Although Johnny is doing the CIE IGCSE Additional Mathematics programme, he is confused at times by questions in regular Mathematics which his teacher deems “

*not so hard*,” such as the following:“Captain Jack has a motorboat that travels at 20 km per hour in still water. Given that the current in the river is 5 km per hour, the boat’s downstream speed (ignoring friction) will be 20 + 5 = 25 km/h while its speed in the upstream direction will be 20 – 5 = 15 km/h. Suppose that Jack has just enough fuel for 3 hours. How far can he go downstream before he has to turn around to get back home?”

The first physics formula Johnny learnt was ‘Distance = Speed × Time’ or D = S × T. For brevity he decides to use subscripts to identity total, average, downstream or upstream variables,

_{T, A, D, }and_{U}_{,}respectively. Thus his formula for the boat’s downstream distance is D_{D}= S_{D}× T_{D }= 25 T_{D. }Similarly his upstream distance formula is D_{U}= S_{U}× T_{U }= 15 T_{U}. Johnny also knows average speed is total distance divided by the total time, giving him the equation S_{A}= D_{T}÷ T_{T}. He reasons that the average speed, S_{A}, is (15 + 25)/2 = 20 and consequently the total distance must be 20 × 3 = 60 km.D

_{D}+ D_{U }= D_{T}S

_{D}× T_{D }+ S_{U}× T_{U}= D_{T}25 T

_{D }+ 15 T_{U }= 6025 T

_{D }+ 15 (3 – T_{D}) = 6025 T

_{D }+ 45 – 15 T_{D }= 6010 T

_{D }= 15T

_{D}= 1.5 hoursConsequently Johnny tells his group the downstream distance will be 25 km/h × 1.5 hours = 37.5 km.

Disagreement

Mary is in the same discussion group. She says to Johnny: That can’t be right. I used ‘Trial and Improvement.’ The two distance variables must be equal, D

_{D}= D_{U}. Given that the total time is 3 hours, I wrote T_{U}= 3 – T_{D}. Obviously the downstream time cannot be less than 0 hours and your own calculations demonstrate that T_{D}must be less than or equal to 1.5 hours. Using the vocabulary of functions, the domain of the downstream distance is 0 < T_{D}< 1.5 hours.Mary’s trials are shown below.

Trial | T _{D} | T _{U} = 3 – T_{D} | D _{D }= 25T_{D} | D _{U }= 15T_{U} | Error |

1st | 1.5 hours | 1.5 hours | 35.5 km | 22.5 km | 15 km |

2nd | 0 hours | 3 hours | 0 km | 45 km | -45 km |

3rd | 1 hour | 2 hours | 25 km | 30 km | -5 km |

4th | 1.25 hour | 1.75 hour | 31.25 km | 26.25 km | +5 km |

5th | 1.125 hour | 1.875 hour | 28.125 km | 28.125 km | 0 km |

Johnny replies:Your table certainly looks convincing. But the algebra I used is perfectly logical. How can there be a mistake?

At this point, as the teacher, I confirm that Mary is correct and ask the group to find the mistake in Johnny’s algebra.

Algebra revisited

Johnny’s experience tells him it is extremely difficult to proof-read for errors so he decides to take a slightly different approach. After discussion with Mary, he now realizes that he does not actually need the total distance. He only requires that:

D

_{D}= D_{U}25 T

_{D }= 15 T_{U}25 T

_{D }= 15 (3 – T_{D})25 T

_{D }= 45 – 15 T_{D}40 T

_{D }= 45T

_{D }= 45 ÷ 40 = 1.125 hoursD

_{D}= 25 km/h × 1.125 h = 28.125 kmWhile he is able to confirm Mary’s answer obtained by ‘Trial and Improvement,’ nevertheless, he is somewhat dissatisfied. Algebra required rather more mental gymnastics and, at least initially, appeared to steer him in the wrong direction. He asks me “Is algebra worth the effort?”

My honest answer is, “We have no need of algebra if we are only seeking the answer to one particular question. Algebra’s real advantage is that it is very easily generalized. Keeping the total time at 3 hours but supposing the current is some other constant than ‘5’, you can readily modify the previous solution as follows.

D

_{D}= D_{U}S

_{D}× T_{D }= S_{U}× T_{U}S

_{D}× T_{D }= S_{U}(3 – T_{D})S

_{D}× T_{D }= 3 S_{U}– S_{U}× T_{D}(S

_{D}+ S_{U}) T_{D }= 3 S_{U}T

_{D }= 3 S_{U}÷ (S_{D}+ S_{U})Would it be significantly harder to suppose we had fuel for H hours instead of 3 hours?

ICT

An approach very similar to Trial and Improvement uses the computer programme Geogebra to model the situation:

Where did Johnny go wrong?

Like many students, Johnny is over-reliant on the arithmetic mean. For the purpose of determining the average speed, the Harmonic Mean is called for. How do you obtain the Harmonic Mean of two numbers

*a*and*b*?Procedure | Algebra | Captain Jack’s speeds |

Start with a and b | a, b | 15, 25 |

Take the reciprocal of each number | 1/ a and 1/b | 1/15, 1/25 |

Rewrite with a common denominator in lowest terms | b/ab, a/ab | 1/15 = 5/75, 1/25 = 3/75 |

Find the arithmetic mean | (a + b)/2/ab | (5+3)/2/75 = 4/75 |

Take the reciprocal of the previous line | 2ab/(a+b) | 75/4 |

The total distance = (harmonic average of speed) × Total time = 75/4 × 3 = 225/4 = 56.25 km.

The downstream distance must be half the upstream distance or 28.125 km, as previously calculated.

(Please note the harmonic mean is

__not__an explicit part of the IGCSE Maths Syllabus and consequently it is unlikely to be found on the examiner’s official mark scheme.)Conclusion

As Grandma used to say, “There’s more than one way “to skin a cat.” (Cat befrienders are advised to replace the offending text with “poach an egg.”) Today’s message is, “There’s more than one way to solve a problem.” In the ‘real world,’ team skills and independent verification of results are important. Used correctly, algebra is indeed logical, but you should appreciate that an apparently small misunderstanding will lead to a false conclusion. There is more to a mathematics education than the syllabus.

## No comments:

## Post a Comment