# Derivative – 27/02/15

$n\ln { x } =y\\ \\ \ln { \left( { x }^{ n } \right) } =y\\ \\ \log _{ e }{ \left( { x }^{ n } \right) } =y\\ \\ { e }^{ y }={ x }^{ n }\\ \\ { e }^{ y }\frac { dy }{ dx } =n{ x }^{ n-1 }\\ \\ \frac { 1 }{ { e }^{ y } } \cdot { e }^{ y }\frac { dy }{ dx } =n{ x }^{ n-1 }\cdot \frac { 1 }{ { e }^{ y } } \\ \\ \frac { dy }{ dx } =n{ x }^{ n-1 }\cdot { x }^{ -n }\\ \\ \frac { dy }{ dx } =n{ x }^{ n-1+\left( -n \right) }\\ \\ \frac { dy }{ dx } =n{ x }^{ -1 }\\ \\ \frac { dy }{ dx } =\frac { n }{ x }$

# Anti-Derivative Proof – 27/02/15

$y=\frac { 1 }{ \left( n+1 \right) } { x }^{ \left( n+1 \right) }\\ \\ \ln { y } =\ln { \left( \frac { { x }^{ \left( n+1 \right) } }{ \left( n+1 \right) } \right) } \\ \\ \ln { y } =\ln { \left( { x }^{ \left( n+1 \right) } \right) } -\ln { \left( \left( n+1 \right) \right) } \\ \\ \ln { y } =\ln { \left( { x }^{ \left( n+1 \right) } \right) } +{ C }_{ 2 }\\ \\ \therefore \quad { C }_{ 2 }=-\ln { \left( \left( n+1 \right) \right) } \\ \\ \ln { y } =\left( n+1 \right) \cdot \ln { x } +{ C }_{ 2 }\\ \\ \frac { 1 }{ y } \cdot \frac { dy }{ dx } =\frac { \left( n+1 \right) }{ x } \\ \\ y\cdot \frac { 1 }{ y } \cdot \frac { dy }{ dx } =\frac { \left( n+1 \right) }{ x } \cdot y\\ \\ \frac { dy }{ dx } =\left( n+1 \right) \cdot { x }^{ -1 }\cdot \frac { 1 }{ \left( n+1 \right) } { x }^{ \left( n+1 \right) }\\ \\ \frac { dy }{ dx } ={ x }^{ -1 }\cdot { x }^{ \left( n+1 \right) }\\ \\ \frac { dy }{ dx } ={ x }^{ -1+\left( n+1 \right) }\\ \\ \frac { dy }{ dx } ={ x }^{ n }\\ \\ \therefore \quad \int { { x }^{ n } } dx=\frac { 1 }{ \left( n+1 \right) } \cdot { x }^{ n+1 }+C$

# Basic Logic Tables

Definitions:

$\vee \quad Or\\ \\ \wedge \quad And\\ \\ \neg \quad Not\\ \\ \forall \quad For\quad All\\ \\ \exists \quad There\quad exists\\ \\ \Rightarrow \quad Implies$

LOGICAL POSSIBILITIES:

 TRUE (T) FALSE (F) TRUE (T) T, T T, F FALSE (F) F, T F, F

“OR” TRUTH TABLE:

 $a$ $b$ $a\vee b$ T T T T F T F T T F F F

“AND” TRUTH TABLE

 $a$ $b$ $a\wedge b$ T T T T F F F T F F F F

“NOT” TRUTH TABLE

 $a$ $\neg a$ T F F T

“IMPLIES” TRUTH TABLE

 $a$ $b$ $a\Rightarrow b$ T T T T F F F T T F F T

# More Mathematical Logic

Axioms of equality are basic rules for using the equals sign…

If a=b and c=d, then a+c=b+d.

If a>b, then a+c>b+c.

Any number added to 0 gives the original number, for instance, n+0=n.

Identity Element For Multiplication:

Any number multiplied by 1 gives the original number, for instance, n*1=n.

Multiplicative Axiom Of Equality:

If a=b and c=d, then ac=bd.

Negative Multiplication Property Of Inequality:

You must reverse the inequality sign when multiplying or dividing by a negative number.

