# How to find a neat version of y when you have the equation of an ellipse

So, you have the equation of the ellipse but you need to completely isolate y. How would you go about doing this? Well, here is a fantastic example…

${ \left( \frac { x }{ a } \right) }^{ 2 }+{ \left( \frac { y }{ b } \right) }^{ 2 }=1\\ \\ \therefore \quad { \left( \frac { y }{ b } \right) }^{ 2 }=1-{ \left( \frac { x }{ a } \right) }^{ 2 }\\ \\ \therefore \quad \frac { { y }^{ 2 } }{ { b }^{ 2 } } =1-\frac { { x }^{ 2 } }{ { a }^{ 2 } } \\ \\ \therefore \quad { y }^{ 2 }={ b }^{ 2 }-\frac { { { b }^{ 2 }x }^{ 2 } }{ { a }^{ 2 } } \\ \\ \therefore \quad { y }^{ 2 }=\frac { { a }^{ 2 }{ b }^{ 2 } }{ { a }^{ 2 } } -\frac { { b }^{ 2 }{ x }^{ 2 } }{ { a }^{ 2 } } \\ \\ \therefore \quad { y }^{ 2 }=\frac { { a }^{ 2 }{ b }^{ 2 }-{ b }^{ 2 }{ x }^{ 2 } }{ { a }^{ 2 } } \\ \\ \therefore \quad { y }^{ 2 }=\frac { { b }^{ 2 }\left( { a }^{ 2 }-{ x }^{ 2 } \right) }{ { a }^{ 2 } } \\ \\ \therefore \quad { y }^{ 2 }=\frac { { b }^{ 2 } }{ { a }^{ 2 } } \cdot \left( { a }^{ 2 }-{ x }^{ 2 } \right) \\ \\ \therefore \quad y=\sqrt { \frac { { b }^{ 2 } }{ { a }^{ 2 } } } \cdot \sqrt { { a }^{ 2 }-{ x }^{ 2 } } \\ \\ \therefore \quad y=\frac { b }{ a } \cdot \sqrt { { a }^{ 2 }-{ x }^{ 2 } }$

This will come in handy if you’re trying to derive the area of an ellipse from absolute scratch.

# Finding the formulas for areas of triangles

In this post I’ll be demonstrating how one can derive the three formulas which can be used to find the areas of triangles.

These formulas are in fact:

$A=\frac { 1 }{ 2 } bc\cdot \sin { \left( A \right) } =\frac { 1 }{ 2 } ac\cdot \sin { \left( B \right) =\frac { 1 }{ 2 } } ab\cdot \sin { \left( C \right) }$

To begin with, let’s start by looking at the diagram below:

Now, if you look at the diagram carefully – you will notice that the area of the triangle is:

$A=\frac { x\cdot CN }{ 2 } +\frac { \left( c-x \right) \cdot CN }{ 2 }$

This can be simplified into:

$\frac { x\cdot CN }{ 2 } +\frac { \left( c-x \right) \cdot CN }{ 2 } \\ \\ =\frac { x\cdot CN+\left( c-x \right) \cdot CN }{ 2 } \\ \\ =\frac { CN\left\{ x+\left( c-x \right) \right\} }{ 2 } \\ \\ =\frac { CN\cdot c }{ 2 }$

Because of SOH CAH TOA, what we can also say is that:

$\sin { \left( A \right) } =\frac { O }{ H } =\frac { CN }{ b } \\ \\ \therefore \quad b\cdot \sin { \left( A \right) } =CN\\ \\ \sin { \left( B \right) =\frac { O }{ H } } =\frac { CN }{ a } \\ \\ \therefore \quad a\cdot \sin { \left( B \right) } =CN$

Now because:

$A=\frac { CN\cdot c }{ 2 }$

This ultimately means that:

