Pupils often have difficulty in translating problems from words to symbols. Here is a very simple technique that I have found quite useful for helping them formulate the problem algebraically. First, they need to understand the importance of defining their variables. At this stage, writing down any values that are given for these variables begins to sort the problem into smaller steps. So, for the problem you have presented, let acceleration = a and force = F.
When a = 18ms-2, F = 150N When a = 4.5ms-2, then F = ?
They should already know the symbol for proportionality as "5".
Their jottings might look like this: This gives a =5 F(note this can also be read or translated as "acceleration varies as force").
The next stage is to write the relationship without the proportional sign.
This is understood easily when students have done work on experimental data and established, graphically and numerically, that "directly proportional" implies a constant scale factor of increase, k (the constant of variation).
The abstract notation can be translated into more recognisable maths when it is known that "= k x " replaces "5". The relationship can thus be written as the equation: a 5 F a = k x F a = kF Next, substitute the first set of values into the equation to find the value of k: a = k xF 18 = k x 150 18150 = k x 150150 325 = k At this stage, it is very important that pupils realise they should now write the general relationship as the equation with the calculated value of k: a = 325 F This equation can now be used to find out what force is needed to produce an acceleration of 4.5ms-2 as required in the question. This is found by substituting 4.5 for a: a = 4.5 = 325 F 4.5 x 25 = 325 F x 25 112.5 = 3F 112.53 = 3F3 37.5 = F giving F = 37.5N When working with indirectly (inversely) proportional relationships - that is, when two quantities increase at opposite rates - the process is very similar. One example is Boyle's law, which states that the volume (V) of a gas is inversely proportional to its pressure (P) when temperature is constant. This is written symbolically as V 5 1P (tell pupils to think of inverse proportion as "one over the variable").
We know that "5" is replaced by "= k x ". Thus the relationship can be written as the equation: V 5 1P V = k x 1P V = kP The "process" for solving word problems that include direct and indirect proportion is: 1. Define the variables.
2. Change words to symbols.
3. Write the relationship in the form of a 5 b or d 5 1c depending on the question.
4. Create the equation by replacing "5" with "= k x ", so aJ= k x b or d = k x 1c 5. Substitute the pair of values given to calculate k.
6. Write out the equation with the calculated value of k included.
7. Use this equation to find the unknown variable.
Make sure that your pupils are not just following a recipe to find an answer, but that they really know what is meant by proportionality.
I particularly like the extension questions that appeared in the book Year 4 of the SMP 11-16 series: "The time for a journey is inversely proportional to the speed. Calculate the percentage increase or decrease in the journey time when the speed is: * increased by 50%;
* decreased by 50%;
* increased by 20%;
* decreased by 20%.
Wendy Fortescue-Hubbard is a teacher and game inventor. She has been awarded a three-year fellowship by the National Endowment for Science, Technology and the Arts (NESTA) to spread maths to the masses.
www.nesta.org.ukEmail your questions to Mathagony Aunt at firstname.lastname@example.org Or write to TES Teacher, Admiral House, 66-68 East Smithfield, London E1W 1BX