:

• use the method of mathematical induction to establish a given result (questions set may involve divisibility tests and inequalities, for example)
• recognise situations where conjecture based on a limited trial followed by inductive proof is a useful strategy, and carry this out in simple cases, e.g. find the nth derivative of $x{e^x}$

Proof by Induction

Many theorems and formulae relating to positive whole numbers can be proved by a process called mathematical induction. The method is summarised thusly:

1. assume that it is true for n = k, and prove that it is also true for n = k + 1
2. prove that the result is true n = 1 (or perhaps for some other small number n = 2 or 3)

If you can prove both (i) and (ii), then you have shown that the theorem is true at start (n = 1, say) and it is therefore true for n = 1 + 1, and then n = 2 + 1, and then n = 3 + 1, and so on for all integral values of n following after the valid starting value ( usually n = 1, but not always). This way of proving the validity of a theorem or formula is the method of mathematical induction. and essential requirement when trying to prove a proposition by induction is that you either know the final result or can surmise it.

Note: In almost all the questions that follow the second step has been left to reader, however in the examinations you will have to show the second step with all its working.

Summation of Series

Consider the following, rather simple example:

Ex. Prove the following:

$\sum\limits_{r = 1}^n {\frac{1}{{r(r + 1)}}}  = \frac{n}{{n + 1}}$

We assume, to begin, that the relationship is correct for n = k, which is to say:

$\frac{1}{{1 \times 2}} + \frac{1}{{2 \times 3}} + \frac{1}{{3 \times 4}} + ... + \frac{1}{{k(k + 1)}} = \frac{k}{{k + 1}}$

We then prove that the sum of terms up to (k + 1) obeys the above relationship. We do this thusly:

$\sum\limits_{r = 1}^{k + 1} {\frac{1}{{r(r + 1)}}}  = \frac{k}{{k + 1}} + \frac{1}{{(k + 1)(k + 2)}}$

$ = \frac{{k(k + 2) + 1}}{{(k + 1)(k + 2)}}$
$ = \frac{{{k^2} + 2k + 1}}{{(k + 1)(k + 2)}}$
$ = \frac{{{{(k + 1)}^2}}}{{(k + 1)(k + 2)}}$
$ = \frac{{k + 1}}{{(k + 2)}}$

Which is exactly the result we would have gotten by the relationship we set out to prove. We can then say with absolute surety that if the relationship holds for n = k, then it will surely stand for n = k + 1. The result obviously holds for n = 1, you may check this by inputting the one into the above relationship.

Ex. Prove that:

$\sum\limits_{r = 1}^n {{r^3}}  = \frac{1}{4}{n^2}{(n + 1)^2}$

Sol.

Let us assume that it holds for n = k, i.e.

$\sum\limits_{r = 1}^k {{r^3}}  = \frac{1}{4}{k^2}{(k + 1)^2}$

For n = k + 1, we have:

$\sum\limits_{r = 1}^{k + 1} {{r^3}}  = \frac{1}{4}{k^2}{(k + 1)^2} + {(k + 1)^3}$
$= \frac{1}{4}{(k + 1)^2}({k^2} + 4k + 4)$
$ = \frac{1}{4}{(k + 1)^2}{(k + 2)^2}$

The case n = 1 is also obviously true.

Hence by the Principle of Mathematical Induction, this result holds.

Let us look at one more application of Mathematical Induction in summation, before we move on to its applications in sequences, divisibility test and inequalities.

Ex. Prove that:

$\sum\limits_{r = 1}^n {r \cdot r!}  = (n + 1)! - 1$
Sol.

Supposing that it holds for n = k, i.e.

$1 \cdot 1! + 2 \cdot 2! + ... + k \cdot k! = (k + 1)! - 1$

For n = k + 1, we have:

$\sum\limits_{r = 1}^{k + 1} {r \cdot r!}  = (k + 1)! - 1 + (k + 1) \cdot (k + 1)!$
$ = (k + 1)!(k + 1 + 1) - 1$ 
$ = (k + 2)(k + 1)! - 1$
$ = (k + 2)! - 1$
$ = \left( {\overline {k + 1}  + 1} \right)! - 1$

N = 1, is easily proved, and therefore, by the Principle of Mathematical Induction, our hypothesis that the sum of this series is given by this expression, is true.

Sequences

The following example involves only proving a sequence to be true.