$A=\frac { 1 }{ 2 } bc\cdot \sin { \left( A \right) } \\ \\ A=\frac { 1 }{ 2 } ac\cdot \sin { \left( B \right) } \\ \\ \therefore \quad A=\frac { 1 }{ 2 } bc\cdot \sin { \left( A \right) =\frac { 1 }{ 2 } ac } \cdot \sin { \left( B \right) }$

Alright, so far so good… Now we must put the icing on the cake and attach the final piece of the jigsaw puzzle to the formula above. In order to find the three equations which can be used to find the areas of triangles, we must now discover the expression for sin(C). We can discover its expression by first saying that:

$C=\left( 90-A \right) +\left( 90-B \right) \\ \\ =90-A+90-B\\ \\ =180-A-B\\ \\ =180-\left( A+B \right) \\ \\ \therefore \quad \sin { \left( C \right) } =\sin { \left( 180-\left( A+B \right) \right) }$

And if we use the trigonometric identity below:

$\sin { \left( \alpha -\beta \right) } =\sin { \left( \alpha \right) \cdot \cos { { \left( \beta \right) } } -\cos { \left( \alpha \right) \cdot \sin { \left( \beta \right) } } }$

We will reach the conclusion:

$\sin { \left( 180-\left( A+B \right) \right) } =\sin { \left( 180 \right) \cdot \cos { \left( A+B \right) -\cos { \left( 180 \right) \cdot \sin { \left( A+B \right) } } } }Â$

But because:

$\sin { \left( 180 \right) =0 } ,\quad \cos { \left( 180 \right) =-1 } \\ \\ \sin { \left( 180-\left( A+B \right) \right) =-\left( -1 \right) \cdot \sin { \left( A+B \right) } } \\ \\ \therefore \quad \sin { \left( C \right) =\sin { \left( A+B \right) } }$

Now, sin(A+B) as a trigonometric identity, is:

$\sin { \left( A+B \right) =\sin { \left( A \right) \cdot \cos { \left( B \right) +\cos { \left( A \right) \cdot \sin { \left( B \right) } } } } }$

And, thanks to SOH CAH TOA…

$\sin { \left( A+B \right) =\sin { \left( C \right) } } \\ \\ \sin { \left( A \right) =\frac { CN }{ b } } \\ \\ \cos { \left( B \right) =\frac { A }{ H } } =\frac { \left( c-x \right) }{ a } \\ \\ \cos { \left( A \right) =\frac { A }{ H } =\frac { x }{ b } } \\ \\ \sin { \left( B \right) =\frac { CN }{ a } }$

Which means that…

$\sin { \left( C \right) =\frac { CN }{ b } \cdot \frac { \left( c-x \right) }{ a } +\frac { x }{ b } \cdot \frac { CN }{ a } } \\ \\ =\frac { CN\left( c-x \right) }{ ab } +\frac { CN\cdot x }{ ab } \\ \\ =\frac { CN\left( c-x \right) +CN\cdot x }{ ab } \\ \\ =\frac { CN\left\{ \left( c-x \right) +x \right\} }{ ab } \\ \\ =\frac { CN\cdot c }{ ab } \\ \\ \therefore \quad ab\cdot \sin { \left( C \right) =CN\cdot c } \\ \\ \therefore \quad \frac { 1 }{ 2 } ab\cdot \sin { \left( C \right) =\frac { CN\cdot c }{ 2 } =A }$

As this is the case, we can conclude that:

$A=\frac { 1 }{ 2 } bc\cdot \sin { \left( A \right) } =\frac { 1 }{ 2 } ac\cdot \sin { \left( B \right) =\frac { 1 }{ 2 } } ab\cdot \sin { \left( C \right) }$

# How to derive the formula for the area of an equilateral triangle

In this post I’ll be showing you how to derive the formula for the area of an equilateral triangle –Â in easy steps. In order to understand this derivation properly, you need to be familiar with Pythagoras’ theorem and also a few algebraic rules. What you’ll also need is a ruler, pair of compasses, a pencil and a sheet of paper.

