- Level: International Baccalaureate
- Subject: Maths
- Word count: 1588
Investigating the Ratios of Areas and Volumes around a Curve
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Introduction
Daniel Bregman IB Mathematics HL Portfolio 14/07/2011
Investigating the Ratios of Areas and Volumes around a Curve
Introduction:
In this investigation I will examine how changing the power of a function changes the ratio of the areas between it and the two axes. I will begin by making assumptions to simplify the mathematics, and then move on to more general cases. The curve used throughout will be
, and I will aim for an expression of an area or volume ratio in terms of n.
Notes:
As the end goal in my investigation is to find a (unitless) ratio, I have left units out of my calculations. For all areas these can be given as ‘units2’, in whatever unit the particular graph is drawn. Volumes would be measured in ‘units3’.
All diagrams have been produced on Autograph V3.20, and modified slightly to add labels/shading in Microsoft Paint.
A simple case:
The curve 
is drawn between x=0 and x=1. The region between y=0, y=1, the curve and the y-axis will be known as A, and the region between x=0, x=1, the curve and the x-axis will be known as B. The ratio A:B can be calculated using the power rule for integration; to calculate A the equation is rearranged into the form 
; the limits remain the same when raised to this power:
Thus for n=2, A:B=2.
Middle
For n=3, between x=-1 and x=0:
Apart from two cases, the correlation (that 
) seems to hold for the new limits. Any proof of the correlation must therefore explain the instances where it seems not to hold. To test further instances I produced a program on my Casio fx-9750G+ graphing calculator. The following program performs the above process automatically:
?→N
?→A
?→B
∫(Y^(1/N),A^N,B^N)/∫(X^N,A,B)
The following table demonstrates the results obtained using this program for several values of n, a, and b. Results which do not correlate with the observed pattern (and cannot be put down to the calculator’s internal rounding errors) are shown in bold.
n | a | b | Ratio |
2 | 0 | 1 | 2.00000000001 |
2 | 0 | 2 | 2 |
2 | -2 | -1 | 2 |
2 | 1 | 2 | 2 |
2 | 1 | 3 | 2 |
2 | 5 | 6 | 2 |
2 | 0.5 | 5.5 | 2 |
2 | 100 | 150 | 2 |
2 | -1 | 1 | 0 |
2 | -1 | 2 | 1.55555555556 |
3 | 0 | 1 | 3 |
3 | 1 | 2 | 3 |
3 | -2 | -1 | 3 |
3 | 3 | 4 | 3 |
3 | 10.5 | 11.5 | 3 |
3 | e | π | 3 |
3 | -1 | 2 | 3 |
3 | -1 | 1 | MATH ERROR |
0.5 | 0 | 1 | 0.499999999997 |
0.5 | 1 | 2 | 0.5 |
0.5 | 2 | 3 | 0.5 |
0.5 | 8 | 9 | 0.5 |
0.5 | 0.1 | 0.9 | 0.5 |
0.5 | -2 | -1 | MATH ERROR |
0.5 | -1 | 0 | MATH ERROR |
-1 | 1 | 2 | -1 |
-1 | 5 | 6 | -1 |
-1 | -2 | -1 | -1 |
-1 | -1 | 1 | MATH ERROR |
-1 | 0 | 1 | MATH ERROR |
-2 | 1 | 2 | -2 |
-2 | 5 | 6 | -2 |
-2 | -2 | -1 | 2 |
-2 | -6 | -5 | 2 |
-2 | 0.5 | 0.8 | -2 |
There is a clear pattern visible, although certain cases do not fit. As well as confirming the pattern, my proof must also give reasons for the ‘anomalous’ values.
CONJECTURE: For a curve of the form, between x=a and x=b, a<b, the ratio A:B of the regions defined below is equal to n.
A: region bounded by the y-axis, the curve,
, and 
.
Conclusion
=a, x=b.
...read more.
PROOF: We proceed through the same process as for the individual examples above.
QED
We can now try a similar process for revolution about the y-axis. The limitations of both proofs will be discussed once this has been done.
For n=2, between x=0 and x=1:
For n=2, between x=1 and x=2:
For n=3, between x=0 and x=1:
CONJECTURE: For a curve of the form, between x=a and x=b, a<b, the ratio A:B of the volumes of revolution about the y-axis defined below is equal to 
.
A: region bounded by the y-axis, the curve,, and .
B: region bounded by the x=axis, the curve, x=a, x=b.
PROOF: We proceed through the same process as for the individual examples above.
QED
The restrictions for this pattern follow similar lines to those for the areas defined above, and the restrictions detailed above still hold. In particular, attention must be paid to values of n<1, which result in cases where a<b but an>bn, so that the formula above for removing volumes from cylinders will sometimes be the ‘wrong way around’. This occurs because the formula used in the proof is based on the arrangement of volumes shown in the diagram on page 11.
In addition, since 2n+1 occurs as a denominator in the first, and 2+n in the second, n must not equal 
or -2.
However, with these restrictions in place, the formula is general and the rule can be said to hold for all 
.
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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