Ex. A sequence, ${u_1},{u_2},{u_3}...$ is defined by the recursive formula:

${u_{n + 1}} = 3 - \frac{2}{{{u_n}}}$

Prove that the nth term of the series. is given by:

${u_n} = \frac{{{2^{n + 1}} - 1}}{{{2^n} - 1}}$

Sol.

Assuming that n = k is true then:

${u_k} = \frac{{{2^{k + 1}} - 1}}{{{2^k} - 1}}$

Using the result given, then:

${u_{k + 1}} = 3 - \frac{2}{{{u_k}}}$

${u_{k + 1}} = 3 - \frac{2}{{\frac{{{2^{k + 1}} - 1}}{{{2^k} - 1}}}}$

$= 3 - \frac{{2({2^k} - 1)}}{{{2^{k + 1}} - 1}}$

$ = \frac{{3 \cdot {2^{k + 1}} - 3 - {2^{k + 1}} + 2}}{{{2^{k + 1}} - 1}}$

$ = \frac{{2 \cdot {2^{k + 1}} - 1}}{{{2^{k + 1}} - 1}}$

$ = \frac{{{2^{\overline {k + 1}  + 1}} - 1}}{{{2^{k + 1}} - 1}}$

${H_k} \Rightarrow {H_{k + 1}}$

This above notation simply reads, that the hypothesis n = k, implies n = k + 1. Instead writing it all over again, you can employ this notation. n = 1 is easily verified.

Ex. (October/November 2012 Paper 12 Q5)

The first part of the question requires the derivation of the reduction formula,

${I_n} = \frac{1}{2}n{I_{n - 1}}$

where $I_n$ is:

$\int_0^\infty  {{x^n}{e^{ - 2x}}dx} $

You may disregard at this point, how we proved this relationship, till we reach the topic reduction formula. The second part is of significance here. Prove by Mathematical Induction that, for all positive integers n, ${I_n} = \frac{{n!}}{{{2^{n + 1}}}}$

Reemploying the strategy from the previous question:

Initial we make some substitutions so that the relationship better conforms with our chosen notation, i.e. replacing n by n + 1 and n - 1 by n.

${I_{n + 1}} = \frac{1}{2}(n + 1){I_n}$

The initial hypothesis is that n = k,

${I_k} = \frac{{k!}}{{{2^{k + 1}}}}$

For n = k + 1

${I_{k + 1}} = \frac{1}{2}(k + 1){I_k}$

${I_{k + 1}} = \frac{1}{2}k\left( {\frac{{k!}}{{{2^{k + 1}}}}} \right)$

${I_{k + 1}} = \frac{{(k + 1)!}}{{{2^{\overline {k + 1}  + 1}}}}$

${H_k} \Rightarrow {H_{k + 1}}$

We prove that n = 1 or $H_1$ is true, by noting:

${I_0} = \int_0^\infty  {{e^{ - 2x}}dx}  = \left[ { - \frac{1}{2}{e^{ - 2x}}} \right]_0^\infty  = \frac{1}{2}$

${I_{n + 1}} = \frac{1}{2}(n + 1){I_n}$ 

${I_1} = \frac{1}{2}(1){I_0} = \frac{1}{2}\left( {\frac{1}{2}} \right) = \frac{{1!}}{{{2^2}}}$

Hence by Principle of Mathematical Induction, this sequence is true for all positive integers.

Divisibility Tests

Now we look at an application of the Inductive proofs in divisibility tests.

Ex. Prove that, for all positive integers, n, ${3^{2n}} + 7$, is divisible by 8.

Sol.

In divisibility tests, we employ a slightly different strategy.

We begin, as before, by assuming $H_k$ ( n = k ) holds, which is to say ${3^{2k}} + 7$ is divisible by 8.

For n = k + 1, the expression becomes

${3^{2(k + 1)}} + 7$

${3^{2k + 2}} + 7$

${3^2} \cdot {3^{2k}} + 7$

We now consider the difference of these two expressions, i.e.,

${3^2} \cdot {3^{2k}} + 7 - ({3^{2k}} + 7) = 9 \cdot {3^{2k}} + 7 - {3^{2k}} - 7 = 8 \cdot {3^{2k}}$

This is obviously divisible by 8, and hence if $H_k$ holds, then $H_{k + 1}$ is also true. $H_1$ is obviously true. Then by Principle of Mathematical Induction, this is true.

In divisibility tests, the general strategy suggested is to consider the either the sum, or difference of any scalar multiple of the expressions derived for n = k +1 and n = k.

