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The Perfect Rugby Kick
This was inspired by the ever excellent Numberphile video which looked at this problem from the perspective of Geogebra. I thought I would look at the algebra behind this.
In rugby we have the situation that when a try is scored, there is an additional kick (conversion kick) which can be taken. This must be in a perpendicular line with where the try was scored, but can be as far back as required.
We can represent this in the diagram above. The line AB represents the rugby goals (5.6 metres across). For a try scored at point D, a rugby player can then take the kick anywhere along the line DC.
Let’s imagine a situation where a player has scored a try at point D – which is a metres from the rugby post at B. For this problem we want to find the distance, x for this value of a such that this maximises the value of θ . The larger the value of θ, the more of the rugby goal the player can aim at and so we are assuming that this is the perfect angle to achieve.
Making an equation:
We can use the diagram to achieve the following equation linking θ and x:
We can use Desmos to plot this graph for different values of a:
We can then find the maximum points from Desmos and record these
This then allows us to find the exponential regression line of the coordinates of the maximum points:
This regression line is given by the equation:
This graph is shown below:
We can also plot the x values of the maximum points against a to give the following linear regression line:
This graph is shown below:
This means that if we know the value of a we can now very easily calculate the value of x which will provide the optimum angle.
Bring in the calculus!
We can then see how close our approximations are by doing some calculus:
We can find the x coordinate of the maximum point in terms of a by differentiating, setting equal to 0 and then solving. This gives:
When we plot this (green) versus our earlier linear approximation we can see a very close fit:
And if we want to find an equation for optimum theta in terms of x we can also achieve this as follow:
When we plot this (green) we can also see a good fit for the domain required:
Conclusion
A really nice investigation – could be developed quite easily to score very highly as an HL IA investigation as it has a nice combination of modelling, trigonometry, calculus and generalised functions. We can see that our approximations are pretty accurate – and so we can say that a rugby player who scores a try a metres from the goal should then take the resultant conversion kick about a+2 metres perpendicular distance from the try line in order to maximise the angle to the goal.
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Finding the volume of a rugby ball (prolate spheroid)
With the rugby union World Cup currently underway I thought I’d try and work out the volume of a rugby ball using some calculus. This method works similarly for American football and Australian rules football. The approach is to consider the rugby ball as an ellipse rotated 360 degrees around the x axis to create a volume of revolution. We can find the equation of an ellipse centered at (0,0) by simply looking at the x and y intercepts. An ellipse with y-intercept (0,b) and x intercept (a,0) will have equation:
Therefore for our rugby ball with a horizontal “radius” (vertex) of 14.2cm and a vertical “radius” (co-vertex) of 8.67cm will have equation:
We can see that when we plot this ellipse we get an equation which very closely resembles our rugby ball shape:
Therefore we can now find the volume of revolution by using the following formula:
But we can simplify matters by starting the rotation at x = 0 to find half the volume, before doubling our answer. Therefore:
Rearranging our equation of the ellipse formula we get:
Therefore we have the following integration:
Therefore our rugby ball has a volume of around 4.5 litres. We can compare this with the volume of a football (soccer ball) – which has a radius of around 10.5cm, therefore a volume of around 4800 cubic centimeters.
We can find the general volume of any rugby ball (mathematically defined as a prolate spheroid) by the following generalization:
We can see that this is very closely related to the formula for the volume of a sphere, which makes sense as the prolate spheroid behaves like a sphere deformed across its axes. Our prolate spheroid has “radii” b, b and a – therefore r cubed in the sphere formula becomes b squared a.
Prolate spheroids in nature
The image above [wiki image NASA] is of the Crab Nebula – a distant Supernova remnant around 6500 light years away. The shape of Crab Nebula is described as a prolate spheroid.
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IB Maths Resources from Intermathematics 