Step 1: Put a point on a blank sheet of paper and name it A.

Step 2: Put the needle of your compass on the point A and draw a circle around it.

Step 3: Add a point B to this circle, on its edge.

Step 4: Put the needle of your compass on the point B and your pencil on the point A.

Step 5: Draw another circle with a radius the length AB.

Step 6: Now add a few extra points to your drawing. Call these points C and D.

Step 7: Connect the points A, B and C forming a triangle.

Step 8: Draw a line going through the points C and D.

Step 9: Where the line going through C and D intersects the triangle, place the point E.

Step 10: Now look at your latest work very carefully… What you will notice is that the lengths AB, AC and BC are all equal to one another. This is because both the circles you drew – are exactly the same size. They each have radiuses equal in proportion. In simple terms, AB=AC=BC.

What you have to do now is name these lengths (r) for radius. Here’s the thing though, because the line going through C and D splits the triangle (equilateral, as each of its sides has the same length) down its middle, the length AE is equal to 1/2 x r, and similarly the length BE is equal to 1/2 x r. Together, the length AE + BE = AB = r.

Step 11: Remember that I said that the line going through C and D splits the triangle down its middle. Also, notice that this exact line is perpendicular to the length AB. Now, because of this, at the point E, you’ve got two right angles. Name these two right angles big R.

[Knowing that these two angles are equal to 90 degrees is vital – because you’ll be able to use Pythagoras’ theorem to find the length CE.]

Step 12: Find the length CE using Pythagoras’ theorem, AdjacentÂ² + OppositeÂ² = HypotenuseÂ². You will need this length to find the area of the equilateral triangle you’ve produced.

*Algebraic skills will be required from this point…

${ AE }^{ 2 }+{ CE }^{ 2 }={ AC }^{ 2 }\\ \\ \Rightarrow \quad { \left( \frac { 1 }{ 2 } r \right) Â }^{ 2 }+{ CE }^{ 2 }={ r }^{ 2 }\\ \\ \Rightarrow \quad { CE }^{ 2 }={ r }^{ 2 }-{ \left( \frac { 1 }{ 2 } r \right) Â }^{ 2 }\\ \\ \Rightarrow \quad { CE }^{ 2 }=\frac { 4r^{ 2 } }{ 4 } -\frac { { r }^{ 2 } }{ 4 } \\ \\ \Rightarrow \quad { CE }^{ 2 }=\frac { 3{ r }^{ 2 } }{ 4 } \\ \\ \Rightarrow \quad CE=\sqrt { \frac { 3{ r }^{ 2 } }{ 4 } Â } \\ \\ \therefore \quad CE=\frac { r\sqrt { 3 } Â }{ 2 }Â$

Step 13: Derive the formula for the area (A) of the equilateral triangle. Remember that the area of a right angled triangle is L x W x 1/2.

$A=\frac { 1 }{ 2 } r\cdot \frac { r\sqrt { 3 } Â }{ 2 } \cdot \frac { 1 }{ 2 } +\frac { 1 }{ 2 } r\cdot \frac { r\sqrt { 3 } Â }{ 2 } \cdot \frac { 1 }{ 2 } \\ \\ =\frac { 1 }{ 8 } { r }^{ 2 }\sqrt { 3 } +\frac { 1 }{ 8 } { r }^{ 2 }\sqrt { 3 } \\ \\ =2\cdot \frac { 1 }{ 8 } { r }^{ 2 }\sqrt { 3 } \\ \\ =\frac { 1 }{ 4 } { r }^{ 2 }\sqrt { 3 }Â$

Presto!!! Keep in mind that you can transform the variable (r) into any variable you wish. This variable (r) is the length of each side of the equilateral triangle you were working with. The formula you’ve derived can be used to find the area of any equilateral triangle.