Inequalities

Now we consider its use in the context of series. Consider the following question from one of your past papers:

Ex. (October/November 2005, Q2)
The sequence ${u_1},{u_2},{u_3}...$ is such that $u_1$ = 1 and:

${u_{n + 1}} =  - 1 + \sqrt {{u_n} + 7}$

(I) Prove by Induction that $u_n$ < 2 for all $n \ge 1$.
(II) Show that if $u_n$ = 2 - $\varepsilon $, where $\varepsilon $ is small, then

${u_{n + 1}} \approx 2 - \frac{1}{6}\varepsilon $

Sol.

(I) We begin by assuming that the relationship holds for n = k, i.e.

${u_k} < 2$

For the case n = k +1

${u_{k + 1}} < 2$

As we know:

${u_{k + 1}} =  - 1 + \sqrt {{u_k} + 7} $

Therefore

$ - 1 + \sqrt {{u_k} + 7}  < 2$

$\sqrt {{u_k} + 7}  < 3$

${u_k} + 7 < 9$

${u_k} < 2$

Therefore n = k implies n = k + 1. It is easily seen that n = 1 is true.

(II) For the second part, the only thing of note was that an approximation was expected.

Given:

${u_n} = 2 - \varepsilon $

Using the recursive function that defines the sequence:

${u_{n + 1}} =  - 1 + {\left( {9 - \varepsilon } \right)^{\frac{1}{2}}}$

At this stage we use a binomial approximation up to the power one of $\varepsilon $

${u_{n + 1}} \approx  - 1 + 3 - \frac{1}{6}\varepsilon  = 2 - \frac{1}{6}\varepsilon $

Proving de Moivre's Theorem using Mathematical Induction

The de Moivre's Theorem is a very important theorem, which enriches the study of trigonometry and complex numbers, even further. The knowledge of it, and some of its uses are required by the syllabus, but they have been deferred till we reach the topic complex numbers. While the only past paper question I could find on it concerned itself with positive integer values of n, I shall for completeness discuss the cases where n is negative and zero.

The de Moivre Theorem states:

${\left( {\cos \theta  + \iota \sin \theta } \right)^n} = \cos (n\theta ) + \iota \sin (n\theta )$, where n is any integer and $\iota  = \sqrt { - 1} $

For the case n is positive

n = 1 is true is quite easily seen.

For the case n = k, we have:

${\left( {\cos \theta  + \iota \sin \theta } \right)^k} = \cos (k\theta ) + \iota \sin (k\theta )$

For the case, n = k + 1, we have that

${\left( {\cos \theta  + \iota \sin \theta } \right)^{k + 1}} = (\cos (k\theta ) + \iota \sin (k\theta ))(cos(\theta ) + \iota sin(\theta ))$

$ = (cos(k\theta )cos(\theta ) + \iota sin(k\theta )cos(\theta ) + \iota \sin (\theta )(cos(k\theta ) + {\iota ^2}\sin (k\theta )sin(\theta ))$

Using the value of ${\iota ^2}$ and the trigonometric identities:

$\cos (A + B) = \cos (A)\cos (B) - \sin (A)sin(B)$
$sin(A + B) = sin(A)cos(B) + sin(B)cos(A)$

We have:

$ = \cos (k + 1)\theta  + i\sin (k + 1)\theta $

For the case that n is a negative number, we begin with the following setting n equal to -m, where m is a positive integer.

${\left( {\cos \theta  + \iota \sin \theta } \right)^{ - m}} = \frac{1}{{{{(\cos \theta  + \iota \sin \theta )}^m}}} = \frac{1}{{\cos (m\theta ) + \iota \sin (m\theta )}}$

$ = \frac{1}{{\cos (m\theta ) + \iota \sin (m\theta )}} \times \frac{{\cos (m\theta ) - i\sin (m\theta )}}{{\cos (m\theta ) - \theta \sin (m\theta )}}$

$ = \frac{{(\cos (m\theta ) - \iota \sin (m\theta ))}}{{{{\cos }^2}(m\theta ) + {{\sin }^2}(m\theta )}} = \cos (m\theta ) - \iota \sin (m\theta )$

$ = \cos ( - m\theta ) + \iota \sin ( - m\theta )$

$ = \cos (n\theta ) + \iota \sin (n\theta )$

These trigonometric identities were used in the third last step:

$\cos (\theta ) = \cos ( - \theta )$
$ - \sin (\theta ) = \sin ( - \theta )$

While I do believe I have covered majority of the possibilities, we should not underestimate the ingenuity of the examiner. There have been questions of induction that involve calculus or some high level algebra. I will get to them, once I have completed all the topics.

