Assorted Proofs of Theorems and Formulae

Proofs:

Note: Any proofs required by your syllabus will be outlined in relevant blog entries; these are simply for some semblance of completeness, and to sate the curious. You will not be required to reproduce these results or derive new ones hence. You will however be expected to understand these formulae and theorems, and use them accordingly.

Chapter 2:

Area enclosed by a polar curve, r = f($ \theta $), and rays, $ \theta = \alpha $ and $ \theta = \beta $.

Consider the following curve:

In the above diagram, one will find the curve drawn from the pole( the point wherefrom all the rays originate, O), the rays, $ \theta = \alpha $ and $ \theta = \beta $. A smaller area in the area considered OPQ. The point P is (r, $ \theta $), and Q is ($r + \delta r$, $\theta  + \delta \theta $). We assume that r increases with $ \theta $ throughout the interval A to B.

Out strategy would be to divide the entire area into such smaller areas, and then summing the individual areas to evaluate the area. By producing OP at Q', we have also constructed two circular sectors. One sector is OPP', the other OQ'Q.

We recall that the area of a circular sector is:

$A = \frac{1}{2}{r^2}\theta $

Hence the areas of the two sectors are 

$Are{a_{OPP'}} = \frac{1}{2}{r^2}\delta \theta $

$Are{a_{OQ'Q}} = \frac{1}{2}{\left( {r + \delta r} \right)^2}\left( { \delta \theta } \right)$

It is not difficult to see that the area of OPQ, lies between, the areas of OPP' and OQ'Q, i.e.

$\frac{1}{2}r\delta \theta  < Area\;of\;OPQ < \frac{1}{2}{\left( {r + \delta r} \right)^2}\left( {\delta \theta } \right)$

It follows, therefore that the area of the required region OAB is the sum of all these smaller regions,  and satisfies the following inequality.

$\sum\limits_{\theta  = \alpha }^{\theta  = \beta } {{\textstyle{1 \over 2}}{r^2}\delta \theta }  < Area\;of\,OAB < \sum\limits_{\theta  = \alpha }^{\theta  = \beta } {{\textstyle{1 \over 2}}{{\left( {r + \delta r} \right)}^2}\delta } \theta $

Now, if I let the number of such smaller segments to tend to infinity, $\delta \theta  \to 0 \Rightarrow \delta r \to 0$

And so,  by the definition of the definite integral:

$\mathop {\lim }\limits_{n \to \infty } \left( {\sum\limits_{\theta  = \alpha }^{\theta  = \beta } {{\textstyle{1 \over 2}}{r^2}\delta \theta } } \right) = \int_\alpha ^\beta  {{\textstyle{1 \over 2}}{r^2}} {\kern 1pt} d\theta $

Similarly:

$\mathop {\lim }\limits_{n \to \infty } \left( {\sum\limits_{\theta  = \alpha }^{\theta  = \beta } {{\textstyle{1 \over 2}}{{\left( {r + \delta r} \right)}^2}\delta \theta } } \right) = \int_\alpha ^\beta  {{\textstyle{1 \over 2}}{r^2}} {\kern 1pt} d\theta $

Therefore, we have that:

$Area\,of{\kern 1pt} OAB = \int_\alpha ^\beta  {{\textstyle{1 \over 2}}{r^2}} {\kern 1pt} d\theta $

Chapter 5

Formula for the co-ordinates of the centroid of an object.

We begin with the fact that the centroid of a uniform object is also its centre of mass.

The method applied here to derive the formula, is used to find the centre of mass of a composite body. Which those who have studied the M2 unit of CIE A Levels Mathematics, must already be aware of. For others, we will study it in the Applied portion of the syllabus.

Consider the curve, y = f(x), with the centroid ( or centre of mass) $C\left( {\bar x,\bar y} \right)$.

Now consider a small almost rectangular element in the curve, with the height y, and width δx. The ordinates that bound the curve, are x = a and x = b, where a < b. If we further take w to be the weight per unit area, and the area of the entire lamina is given by A, we have that.