# Messing Around With Exponentials

${ \sqrt { 2 } }^{ { \sqrt { 2 } }^{ { \sqrt { 2 } }^{ \sqrt { 2 } } } }\\ \\ ={ \sqrt { 2 } }^{ { \sqrt { 2 } }^{ { \left( { 2 }^{ \frac { 1 }{ 2 } } \right) }^{ \sqrt { 2 } } } }\\ \\ ={ \sqrt { 2 } }^{ { \sqrt { 2 } }^{ { 2 }^{ \frac { 1 }{ 2 } \sqrt { 2 } } } }\\ \\ ={ \sqrt { 2 } }^{ { \sqrt { 2 } }^{ { 2 }^{ \frac { \sqrt { 2 } }{ 2 } } } }$

$\\ \\ ={ \sqrt { 2 } }^{ { \sqrt { 2 } }^{ \sqrt { { 2 }^{ \sqrt { 2 } } } } }\\ \\ ={ \sqrt { 2 } }^{ { \left( { 2 }^{ \frac { 1 }{ 2 } } \right) }^{ \sqrt { { 2 }^{ \sqrt { 2 } } } } }\\ \\ ={ \sqrt { 2 } }^{ { 2 }^{ \frac { 1 }{ 2 } \sqrt { { 2 }^{ \sqrt { 2 } } } } }$

$\\ \\ ={ \sqrt { 2 } }^{ { 2 }^{ \frac { \sqrt { { 2 }^{ \sqrt { 2 } } } }{ 2 } } }\\ \\ ={ \sqrt { 2 } }^{ \sqrt { { 2 }^{ \sqrt { { 2 }^{ \sqrt { 2 } } } } } }\\ \\ ={ \left( { 2 }^{ \frac { 1 }{ 2 } } \right) }^{ \sqrt { { 2 }^{ \sqrt { { 2 }^{ \sqrt { 2 } } } } } }$

$\\ \\ ={ 2 }^{ \frac { 1 }{ 2 } \sqrt { { 2 }^{ \sqrt { { 2 }^{ \sqrt { 2 } } } } } }\\ \\ ={ 2 }^{ \frac { \sqrt { { 2 }^{ \sqrt { { 2 }^{ \sqrt { 2 } } } } } }{ 2 } }\\ \\ =\sqrt { { 2 }^{ \sqrt { { 2 }^{ \sqrt { { 2 }^{ \sqrt { 2 } } } } } } }$

# Measuring the time it takes for the Sun’s light to reach us…

The Sun is a luminous object; by definition:

“A luminous object is one that gives off light. In other words, it glows of its own accord. To be able to glow, the object must have its own source of energy. A torch shines because of the energy stored in its batteries, whereas all stars shine using energy created by nuclear fusion.” – Source: http://www.schoolsobservatory.org.uk/astro/stars/luminous

To find out how long it would take for the Sun’s light to reach us, we must use a few mathematical tools.

We must know that:

$speed=\frac { distance }{ time }$

We then manipulate this formula algebraically to get:

$speed\cdot time=distance\\ \\ \therefore \quad time=\frac { distance }{ speed }$

Now it is said that the Earth’s distance from the Sun is 149,600,000 km (Source: Google) and that the speed of light in a vacuum is 299,792 km per second (Source: Google). As space is mostly empty we’ll be using the speed of light in a vacuum to measure how long it would take for the Sun’s light to reach us.

So knowing that:

Distance = 149,600,000 km

Speed = 299,792 km / s

Let’s plug these values into the time formula in order to know how long it would take for the Sun’s light to reach us:

$time=\frac { distance }{ speed } \\ \\ =\frac { 149,600,000\quad km }{ 299,792\quad km/s } \\ \\ =\frac { 149,600,000 }{ 299,792 } \cdot \frac { km }{ \frac { km }{ s } } \\ \\ =\frac { 149,600,000 }{ 299,792 } \cdot \frac { km }{ 1 } \cdot \frac { s }{ km } \\ \\ =\frac { 149,600,000 }{ 299,792 } \cdot s\cdot \frac { km }{ km } \\ \\ =\frac { 149,600,000 }{ 299,792 } s\cdot 1\\ \\ =\frac { 149,600,000 }{ 299,792 } s\\ \\ \approx 499.0126488\quad seconds$

Now what is 499.0126488 seconds in minutes? Let’s find out:

$1\quad minute\quad =\quad 60\quad seconds\\ \\ \therefore \quad \frac { 1 }{ 60 } \quad minute\quad =\quad 1\quad second$

Knowing this we can say that:

$\frac { 149,600,000 }{ 299,792 } \cdot \frac { 1 }{ 60 } \quad minute\quad =\quad \frac { 149,600,000 }{ 299,792 } \quad second\\ \\ 8.316877479\quad minutes\quad =\quad 499.0126488\quad seconds$

So it takes approximately 8.32 (to 2 decimal places) minutes for the Sun’s light to reach us. We are in fact seeing the Sun how it was approximately 8 minutes ago. When we look into space or at any distant object for that matter, we are actually seeing the past.