Further Pure | Topic 3: Summation of Series

Curriculum Objectives:

• use the standard results for $\sum r $, $\sum r^2 $, $\sum r^3 $ to find related sums;
• use the method of differences to obtain the sum of a finite series, e.g. by expressing the general term in partial fractions;
• recognise, by direct consideration of a sum to n terms, when a series is convergent, and find the sum to infinity in such cases.

'$\sum $' notation and properties thereof:

A very useful shorthand for the sum of series containing a large number of terms is the sigma notation. The symbol $\sum\limits_{k = 1}^n {{a_k}} $ denotes the sum ${a_1} + {a_2} + ... + {a_n}$. The a's are terms of the sum: ${a_1}$ is the initial term, ${a_2}$ is the second term, ${a_k}$ is the kth term, and ${a_n}$ is the nth and last term of the sum. The variable k is called the index of summation. The values of k include all values from 1 to n.

Consider the following summations:

$\sum\limits_{k = 1}^5 k  = 1 + 2 + 3 + 4 + 5 = 15$
$\sum\limits_{k = 1}^3 {{{( - 1)}^k}k}  = {( - 1)^1}(1) + {( - 1)^2}(2) + {( - 1)^3}(3) =  - 2$

You will be required to have a firm grasp of the alegbra with these sums. Summarised below are some properties of this notation.

1. Sum Rule:

$\sum\limits_{k = 1}^n {({a_k} + {b_k})}  = \sum\limits_{k = 1}^n {{a_k}}  + \sum\limits_{k = 1}^n {{b_k}} $

2. Difference Law

$\sum\limits_{k = 1}^n {({a_k} - {b_k})}  = \sum\limits_{k = 1}^n {{a_k}}  - \sum\limits_{k = 1}^n {{b_k}} $

3. Constant Multiple Rule

$\sum\limits_{k = 1}^n {c{a_k}}  = c\cdot\sum\limits_{k = 1}^n {{a_k}} $ (for any number c)

4. Constant Value Rule

$\sum\limits_{k = 1}^n c  = n \cdot c$    (c is any constant value)

You may also need to use your knowledge of arithmetic and geometric series, to solve these questions. Formulae pertaining to those series, are thusly summarised:
For an arithmetic series:

${a_n} = {a_1} + (n - 1)d$, where $a_n$ is the nth term, $a_1$ is the first term, and d is the common difference.

${a_n} = {a_m} + (n - m)d$, for any general term, m.

The sum is given by:

${S_n} = \frac{n}{2}\left( {2{a_1} + (n - 1)d} \right)$

For a geometric progression:

${a_n} = {a_1}{r^{n - 1}}$, for the general term n, where $a_n$ is the nth term, $a_1$ the first term, and r is the common ratio.

The sum of a geometric progression, say $a{r^0} + a{r^1} + a{r^2} + ... + a{r^{n - 1}}$ is given by:

${S_n} = \frac{{a(1 - {r^n})}}{{1 - r}}$

The following general results are quoted in your MF10 formula sheet, and you should be aware of them.

$\sum\limits_{r = 1}^n r  = \frac{1}{2}n(n + 1)$

 $\sum\limits_{r = 1}^n {{r^2}}  = \frac{1}{6}n(n + 1)(2n + 1)$

$\sum\limits_{r = 1}^n {{r^3}}  = \frac{1}{4}{n^2}{(n + 1)^2}$

The following examples highlight the level of algebra you are expected to have mastered.

Ex. Find the following sums:

$a)\sum\limits_{r = 7}^{20} {{r^2}} $
$b)\sum\limits_{r = 12}^{25} {{r^3}} $

Sol:

The initial value here is not the usual 1 (or 0). To evaluate these summations, we evaluate two different summations, one from 1 to one less than the initial value and the other from 1 to the final value, and then subtract one from the other to arrive at the answer.

$\sum\limits_{r = 7}^{20} {{r^2}}  = \sum\limits_{r = 1}^{20} {{r^2}}  - \sum\limits_{r = 1}^6 {{r^2}}  = \frac{{20}}{6}(21)(41) - \frac{6}{6}(7)(13) = 2870 - 91 = 2779$

$\sum\limits_{r = 12}^{25} {{r^3}}  = \sum\limits_{r = 1}^{25} {{r^3}}  - \sum\limits_{r = 1}^{11} {{r^3}}  = \frac{{{{25}^2}}}{4}({26^2}) - \frac{{{{11}^2}}}{4}({12^2}) = 105625 - 4356 = 101269$

Ex. Prove that

$\sum\limits_{r = 1}^n {r(r + 1) \equiv } \frac{1}{3}n(n + 1)(n + 2)$

Sol.