By consideration of moments we see that the sum of the moments about the origin ( which I forgot to mark, but is at the intersection of the two axes), of all the small elements, should be approximately, equal to moment produced about the origin by the mass of the entire lamina through the centroid. The centroid of the elements is simply (x, (1/2)y) as the element, is to an adequate approximation, a rectangle. Which is to say:

$\sum\limits_{x = a}^{x = b} {\left( {y\,\delta x} \right)wx}  = Aw\bar x$

Cancelling w, and then letting the number such elements approach infinity, I get:

$A\bar x = \int_a^b {xy\,dx} $

For the y co-ordinate, we have:

$\sum\limits_{x = a}^{x = b} {\left( {y\,\delta x} \right)w({\textstyle{1 \over 2}}y)}  = Aw\bar y$

By the reasoning already stated, we are led to:

$A\bar y = {\textstyle{1 \over 2}}\int_a^b {{y^2}\,dx} $

A is the area of the curve, which you may remember from elementary calculus is:

$A = \int_a^b {y\,dx} $

Mean value of a function

The mean value of n numbers, is the their sum, divided by n. We adopt this to a continuous function, y = f(x), defined over an interval of, say [a, b]. We cut it up into subintervals, of number n, and of lengths:

$\delta x = \frac{{b - a}}{n}$

We then select from every interval, a value of x. The choice of x from within interval is arbitrary; it is only required that the value of x come from that interval, say ${x_1},{x_2},...,{x_n}$, and the images of the functions then are, $f({x_1}),f({x_2}),...,f({x_n})$. The mean of the function is then:

$\frac{{f({x_1}) + f({x_2})+...+f({x_n})}}{n}$

Substituting in the value of n, we get:

$\frac{{(f({x_1}),f({x_2}),...,f({x_n}))\delta x}}{{b - a}}$

$\frac{{(f({x_1}),f({x_2}),...,f({x_n}))\delta x}}{{b - a}} = \frac{1}{{b - a}}\left[ {\left( {f({x_1}) + f({x_2}) + ... + f({x_n})} \right)\delta x} \right]$

$ = \frac{1}{{b - a}}\sum\limits_{k = 1}^n {f({x_k})} \,\delta x$

If we now let the value of n increase, as it approaches infinity, we get:

$\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{{b - a}}\sum\limits_{k = 1}^n {f({x_k})} \,\delta x} \right) = \frac{1}{{b - a}}\mathop {\lim }\limits_{n \to \infty } \left[ {\sum\limits_{k = 1}^n {f({x_k}){\kern 1pt} \delta x} } \right] = \frac{1}{{b - a}}\int_a^b {f(x)\,dx} $

Arc Length

For an infinitesimal piece of the curve, $\delta s$ , we have by the Pythagoras' theorem:

$\begin{array}{l} {\left( {\delta s} \right)^2} = {\left( {\delta x} \right)^2} + {\left( {\delta y} \right)^2}\\ {\left( {\frac{{\delta s}}{{\delta x}}} \right)^2} = 1 + {\left( {\frac{{\delta y}}{{\delta x}}} \right)^2}\\ \left( {\frac{{\delta s}}{{\delta x}}} \right) = \sqrt {1 + {{\left( {\frac{{\delta y}}{{\delta x}}} \right)}^2}} \\ \delta s = \sqrt {1 + {{\left( {\frac{{\delta y}}{{\delta x}}} \right)}^2}} {\kern 1pt} \delta x \end{array}$

Or alternatively, we can employ the mean value theorem.

$\begin{array}{l} {\left( {\delta s} \right)^2} = {\left( {\delta x} \right)^2} + {\left( {\delta y} \right)^2}\\ {\left( {\delta s} \right)^2} = {\left( {\delta x} \right)^2} + {\left( {\delta x \cdot \frac{{dy}}{{dx}}} \right)^2}\\ {\left( {\delta s} \right)^2} = \left( {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right){\left( {\delta x} \right)^2}\\ \delta s = \sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} {\kern 1pt} \delta x \end{array}$

Now by the definition of the integral, we know that the sum of these infinitesimals is given by:

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

Similarly you can deduce a proof involving the integration with respect to y. Also, with little ingenuity you can extend this to the parametric case. For the polar one, consider the following argument:

As already stated:

$x = r\cos \theta $  and   $y = r\sin \theta $

Knowing that r is a function of $\theta$ we have that:

$\frac{{dx}}{{d\theta }} = \frac{{dr}}{{d\theta }}\cos \theta  - r\sin \theta $  $\frac{{dy}}{{d\theta }} = \frac{{dr}}{{d\theta }}\sin \theta  + r\cos \theta $

${\left( {\frac{{dx}}{{d\theta }}} \right)^2} + {\left( {\frac{{dy}}{{d\theta }}} \right)^2} = {\left( {\frac{{dr}}{{d\theta }}} \right)^2}{\cos ^2}\theta  - 2r\frac{{dr}}{{d\theta }}\sin \theta \cos \theta  + {r^2}{\sin ^2}\theta  + {\left( {\frac{{dr}}{{d\theta }}} \right)^2}{\sin ^2}\theta  + 2r\frac{{dr}}{{d\theta }}\sin \theta \cos \theta  + {r^2}{\cos ^2}\theta $

