91Ó°ÊÓ

(a) (i) A cycling road race requires one to complete 3 laps of a long road circuit. On the first lap I average \(40 \mathrm{~km} / \mathrm{h} ;\) on the second lap I average \(30 \mathrm{~km} / \mathrm{h} ;\) and on the third lap I only average \(20 \mathrm{~km} / \mathrm{h}\). What is my average speed for the whole race? (ii) I cycle for 3 hours round the track of a velodrome, averaging \(40 \mathrm{~km} / \mathrm{h}\) for the first hour, \(30 \mathrm{~km} / \mathrm{h}\) for the second hour, and \(20 \mathrm{~km} / \mathrm{h}\) for the final hour. What is my average speed over the whole 3 hours? (b) Two cyclists compete in an endurance event. (i) The first cyclist pedals at \(60 \mathrm{~km} / \mathrm{h}\) for half the time and then at 40 \(\mathrm{km} / \mathrm{h}\) for the other half. The second cyclist pedals at \(60 \mathrm{~km} / \mathrm{h}\) for half of the total distance and then at \(40 \mathrm{~km} / \mathrm{h}\) for the remaining half. Who wins? (ii) In a two hour event, the first cyclist pedals at \(u \mathrm{~km} / \mathrm{h}\) for the first hour and then at \(v \mathrm{~km} / \mathrm{h}\) for the second hour. The second cyclist pedals at \(u\) \(\mathrm{km} / \mathrm{h}\) for half of the total distance and then at \(v \mathrm{~km} / \mathrm{h}\) for the remaining half. Who wins? (c)(i) Apply your argument in (b)(ii) to prove an inequality between \(*\) the arithmetic mean $$ \frac{u+v}{2} $$ of two positive quantities \(u, v,\) and \(*\) the harmonic mean $$ \frac{2}{\frac{1}{u}+\frac{1}{v}} $$ (ii) Give a purely algebraic proof of your inequality in (i).

Step by step solution

01

Calculate the total distance for the road race (a)(i)

Assume the lap distance is \(d\) km. The first lap at 40 km/h takes time \(\frac{d}{40}\) hours, the second lap at 30 km/h takes \(\frac{d}{30}\) hours, and the third lap at 20 km/h takes \(\frac{d}{20}\) hours. The total time for 3 laps is then \(\frac{d}{40} + \frac{d}{30} + \frac{d}{20}\) hours, and the total distance is \(3d\) km.

02

Calculate average speed for the road race

The average speed is the total distance divided by the total time. Using the expression from Step 1, the average speed is \(\frac{3d}{\frac{d}{40} + \frac{d}{30} + \frac{d}{20}} = \frac{3}{\frac{1}{40} + \frac{1}{30} + \frac{1}{20}}\). Calculating gives \(\frac{3}{\frac{3+4+6}{120}} = \frac{3 \times 120}{13} = \frac{360}{13} \approx 27.69 \) km/h.

03

Calculate average speed for the velodrome race (a)(ii)

Over the 3 hours, the total distance cycled is \(40 + 30 + 20 = 90\) km. The average speed is the total distance divided by the total time, so \(\frac{90}{3} = 30\) km/h.

04

Determine winner of endurance event (b)(i)

The first cyclist's time is split equally, with speeds of 60 km/h and 40 km/h, leading to an average speed of \(\frac{60 + 40}{2} = 50\) km/h. The second cyclist divides distance equally, meaning the harmonic mean is given by \(\frac{2}{\frac{1}{60} + \frac{1}{40}} = \frac{240}{5} = 48\) km/h. The first cyclist wins.

05

Determine winner of theoretical race (b)(ii)

First cyclist's average speed is arithmetic mean \(\frac{u+v}{2}\). Second cyclist's average speed is harmonic mean \(\frac{2}{\frac{1}{u}+\frac{1}{v}}\). Since arithmetic mean \(\geq\) harmonic mean, first cyclist wins.