We begin with the following consideration:

$\sum\limits_{r = 1}^n {r(r + 1) = \sum\limits_{r = 1}^n {({r^2} + r)} }  = \sum\limits_1^n {{r^2}}  + \sum\limits_1^n r $
$ = \frac{1}{6}n(n + 1)(2n + 1) + \frac{1}{2}n(n + 1)$
$ = \frac{1}{6}n(n + 1)((2n + 1) + 3))$
$ = \frac{1}{6}n(n + 1)(2n + 4)$
$ = \frac{1}{3}n(n + 1)(n + 2)$

The Method of Differences

While there does not exist, a single method to sum a finite series, the method of differences is a particularly elegant one for some. The general approach is to find a function, f(r), such that the rth term, $u_r$, of a series can be expressed as

${u_r} = f(r + 1) - f(r)$

Then we add all the terms, ${u_1},{u_2},...,{u_n}$, making cancellations as we go along to arrive at the final answer. The following example elucidates this:

Say we wanted to prove the sum 1 + 2 + 3 + ... + n, i.e.,

$\sum\limits_{r = 1}^n r $

Consider the identity:

$2r \equiv r(r + 1) - (r - 1)r$

By taking consecutive terms, 1, 2,..., n for r we get:

$2(1) = 1(2) - (0)1$
$2(2) = 2(3) - (1)(2)$
$2(3) = 3(4) - (2)(3)$
.
.
$2(n - 1) = (n - 1)(n) - (n - 2)(n - 1)$
$2n = n(n + 1) - (n - 1)(n)$

Now if we were to sum every term, we find that almost all terms cancel out leaving:

$2(1 + 2 + 3 + ... + (n - 1) + n) = n(n + 1)$
$2\sum\limits_{r = 1}^n r  = n(n + 1)$
$\sum\limits_{r = 1}^n r  = \frac{1}{2}n(n + 1)$

In an examination, you will either be given the function, or information, wherewith you maybe derive it yourself.

Ex. (May/June 2013 Paper 12 Q5.)
Use the method of differences to show that $\sum\limits_{r = 1}^N {\frac{1}{{(2r + 1)(2r + 3)}}}  = \frac{1}{6} - \frac{1}{{2(2N + 3)}}$.

Sol.

We use partial decomposition:

$\frac{1}{{(2r + 1)(2r + 3)}} = \frac{A}{{(2r + 1)}} + \frac{B}{{(2r + 3)}}$

Multiplying across by (2r + 1) and then putting $r =  - \frac{1}{2}$, we find A is equal to (1/2) and by multiplying across by (2r + 3) and then putting in $r =  - \frac{3}{2}$, to find B is equal to (-1/2). Thus:

$\frac{1}{{(2r + 1)(2r + 3)}} = \frac{1}{{2(2r + 1)}} - \frac{1}{{2(2r + 3)}}$

For different values of r we have:

r = 1

$\frac{1}{{2(3)}} - \frac{1}{{2(5)}}$

r = 2

$\frac{1}{{2(5)}} - \frac{1}{{2(7)}}$

r = N - 1

$\frac{1}{{2(2N - 1)}} - \frac{1}{{2(2N + 1)}}$

r = N

$\frac{1}{{2(2N + 1)}} - \frac{1}{{2(2N + 3)}}$

Adding all the terms, the cancellations are apparent and we are left with the following result:

$\sum\limits_{r = 1}^N {\frac{1}{{(2r + 1)(2r + 3)}}}  = \frac{1}{6} - \frac{1}{{2(2N + 3)}}$

While the question doesn't ask for it, consider what would happen if n approached infinity. The fraction, would become smaller and smaller to the point that it no longer matters to the calculation at hand to any good approximation. Therefore:

$\sum\limits_{r = 1}^\infty  {\frac{1}{{(2r + 1)(2r + 3)}}}  = \frac{1}{6}$

 Deduce then $\sum\limits_{r = N + 1}^{2N} {\frac{1}{{(2r + 1)(2r + 3)}}}  < \frac{1}{{8N}}$

Sol.