It is not difficult to see:

${\left( {\frac{{dx}}{{d\theta }}} \right)^2} + {\left( {\frac{{dy}}{{d\theta }}} \right)^2} = {\left( {\frac{{dr}}{{d\theta }}} \right)^2} + {r^2}$

Now we may substitute this into the result proven for the parametric case ( I have not, but it is not difficult to extend it from the cartesian one )

Area of Surface

For the proof we divide the curve into fulcrums ( fulcrums in the above diagram are of different colours ). The surface area of a fulcrum is given by:

\[A = 2\pi ({r_1} + {r_2})l\]

We know that ${r_1}$ and ${r_2}$ are the radii on the either ends of the fulcrum, and l is the slanted height. For the curve these radii are the value of the function defining the curve.

Now for an infinitesimal fulcrum in the entire surface, both end should be to a great approximation equal. Therefore:

${r_1} \approx y;\quad {r_2} \approx y$

We already know from the previous proof that the slanted height:

$l \approx \sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \,dx$

The surface area of the infinitesimal, approximately, is then given  by:

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

The sum of these infinitesimal can then be considered:

$S \approx \sum\limits_{i = 1}^n {2\pi {y_i}\sqrt {1 + {{\left( {\frac{{d{y_i}}}{{d{x_i}}}} \right)}^2}} {\kern 1pt} dx} $

$\begin{array}{l} \mathop {\lim }\limits_{n \to \infty } \left( S \right) = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {2\pi {y_i}\sqrt {1 + {{\left( {\frac{{d{y_i}}}{{d{x_i}}}} \right)}^2}} {\kern 1pt} dx} \\ = \int_a^b {2\pi y\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} {\kern 1pt} dx} \end{array}$

Complex Numbers

i) $\overline {\left( {{z_1} + {z_2}} \right)}  = \overline {{z_1}}  + \overline {{z_2}} $

$\begin{array}{l} Let\quad {z_1} = {x_1} + \iota {y_1};\,{z_2} = {x_2} + \iota {y_2}\\ \overline {\left( {{z_1} + {z_2}} \right)} = \overline {\left( {{x_1} + \iota {y_1} + {x_2} + \iota {y_2}} \right)} \\ = \overline {\left( {{x_1} + {x_2}} \right) + \iota \left( {{y_1} + {y_2}} \right)} \\ = \left( {{x_1} + {x_2}} \right) - \iota \left( {{y_1} + {y_2}} \right)\\ = \left( {{x_1} - \iota {y_1}} \right) + \left( {{x_2} - i{y_2}} \right)\\ = \overline {{z_1}} + \overline {{z_2}} \end{array}$

iii) $\overline {\left( {\frac{{{z_1}}}{{{z_2}}}} \right)}  = \frac{{\overline {{z_1}} }}{{\overline {{z_2}} }}$

$\begin{array}{l} \frac{{{z_1}}}{{{z_2}}} = \frac{{{x_1} + \iota {y_1}}}{{{x_2} + \iota {y_2}}} \times \frac{{{x_2} - \iota {y_2}}}{{{x_2} - \iota {y_2}}}\\ = \frac{{\left( {{x_1} + \iota {y_1}} \right)\left( {{x_2} - \iota {y_2}} \right)}}{{\left( {{x_2} + \iota {y_2}} \right)\left( {{x_2} - \iota {y_2}} \right)}}\\ = \frac{{\left( {{x_1}{x_2} + {y_1}{y_2}} \right) + \iota \left( {{x_2}{y_1} - {x_1}{y_2}} \right)}}{{{{\left( {{x_2}} \right)}^2} + {{\left( {{y_2}} \right)}^2}}}\\ \therefore \;\overline {\,\left( {\frac{{{z_1}}}{{{z_2}}}} \right)} = \frac{{\left( {{x_1}{x_2} + {y_1}{y_2}} \right) - \iota \left( {{x_2}{y_1} - {x_1}{y_2}} \right)}}{{{{\left( {{x_2}} \right)}^2} - {{\left( {\iota {y_2}} \right)}^2}}}\\ = \frac{{\left( {{x_1} - \iota {y_1}} \right)\left( {{x_2} + \iota {y_2}} \right)}}{{\left( {{x_2} - \iota {y_2}} \right)\left( {{x_2} + \iota {y_2}} \right)}} = \frac{{\left( {{x_1} - \iota {y_1}} \right)}}{{\left( {{x_2} - \iota {y_2}} \right)}} = \frac{{\overline {{z_1}} }}{{\overline {{z_2}} }} \end{array}$

