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This post is inspired by the Quora thread on interesting functions to plot.

**The butterfly**

This is a slightly simpler version of the butterfly curve which is plotted using polar coordinates on Desmos as:

Polar coordinates are an alternative way of plotting functions – and are explored a little in HL Maths when looking at complex numbers. The theta value specifies an angle of rotation measured anti-clockwise from the x axis, and the r value specifies the distance from the origin. So for example the polar coordinates (90 degrees, 1) would specify a point 90 degrees ant clockwise from the x axis and a distance 1 from the origin (i.e the point (0,1) in our usual Cartesian plane).

2. **Fermat’s Spiral**

This is plotted by the polar equation:

The next 3 were all created by my students.

3. **Chaotic spiral (by Laura Y9)**

I like how this graph grows ever more tangled as it coils in on itself. This was created by the polar equation:

4. **The flower (by Felix Y9)**

Some nice rotational symmetries on this one. Plotted by:

5. **The heart (by Tiffany Y9)**

Simple but effective! This was plotted using the usual x,y coordinates:

You can also explore how to draw the Superman and Batman logos using Wolfram Alpha here.

**A geometric proof for the Arithmetic and Geometric Mean**

There is more than one way to define the mean of a number. The arithmetic mean is the mean we learn at secondary school – for 2 numbers a and b it is:

(a + b) /2.

The geometric mean on the other hand is defined as:

(x_{1}.x_{2}.x_{3}…x_{n})^{1/n}

So for example with the numbers 1,2,3 the geometric mean is (1 x 2 x 3)^{1/3}.

With 2 numbers, a and b, the geometric mean is (ab)^{1/2}.

We can then use the above diagram to prove that (a + b) /2 ≥ (ab)^{1/2} for all a and b. Indeed this inequality holds more generally and it can be proved that the Arithmetic mean ≥ Geometric mean.

Step (1) We draw a triangle as above, with the line MQ a diameter, and therefore angle MNQ a right angle (from the circle theorems). Let MP be the length a, and let PQ be the length b.

Step (2) We can find the length of the green line OR, because this is the radius of the circle. Given that the length a+b was the diameter, then (a+b) /2 is the radius.

Step (3) We then attempt to find an equation for the length of the purple line PN.

We find MN using Pythagoras: (MN)^{2} = a^{2} +x^{2}

We find NQ using Pythagoras: (NQ)^{2} = b^{2} +x^{2}

Therefore the length MQ can also be found by Pythagoras:

(MQ)^{2} = (MN)^{2 } + (NQ)^{2}

(MQ)^{2 } = a^{2} +x^{2} + b^{2} +x^{2}

But MQ = a + b. Therefore:

(a + b)^{2 } = a^{2} +x^{2} + b^{2} +x^{2}

a^{2}+ b^{2} + 2ab = a^{2} +x^{2} + b^{2} +x^{2}

2ab = x^{2} +x^{2}

ab = x^{2}

x = (ab)^{1/2}

Therefore our green line represents the arithmetic mean of 2 numbers (a+b) /2 and our purple line represents the geometric mean of 2 numbers (ab)^{1/2}. The green line will always be greater than the purple line (except when a = b which gives equality) therefore we have a geometrical proof of our inequality.

There is a more rigorous proof of the general case using induction you may wish to explore as well.

This is an example of how an investigation into area optimisation could progress. The problem is this:

A farmer has 40m of fencing. What is the maximum area he can enclose?

**Case 1: The rectangle:**

Reflection – the rectangle turns out to be a square, with sides 10m by 10m. Therefore the area enclosed is 100 metres squared.

**Case 2: The circle:**

Reflection: The area enclosed is greater than that of the square – this time we have around 127 metres squared enclosed.

**Case 3: The isosceles triangle:**

Reflection – our isosceles triangle turns out to be an equilateral triangle, and it only encloses an area of around 77 metres squared.

**Case 4, the n sided regular polygon**

Reflection: Given that we found the cases for a 3 sided and 4 sided shape gave us the regular shapes, it made sense to look for the n-sided regular polygon case. If we try to plot the graph of the area against n we can see that for n ≥3 the graph has no maximum but gets gets closer to an asymptote. By looking at the limit of this area (using Wolfram Alpha) as n gets large we can see that the limiting case is the circle. This makes sense as regular polygons become closer to circles the more sides they have.

**Proof of the limit using L’Hospital’s Rule**

Here we can prove that the limit is indeed 400/pi by using L’Hospital’s rule. We have to use it twice and also use a trig identity for sin(2x) – but pleasingly it agrees with Wolfram Alpha.

So, a simple example of how an investigation can develop – from a simple case, getting progressively more complex and finishing with some HL Calculus Option mathematics.