Amazing!

# Integration Problem – Finding the area of quarter of a circle by integrating…

Integrate:

$\int _{ 0 }^{ r }{ \sqrt { { r }^{ 2 }-{ x }^{ 2 } } } dx$

Say that:

$x=r\sin { \theta }$

So…

$x=r\sin { \theta } \\ \\ x=ru\\ \\ \frac { dx }{ du } =r\\ \\ u=\sin { \theta } \\ \\ \frac { du }{ d\theta } =\cos { \theta } \\ \\ \therefore \quad \frac { dx }{ d\theta } =r\cos { \theta } \\ \\ \therefore \quad dx=r\cos { \theta } d\theta$

Now…

$When\quad x=r,\\ \\ r\sin { \theta =r } \\ \\ \sin { \theta } =1\\ \\ \therefore \quad \theta =\frac { \pi }{ 2 } \\ \\ As\quad \left\{ 0\le \theta \le \frac { \pi }{ 2 } \right\} \\ \\ When\quad x=0,\\ \\ r\sin { \theta } =0\\ \\ \sin { \theta } =0\\ \\ \therefore \quad \theta =0$

So you now have to integrate:

$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt { { r }^{ 2 }-{ { r }^{ 2 }\sin ^{ 2 }{ \theta } } } } \cdot r\cos { \theta } d\theta \\ \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt { { r }^{ 2 }\left( 1-\sin ^{ 2 }{ \theta } \right) } } \cdot r\cos { \theta } d\theta \\ \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt { { r }^{ 2 }\cos ^{ 2 }{ \theta } } } \cdot r\cos { \theta } d\theta \\ \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \left( { r }^{ 2 } \right) }^{ \frac { 1 }{ 2 } } } { \left( \cos ^{ 2 }{ \theta } \right) }^{ \frac { 1 }{ 2 } }\cdot r\cos { \theta } d\theta \\ \\ =\int _{ o }^{ \frac { \pi }{ 2 } }{ r\cos { \theta } } \cdot r\cos { \theta } d\theta \\ \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { r }^{ 2 } } \cos ^{ 2 }{ \theta } d\theta$

But first, realise that:

$\cos { \left( \theta +\theta \right) } =\cos { \theta } \cos { \theta } -\sin { \theta } \sin { \theta } \\ \\ \cos { 2\theta } =\cos ^{ 2 }{ \theta } -\sin ^{ 2 }{ \theta } \\ \\ \cos { 2\theta } =\cos ^{ 2 }{ \theta } -\left( 1-\cos ^{ 2 }{ \theta } \right) \\ \\ \cos { 2\theta } =\cos ^{ 2 }{ \theta } -1+\cos ^{ 2 }{ \theta } \\ \\ \cos { 2\theta } =2\cos ^{ 2 }{ \theta } -1\\ \\ 2\cos ^{ 2 }{ \theta } =\cos { 2\theta } +1\\ \\ \cos ^{ 2 }{ \theta } =\frac { 1 }{ 2 } \cos { 2\theta } +\frac { 1 }{ 2 }$

So you now integrate:

$=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { r }^{ 2 } } \left( \frac { 1 }{ 2 } \cos { 2\theta } +\frac { 1 }{ 2 } \right) d\theta \\ \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { 1 }{ 2 } } { r }^{ 2 }\cos { 2\theta } +\frac { 1 }{ 2 } { r }^{ 2 }\quad d\theta$

And:

$If\quad p=\frac { 1 }{ 4 } { r }^{ 2 }\sin { 2\theta } =\frac { 1 }{ 4 } { r }^{ 2 }u\\ \\ \frac { dp }{ du } =\frac { 1 }{ 4 } { r }^{ 2 }\\ \\ u=\sin { 2\theta } =\sin { q } \\ \\ \frac { du }{ dq } =\cos { q } =\cos { 2\theta } \\ \\ q=2\theta ,\quad \frac { dq }{ d\theta } =2\\ \\ \frac { du }{ d\theta } =2\cos { 2\theta } \\ \\ \therefore \quad \frac { dp }{ d\theta } =\frac { 1 }{ 4 } { r }^{ 2 }\cdot 2\cos { 2\theta } \\ \\ =\frac { 1 }{ 2 } { r }^{ 2 }\cos { 2\theta }$

Therefore:

Now, to get the area of a circle, you multiply the final result by 4.

$4\cdot \frac { 1 }{ 4 } \pi { r }^{ 2 }\\ \\ =\pi { r }^{ 2 }\\ \\ \therefore \quad A=\pi { r }^{ 2 }$