# How to quickly double the area of a square (simple geometry lesson)

In this post, I’ll be demonstrating how you can quickly double the area of a square using a simple geometrical trick.

Let’s say you have an ordinary square, like the one below…

Firstly, what you have to do is name the area of this square “A”…

Then, what you do next is divide this square (diagonally) into 4 equal parts…

After you have done this, you then name each part of this square “1/4 x A”…

Notice now, that to double the area of this square, all you have to do, is double the number of the 1/4 x A right angled triangles which currently exist – then configure them – like this…

As you can see, you’ve now got eight of these 1/4 x A right angled triangles neatly configured…

Not only are you left with a new square, double the size of your original square (follow the lines on the outside of the shape), but a handy equation, which proves that you doubled the area of the square you started off with…

$8\times \frac { 1 }{ 4 } A=2A$

# Derive the formula to find areas underneath curves

In this post I’ll be revealing how you can derive the formula which can be used to find areas underneath curves, from absolute scratch. Now, just below, what you will find is the diagram that will help us produce this formula…

In this diagram what you will discover is that:

• A length a exists, which starts at the origin O and ends at a;
• A lengthÂ xÂ exists, which starts at the origin O and ends at x;
• A lengthÂ x+?x exists, which starts at the origin O and ends at x+?x;
• A lengthÂ ?x exists, which starts at x and ends at x+?x;
• A height y exists, which starts at the origin O and ends at y;
• A height y+?y exists, which starts at the origin O and ends at y+?y;
• A heightÂ ?y exists, which starts at y and ends at y+?y;
• There is a curve called y=f(x);
• There is an area underneath the curve called A which commences at a and ends at x;
• There is an area underneath the curve calledÂ ?A whichÂ commences at x and ends at x+?xÂ (Note: If you extend the distance from a to x what you get is a larger area, and the change in area can be measured. This change or difference is called ?A);
• There is a rectangle that exists called QRUT. It has an area which is y?x;
• There is a rectangle that exists called PRUS. It has an area which is (y+?y)?x;
• ?A has an area larger than that of the rectangle QRUT, but smaller than that of the rectangle PRUS.

Producing the formula with the information we’ve discovered…

Ok, so we want to produce the formula which will help us find areas underneath curves from absolute scratch. At our disposal we have a helpful diagram (which we’ve looked at and analysed carefully) and we’ve been able to discover a few facts about it. I think we can now get to work…

Let’s start off by saying that:

Area QRUT <Â ?A < Area PRUS

Which is something we already discovered.

If this is the case, we can say that:

y?x <Â ?A <Â (y+?y)?x

Now, check out what happens when we divide all the elements of this expression byÂ ?x:

What we end up with is…

Alright, now you may be saying to yourself, why do I need to know this? Well, it turns out that:

This is because asÂ ?x approaches 0, Â ?y approaches 0 leavingÂ (?A)/(?x) sandwiched between y and y+0.000000000000000001 which is virtually y.

And, also…

As a consequence, this ultimately means that:

$y=\frac { dA }{ dx }$

This is incredibly significant, because if we then integrate both sides of this equation, we get:

$\int { ydx=\int { \frac { dA }{ dx } } } dx\quad \Rightarrow \quad A=\int { ydx }$

And…

$A=\int { ydx } =F\left( x \right) +C$

Now, this equation can actually be used to find the area A underneath the curve from a to x. What we’re basically saying is that this area is equal to some function of x plus a constant. This ‘some function of x’ occurs when we integrate y which is a function of x.