$\sum\limits_{r = N + 1}^{2N} {\frac{1}{{(2r + 1)(2r + 3)}}}  = \sum\limits_{r = 1}^{2N} {\frac{1}{{(2r + 1)(2r + 3)}}}  - \sum\limits_{r = 1}^N {\frac{1}{{(2r + 1)(2r + 3)}}} $ 
$ = \frac{1}{6} - \frac{1}{{2(4N + 3)}} - \left( {\frac{1}{6} - \frac{1}{{2(2N + 3)}}} \right)$
$ = \frac{1}{6} - \frac{1}{{2(4N + 3)}} - \frac{1}{6} + \frac{1}{{2(2N + 3)}}$
$ = \frac{{ - 2N - 3 + 4N + 3}}{{2(4N + 3)(2N + 3)}}$
$ = \frac{{2N}}{{2(4N + 3)(2N + 3)}}$
$ = \frac{N}{{8{N^2} + 12N + 6N + 9}}$
$ = \frac{N}{{8{N^2} + 18N + 9}}$
$ = \frac{1}{{8N + 18 + \frac{9}{N}}}$

As N is greater than or equal to 1:

$18 + \frac{9}{N} > 0$
$\frac{1}{{8N + 18 + \frac{9}{N}}} < \frac{1}{{8N}}$

Hence proved.

Ex.(October/November 2012 Paper 12 Q4)

Let $f(r) = r(r + 1)(r + 2)$. Show that:

$f(r) - f(r - 1) = 3r(r + 1)$

Sol.

$f(r) - f(r - 1) = r(r+1)(r+2)-(r-1)r(r+1)$
$= r(r + 1)(r + 2 - (r - 1))$
$= r(r + 1)(r + 2 - r + 1)$
$= 3r(r + 1)$

Hence show that:

$\sum\limits_{r = 1}^n {r(r + 1)}  = \frac{1}{3}n(n + 1)(n + 2)$

Using the identity proven above we have:

For r = 1

$1(2)(3) - (0)(1)(2)$

r = 2

$2(3)(4) - (1)(2)(3)$

r = n - 1

$(n - 1)(n)(n + 1) - (n - 2)(n - 1)(n)$

r = n

$n(n + 1)(n + 2) - (n - 1)(n)(n + 1)$

Adding and cancelling terms, we get:

$\sum\limits_{r = 1}^n {3r(r + 1)}  = n(n + 1)(n + 2)$
$\sum\limits_{r = 1}^n {r(r + 1)}  = \frac{1}{3}n(n + 1)(n + 2)$

Using the standard result for the $\sum r $, deduce the general formula for $\sum r^2 $

Sol.

We know that:

$\sum\limits_{r = 1}^n {r(r + 1)}  = \frac{1}{3}n(n + 1)(n + 2)$
$\sum\limits_{r = 1}^n {{r^2} + r}  = \frac{1}{3}n(n + 1)(n + 2)$
$\sum\limits_{r = 1}^n {{r^2}}  = \frac{1}{3}n(n + 1)(n + 2) - \sum\limits_{r = 1}^n r $
$\sum\limits_{r = 1}^n {{r^2}}  = \frac{1}{3}n(n + 1)(n + 2) - \frac{1}{2}n(n + 1)$
$\sum\limits_{r = 1}^n {{r^2}}  = \frac{1}{6}n(n + 1)(2(n + 2) - 3)$
$\sum\limits_{r = 1}^n {{r^2}}  = \frac{1}{6}n(n + 1)(2n + 1)$

Find the sum of the series:

${1^2} + 2 \times {2^2} + {3^2} + 2 \times {4^2} + {5^2} + 2 \times {6^2} + ... + 2{(n - 1)^2} + {n^2}$

where n is odd

Sol.

For this question we rearrange the sequence thusly:

$({1^2} + {2^2} + {3^2} + {4^2} + {6^2} + ... + {(n - 1)^2} + {n^2}) + ({2^2} + {4^2} + {5^2} + {6^2} + ... + {(n - 1)^2})$

The first series is simply the sum of the squares of natural numbers from one to the odd number n, i.e.

$\sum\limits_1^n {{r^2}}  = \frac{1}{6}n(n + 1)(2n + 1)$

The second series is the sum of the square of even numbers up to the even number (n -1).

An even number is simply twice a natural number, therefore substituting 2n into $\sum\limits_1^n r^2 $, should give us an expression for the sum of even numbers i.e.