iv) $z\bar z = \left( {x + \iota y} \right)\left( {x - \iota y} \right) = {x^2} - {\iota ^2}{y^2} - x\iota y + x\iota y = {x^2} + {y^2} = {\left| z \right|^2}$

v) $\left| {{z_1}{z_2}} \right| = \left| {{z_1}} \right|\left| {{z_2}} \right|$

$\begin{array}{l} \left| {{z_1}{z_2}} \right| = \left| {\left( {{x_1} + \iota {y_1}} \right)\left( {{x_2} + \iota {y_2}} \right)} \right|\\ = \left| {\left( {{x_1}{x_2} - {y_1}{y_2}} \right) + \iota \left( {{x_1}{y_2} + {x_2}{y_1}} \right)} \right|\\ = \sqrt {{{\left( {{x_1}{x_2} - {y_1}{y_2}} \right)}^2} + {{\left( {{x_1}{y_2} + {x_2}{y_1}} \right)}^2}} \\ = \sqrt {{x_1}^2{x_2}^2 + {y_1}^2{y_2}^2 + {x_1}^2{y_2}^2 + {x_2}^2{y_1}^2} \\ = \sqrt {\left( {{x_1}^2 + {y_1}^2} \right)\left( {{x_2}^2 + {y_2}^2} \right)} \\ = \sqrt {\left( {{x_1}^2 + {y_1}^2} \right)} \sqrt {\left( {{x_2}^2 + {y_2}^2} \right)} \\ = \left| {{z_1}} \right|\left| {{z_2}} \right| \end{array}$

vi) $\left| {{z_1}} \right| - \left| {{z_2}} \right| \le \left| {{z_1} + {z_2}} \right| \le \left| {{z_1}} \right| + \left| {{z_2}} \right|$

$\begin{array}{l} {\left| {{z_1} + {z_2}} \right|^2} = {\left| {\left( {{x_1} + \iota {y_1}} \right) + \left( {{x_2} + \iota {y_2}} \right)} \right|^2}\\ = {\left( {{x_1} + {x_2}} \right)^2} + {\left( {{y_1} + {y_2}} \right)^2}\\ = \left( {{x_1}^2 + {x_2}^2} \right) + \left( {{y_1}^2 + {y_2}^2} \right) + 2\left( {{x_1}{x_2} + {y_1}{y_2}} \right)\\ = \left( {{x_1}^2 + {x_2}^2} \right) + \left( {{y_1}^2 + {y_2}^2} \right) + 2\sqrt {{{\left( {{x_1}{x_2} + {y_1}{y_2}} \right)}^2}} \\ \le \left( {{x_1}^2 + {x_2}^2} \right) + \left( {{y_1}^2 + {y_2}^2} \right) + 2\sqrt {{{\left( {{x_1}{x_2} + {y_1}{y_2}} \right)}^2} + {{\left( {{x_1}{y_2} - {x_2}{y_1}} \right)}^2}} \\ = \left( {{x_1}^2 + {y_2}^2} \right) + \left( {{x_1}^2 + {y_2}^2} \right) + 2\sqrt {{x_1}^2{x_2}^2 + {y_1}^2{y_2}^2 + {x_1}^2{y_2}^2 + {x_2}^2{y_1}^2} \\ = \left( {{x_1}^2 + {y_2}^2} \right) + \left( {{x_1}^2 + {y_2}^2} \right) + 2\sqrt {\left( {{x_1}^2 + {y_1}^2} \right)\left( {{x_2}^2 + {y_2}^2} \right)} \\ = \left( {\sqrt {{x_1}^2 + {y_2}^2} + \sqrt {{x_1}^2 + {y_2}^2} } \right)\\ = {\left( {\left| {{z_1}} \right| + \left| {{z_2}} \right|} \right)^2}\\ \therefore \left| {{z_1} + {z_2}} \right| \le \left| {{z_1}} \right| + \left| {{z_2}} \right|\\ \left| {{z_1}} \right| = \left| {\left( {{z_1} - {z_2}} \right) + {z_2}} \right| \Rightarrow \left| {{z_1} - {z_2}} \right| + \left| {{z_2}} \right|\\ \left| {{z_1}} \right| - \left| {{z_2}} \right| \le \left| {{z_1} - {z_2}} \right|\\ Taking\, - {z_2}\,in\,place\,of\,{z_2}\\ \left| {{z_1}} \right| - \left| {{z_2}} \right| \le \left| {{z_1} + {z_2}} \right| \end{array}$