06

Proving AM-HM inequality (c)(i)

Using (b)(ii) logic, AM \(\frac{u+v}{2}\) is the first cyclist's average speed, and HM \(\frac{2}{\frac{1}{u}+\frac{1}{v}}\) is the second's. Since the first wins, \(\frac{u+v}{2} \geq \frac{2}{\frac{1}{u}+\frac{1}{v}}\).

07

Algebraic proof of AM-HM inequality (c)(ii)

First, simplify the harmonic mean: \(\frac{2uv}{u+v}\). To prove \(\frac{u+v}{2} \geq \frac{2uv}{u+v}\), multiply both sides by \(u+v\), leading to \((u+v)^2 \geq 4uv\). Simplifying gives \(u^2 + 2uv + v^2 \geq 4uv\), simplifying further to \(u^2 - 2uv + v^2 \geq 0\). This is valid as it is \((u-v)^2 \geq 0\). Therefore, the inequality holds.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Arithmetic Mean

The arithmetic mean is a way of finding the average or central value of a set of numbers. You simply add up all the numbers and then divide by how many numbers there are. For example, if you rode three laps with speeds of 40 km/h on the first, 30 km/h on the second, and 20 km/h on the third, the arithmetic mean is calculated by:

• Adding the speeds together: 40 + 30 + 20 = 90 km/h
• Dividing by the total number of laps: 90 km/h ÷ 3 = 30 km/h

This method works well when you want to find an average over equal durations, like over the same number of laps in our cycling example.
The arithmetic mean is easy to calculate and gives a quick sense of the overall average speed or value of a set. It's used often in many everyday calculations.

Harmonic Mean

The harmonic mean is a different type of average, often used when dealing with speeds, especially when the distances are the same but the speeds vary over the same distance. Unlike the arithmetic mean, it puts less weight on very high values and more on lower values, which is quite useful in some contexts.

• Calculate by:
  \( \frac{2}{\frac{1}{a} + \frac{1}{b}} \) for two speeds \(a\) and \(b\).

Consider the endurance event mentioned in the exercise, where the second cyclist rides at 60 km/h for half the distance and 40 km/h for the other half. Here, the harmonic mean gives a more accurate measure of average speed because it takes into account the variation over equal distances, resulting in \( \frac{2}{\frac{1}{60} + \frac{1}{40}} = 48\) km/h.
The harmonic mean is particularly useful in situations involving rates or ratios, like speed or density, where you want a more balanced average.

Inequality Proof

The AM-HM inequality states that for any set of positive numbers, the arithmetic mean is always greater than or equal to the harmonic mean. This property is fundamental to understanding rate-based calculations like in cycling races.

• Begin with: Arithmetic Mean \( \frac{u+v}{2} \)
• Harmonic Mean: \( \frac{2uv}{u+v} \)
• Prove by showing \( (u+v)^2 \geq 4uv \)

This inequality is derived by algebraic manipulation: expanding and simplifying leads to \( (u-v)^2 \geq 0 \), which is always true. Therefore, since the algebra supports the argument, we can conclude that the AM is at least the same as HM, if not greater, which explains why in certain cycling scenarios, one competitor may seem advantaged depending on differing strategies.

Cycling Race Problem

Cycling races can be used to perfectly illustrate concepts like average speed and mean inequalities. In multi-lap races, understanding how to compute the average speed using arithmetic or harmonic means helps determine which competitor might win under different conditions.

• Example: Different average speeds for separate laps or hours.
• Applying the arithmetic mean calculates the average when time duration is key.
• The harmonic mean is used for fixed distances, showing another strategic angle.

For instance, in the exercise's scenario (b) comparing two cyclists, one uses the arithmetic mean and the other the harmonic mean. This choice affects the outcome depending on whether they are averaging over consistent times or consistent distances, respectively. With these calculations in mind, competitors can optimize their strategies based on lap and endurance events, potentially changing their cycling tactics to ensure the best results.