$\sum\limits_{r = 1}^{2n} {{{\left( {2r} \right)}^2} = 4\sum\limits_{r = 1}^{2n} {{r^2}} } $

Another important thing to be taken note of is that this expression would give us the sum of the squares of even numbers from one to 2n. We however require that up to n - 1. For this we substitute $\frac{{n - 1}}{2}$ into our expression of the sum of the squares of the even numbers, i.e.

$4\sum\limits_{r = 1}^{2\left( {\frac{{n - 1}}{2}} \right)} {{r^2}}  = 4\left( {\frac{1}{6}} \right)\left( {\frac{{n - 1}}{2}} \right)\left( {\frac{{n - 1}}{2} + 1} \right)\left( {2\left( {\frac{{n - 1}}{2}} \right) + 1} \right)$

$4\sum\limits_{r = 1}^{n - 1} {{r^2}}  = 4\left( {\frac{1}{6}} \right)\left( {\frac{{n - 1}}{2}} \right)\left( {\frac{{n + 1}}{2}} \right)\left( n \right)$

Adding these two expressions:

$\sum\limits_{r = 1}^n {{r^2}}  + 4\sum\limits_{r = 1}^{n - 1} {{r^2}}  = \frac{1}{6}n(n + 1)(2n + 1) + 4\left( {\frac{1}{6}} \right)\left( {\frac{{n - 1}}{2}} \right)\left( {\frac{{n + 1}}{2}} \right)n = \frac{1}{2}{n^2}(n + 1)$ 

(Optional) Double Summations

While not necessarily required for the course you may be interested in the use of double or even more summations performed concurrently. The following examples exhibit a few double summations for the reader to mull over.

$\sum\limits_{j = 1}^2 {\sum\limits_{i = 1}^3 {{a_{ij}}} }  = \sum\limits_{j = 1}^2 {\left( {\sum\limits_{i = 1}^3 {{a_{ij}}} } \right) = \sum\limits_{j = 1}^2 {\left( {{a_{1j}} + {a_{2j}} + {a_{3j}}} \right) = \sum\limits_{j = 1}^2 {{a_{1j}} + \sum\limits_{j = 1}^2 {{a_{2j}} + \sum\limits_{j = 1}^2 {{a_{3j}}} } } } }  = {a_{11}} + {a_{12}} + {a_{21}} + {a_{22}} + {a_{31}} + {a_{32}}$

Further Pure | Topic 2: Polar Coordinates

Curriculum Objectives:

• Understand the relations between cartesian and polar coordinates (using the convention $r \ge 0$), and convert equations of curves from cartesian to polar form and vice versa;
• sketch simple polar curves, for 0 < θ < 2π or −π < θ < π or a subset of either of these intervals (detailed plotting of curves will not be required, but sketches will generally be expected to show significant features, such as symmetry, the form of the curve at the pole and least/greatest values of r );
• recall the formula $\frac{1}{2}\int_\alpha ^\beta  {{r^2}d\theta } $ for the area of a sector, and use this formula in simple cases. 

Polar Coordinates:

I can say this, without exaggerating, that the cartesian coordinates are the most used of any coordinate system. However there are curves, the representation whereof in the cartesian system is an onerous task. Fortunately though, we have other systems of coordinates available, which can conveniently simplify the task. One such is the Polar Coordinate System.

Consider the point (3,4) on the cartesian plane. With the origin defined, you can without ambiguity fix this point to a plane. If you were to relay these coordinates to someone else, they too would be able to identify this unique point on the plane. But is that the only way to define that point? No, one alternative would be tell whoever is to plot the point that it is a distance a from some fixed point, let us in this example assume that it is the origin of the cartesian plane considered previously, and furthermore tell him that it makes an angle of say b radians, with some given line, which we may most unimaginatively denote by the initial line. Carrying our analogy of cartesian plane further, let the initial line be the x-axis. In this example a would be 5 and b would be 0.927 rad.

Something to this effect. (Disregard the crudeness of the plot)

This system at the moment does have some problems, that can be adequately overcome, by agreeing on the following conventions:

1) The length from the fixed point, r, shall always be equal to or greater than zero. 

2) The angle made with the initial line, $\theta $, will remain in the confines of the following inequality: $ - \pi  < \theta \le \pi $

3) The angle measured counter-clockwise from the initial line is positive, and that measured clockwise is negative.

If we agree upon these rules, then (r, $\theta $) are the polar coordinates of a point. 

