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Zeno (circa 490 BC - circa 430 BC) was a pre-Socratic Greek philosopher of southern Italy who is famous for his paradoxes.
One of Zeno's paradoxes can be summarised by:
Achilles and a tortoise agree to a race, but the tortoise is unhappy because Achilles is very fast. So, the tortoise asks Achilles for a head-start. Achilles agrees to give the tortoise a 1 000 m head start. Does Achilles overtake the tortoise?
We know how to solve this problem. We start by writing:
where
${x}_{A}$ | distance covered by Achilles |
${v}_{A}$ | Achilles' speed |
$t$ | time taken by Achilles to overtake tortoise |
${x}_{t}$ | distance covered by the tortoise |
${v}_{t}$ | the tortoise's speed |
If we assume that Achilles runs at 2 m $\xb7$ s ${}^{-1}$ and the tortoise runs at 0,25 m $\xb7$ s ${}^{-1}$ then Achilles will overtake the tortoise when both of them have covered the same distance. This means that Achilles overtakes the tortoise at a time calculated as:
However, Zeno (the Greek philosopher who thought up this problem) looked at it as follows: Achilles takes $t=\frac{1000}{2}=500\mathrm{s}$ to travel the 1 000 m head start that the tortoise had. However, in this 500 s, the tortoise has travelled a further $x=\left(500\right)(0,25)=125\mathrm{m}.$ Achilles then takes another $t=\frac{125}{2}=62,5\mathrm{s}$ to travel the 125 m. In this 62,5 s, the tortoise travels a further $x=(62,5)(0,25)=15,625\mathrm{m}.$ Zeno saw that Achilles would always get closer but wouldn't actually overtake the tortoise.
So what does Zeno, Achilles and the tortoise have to do with calculus?
Well, in Grades 10 and 11 you studied sequences. For the sequence $0,\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},...$ which is defined by the expression ${a}_{n}=1-\frac{1}{n}$ the terms get closer to 1 as $n$ gets larger. Similarly, for the sequence $1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},...$ which is defined by the expression ${a}_{n}=\frac{1}{n}$ the terms get closer to 0 as $n$ gets larger. We have also seen that the infinite geometric series has a finite total. The infinite geometric series is
where ${a}_{1}$ is the first term of the series and $r$ is the common ratio.
We see that there are some functions where the value of the function gets close to or approaches a certain value.
Similarly, for the function: $y=\frac{{x}^{2}+4x-12}{x+6}$ The numerator of the function can be factorised as: $y=\frac{(x+6)(x-2)}{x+6}.$ Then we can cancel the $x-6$ from numerator and denominator and we are left with: $y=x-2.$ However, we are only able to cancel the $x+6$ term if $x\ne -6$ . If $x=-6$ , then the denominator becomes 0 and the function is not defined. This means that the domain of the function does not include $x=-6$ . But we can examine what happens to the values for $y$ as $x$ gets close to -6. These values are listed in [link] which shows that as $x$ gets closer to -6, $y$ gets close to 8.
$x$ | $y=\frac{(x+6)(x-2)}{x+6}$ |
-9 | -11 |
-8 | -10 |
-7 | -9 |
-6.5 | -8.5 |
-6.4 | -8.4 |
-6.3 | -8.3 |
-6.2 | -8.2 |
-6.1 | -8.1 |
-6.09 | -8.09 |
-6.08 | -8.08 |
-6.01 | -8.01 |
-5.9 | -7.9 |
-5.8 | -7.8 |
-5.7 | -7.7 |
-5.6 | -7.6 |
-5.5 | -7.5 |
-5 | -7 |
-4 | -6 |
-3 | -5 |
The graph of this function is shown in [link] . The graph is a straight line with slope 1 and intercept -2, but with a missing section at $x=-6$ .
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