Finalising the formula…

Alright so we’ve managed to latch on to something incredibly significant… We’ve got an important equation:

$A=\int { ydx } =F\left( x \right) +C$

However, it is not complete. We need to know what the constant C is. So…

If we say that at x=a the area A underneath the curveÂ is 0, watch what happens… Look at what we get…

$O=F\left( a \right) +C$

Which means that:

$C=-F\left( a \right)Â$

Hence, we can conclude that:

$A=\int { ydx=F\left( x \right) Â } -F\left( a \right)Â$

And this formula can be transformed into something more fancy if we are measuring an area underneath a curve from x=a to x=b

This is probably the formula you’re most familiar with…

$A=\int _{ a }^{ b }{ ydx=F\left( b \right) Â } -F\left( a \right) ={ \left[ F\left( x \right) Â \right] Â }_{ a }^{ b }$

Which is the formula which can be used to find areas underneath curves.

If you are still confused and would like to go through this proof once again, please watch my video below…

Related:

Trapezium Rule Formula – Derivation

# Latest Mathematics Proofs: August 2016

Recently I discovered a few more proofs, some related to A Level mathematics. Â You can access these proofs by clicking on the links below.

Derive the formula of an ellipse:Â https://plus.google.com/+mathsvideosforweb/posts/ZF8D3ghSNRD

Discover the distance between point A and B on the edge of a circle:Â https://plus.google.com/+mathsvideosforweb/posts/9fmuEJ4qeXY

Find the area of a sector within a circle:Â https://plus.google.com/+mathsvideosforweb/posts/cHt9t9PeSfg

Formula for a torus, derived from scratch:Â https://plus.google.com/+mathsvideosforweb/posts/TS84TL44BYd

Create the mathematical singularity shown in movie documentaries:Â https://plus.google.com/+mathsvideosforweb/posts/7p2onSzNYte

Parameterised formula for a torus derived from scratch:Â https://plus.google.com/+mathsvideosforweb/posts/AZrVMiPVdbv

Derive the formula for an ellipsoid (normal and parameterised) from scratch:Â https://plus.google.com/+mathsvideosforweb/posts/Wijf94ztn3v

Other news:

If you love mathematical proofs, please feel free to join my ‘mathematics proofs’ Google community at:Â https://plus.google.com/communities/106007058741903558109

I’ve also got a new Facebook page related to stuff about the universe and also mathematics. You can add me as a friend by accessing this link: http://www.facebook.com/tiago.hands

# 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.

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

# Sine Rule Derivation

Use the formulas you’d use to calculate the area of a triangle. See the magic emerge.

$\frac { 1 }{ 2 } bcsinA=\frac { 1 }{ 2 } acsinB\\ \\ bcsinA=acsinB\\ \\ bsinA=asinB\\ \\ \frac { bsinA }{ b } =\frac { asinB }{ b } \\ \\ \frac { sinA }{ 1 } =\frac { asinB }{ b } \\ \\ \frac { sinA }{ 1 } \cdot \frac { 1 }{ a } =\frac { asinB }{ b } \cdot \frac { 1 }{ a } \\ \\ \frac { sinA }{ a } =\frac { sinB }{ b } \\ \\ \\ OR:\\ \\ bsinA=asinB\\ \\ \frac { bsinA }{ sinB } =\frac { asinB }{ sinB } \\ \\ \frac { bsinA }{ sinB } =\frac { a }{ 1 } \\ \\ \frac { bsinA }{ sinB } \cdot \frac { 1 }{ sinA } =\frac { a }{ 1 } \cdot \frac { 1 }{ sinA } \\ \\ \frac { b }{ sinB } =\frac { a }{ sinA }$

# (a+b)(a-b)=a^2-b^2 (The Real Proof) (Difference Of Two Squares Demystified)

Ever wondered why (a+b)(a-b)=a^2-b^2? The videos below will reveal to you why we accept this fact. Enjoy!

1) Why (a+b)(a+b)=a^2+2ab+b^2

2) Why (a+b)(a-b)=a^2-b^2

# Trapezium Rule Formula – Derivation

Find out how to come up with the Trapezium Rule formula from scratch.

1) Derive the formula for the area of trapeziums:

2) Use the area of trapeziums formula to come up with the Trapezium Rule formula:

Related:

Derive the formula to find areas underneath curves