The following graph might help in visualising the system:

Conversion of a curve from Polar to Cartesian form

Using basic trigonometry we can derive the relationship between a polar and cartesian system. Consider the follow the plot:

It is evident from above the sketch, that if we positioned the pole (the fixed point wherefrom the distance is measured) at the origin of a cartesian plane, and take the positive x axis as the initial line, we have the following relationships:

$r\cos \theta  = x$

$r\sin \theta  = y$

${r^2} = {x^2} + {y^2}$

The general method is explained through the following examples:

Ex. Convert the following polar equation into its cartesian form:

${r^2} = 4\cos (2\theta )$

Sol: We consider the following relationships:

${r^2} = {x^2} + {y^2}$

$\cos (2\theta ) = {\cos ^2}\theta  - {\sin ^2}\theta $

$\sin \theta  = \frac{y}{r}$

$\cos \theta  = \frac{x}{r}$

Using these it is not difficult to see that the cartesian equation can be derived thusly:

${x^2} + {y^2} = 4\left( {\frac{{{x^2}}}{{{r^2}}} - \frac{{{y^2}}}{{{r^2}}}} \right)$

Then taking $r^2$ common, and using the suitable relationship, we arrive at the cartesian equation:

$r\left( {{x^2} + {y^2}} \right) = 4\left( {{x^2} - {y^2}} \right)$

${\left( {{x^2} + {y^2}} \right)^2} = 4\left( {{x^2} - {y^2}} \right)$

Ex. Oct/Nov 2012 Paper 12, Question 1:

Find the cartesian equation corresponding to the polar equation 

$r = \sqrt 2 \sec \left( {\theta  - \frac{\pi }{4}} \right)$

Sol: 

We rearrange the equation as follows:

$r\left( {\cos \theta  - \frac{\pi }{4}} \right) = \sqrt 2 $

$r\cos \theta cos\left( {\frac{\pi }{4}} \right) + r\sin \theta sin\left( {\frac{\pi }{4}} \right) = \sqrt 2 $

$\frac{x}{{\sqrt 2 }} + \frac{y}{{\sqrt 2 }} = \sqrt 2 $

$y = 2 - x$

Curve Sketching

The most effective strategy in sketching a curve, $r = f(\theta )$, is to find the polar coordinates of a number of points on the curve. Following are plots of curves of the type expected by your examiners and some points of note on the matter.

Ex. Sketch the curve 

$r = 2a(1 + \cos \theta )$

Before the actual plotting, we find by inspection alone, that the function is an even function, which means that the function yields the same value of $\theta $, as it does for - $\theta $. This is because of the well known identity:

$\cos ( - \theta ) = \cos (\theta )$

This means the function will be symmetrical about the initial line, and you will need only the values in the region $0 \le \theta  \le \pi $

We then put in values of $\theta $ with the increments of $\frac{\pi }{6}$ in the range given above.

Using this information you can get the plot of the following shape.

This plot was made by taking a = 2. You however would plot it by using
values of u, 2u, 3u... so forth on the axes. 

Area of a Sector

For any curve, the area bound by the curve and the lines, $\theta  = \alpha $ and $\theta = \beta $, is given by:

 $\frac{1}{2}\int_\alpha ^\beta  {{r^2}d\theta } $

Perhaps the above graph will hep you visualise the result. The formula will give you the area shaded in the above diagram.

Ex. Find the area of the cardioid  $r = 2(1 + \cos \theta )$

We note that:

${r^2} = {\left( {2(1 + \cos \theta )} \right)^2}$
${r^2} = 4(1 + 2\cos \theta  + {\cos ^2}\theta )$
${r^2} = 4 + 8\cos \theta  + 4{\cos ^2}\theta $

Further considering:

$\cos (2\theta ) = 2{\cos ^2}\theta  - 1$
$2{\cos ^2}\theta  = \cos (2\theta ) + 1$
$4co{s^2}\theta  = 2\cos (2\theta ) + 2$

We have then:

${r^2} = 4 + 8\cos \theta  + 2\cos (2\theta ) + 2 = 2\cos (2\theta ) + 8\cos (\theta ) + 6$
$\frac{1}{2}\int_0^{2\pi } {{r^2}d\theta } $ = $\frac{1}{2}\int_0^{2\pi } {\left( {2\cos (2\theta ) + 8\cos (\theta ) + 6} \right)d\theta } $ $ = 6\pi $

Because the cardioid is symmetrical about the x-axis, you could evaluated the area by using the formula from zero to $\pi $ instead of $2 \pi $ and then multiplying the resulting area by two. But as this makes the evaluation only marginally less tedious, and introduces room for error, it is not recommended.