Q I am teaching maths after a long time away. Being "able to do" and "having to teach" the subject are not always well matched.
Have you suggestions for the best way to introduce calculus?
A First it is important that you "refresh" yourself. There is nothing worse for your confidence as a teacher than to not have a "feel" for the area you are teaching. A good place to begin is to read Understanding Calculus by David Tall (www. warwick.ac.ukstaffDavid.Tall) published by ATM.
This series of articles is about using graphs to introduce calculus. How many of us take it for granted that pupils really understand what we are doing when we draw a tangent to a graph to look at a moment in time? Or that pupils are happy when we use a straight line to find the gradient of a curve.
Using a graphical calculator, ask pupils to take a "curved" graph and magnify it over very small ranges. As the range becomes smaller, they will see the curve approximating to a straight-line. This will provide much interesting discussion. Next the class can have a discussion on the idea of things changing with time. For instance, temperature varies throughout the day, fish stocks increase and decrease over years, the amount of a substance produced in a chemical reaction varies with time.
Differential calculus is about finding out how fast some quantity varies as some other quantity (usually time) changes:it is a method for finding rate of change. Examples can be found in text-books, but I like this one from Dr Dave Wilton at the University of Plymouth, as pupils can collect their own data using a data logger and a motion sensor (as suggested in last week's issue), and then check their calculated results with their own graphs.
A ball is thrown vertically upwards with a speed of 10ms-1. It is known that after t seconds the height of the ball in metres, h(t), is given approximately by the formula h(t) = 10t - 5t2. Find the average speed of the ball during the following time intervals: t = 0.25 to: (i)t =1 (0.25 + 0.75);
(ii)t = 0.5 (0.25 + 0.25);
(iii) t = 0.375 (0.25 + 0.125);
(iv)t = 0.3125 (0.25 + 0.0625);
(v) t = 0.251 (0.25 + 0.001);
(vi) t = 0.2501 (0.25 + 0.0001).
Explain the notation as you go along. The values for each height are found by substitution into the formula: h(t) = 10t - 5t2. At 0.25 seconds it is calculated as:h(0.25) = 10 X 0.25-5 X 0.252= 2.5 - 0.3425 = 2.1875m.
After 1 second, the height h(1) = 10 X 1 - 5 X 12 = 10 - 5 = 5m.
The other related heights are:
h(0.5) = 3.75m;
h(0.375) = 3.046875m;
h(0.3125) = 2.636719m;
h(0.251) = 2.194995m;
h(0.2501) = 2.18825m.
The average speed is calculated using the formula: speed = distance
= h(1) - h(0.25) = 5 - 2.1875
JJJ1 - 0.25 JJJ 0.75
Average speed is calculated for each time interval. For the smallest, it is approximately 7.5ms-1. From the calculations, it seems that as the time interval shrinks the average speed gets closer to 7.5ms-1. Ask pupils to try smaller time limits, from below t=0.25. It then appears reasonable to take 7.5ms-1 as the speed of the ball at the instant t=0.25s, an instantaneous speed at a particular time.This is a long process, and calculus offers us a better way.
Suppose at a time (t) the height of the ball is h(t). A very short time later, at time t + dt, the height will be h(t + dt). Using the formula h(t) = 10t - 5t2 we have h(t + dt) = 10(t + dt) - 5(t + dt)2= 10t + 10dt - 5(t + dt)(t + dt)= 10t + 10dt - 5(t2 + 2tdt + (dt)2 = 10t + 10dt - 5t2 - 10tdt - 5(dt)2J From the formula h(t) = 10t - 5t2, h(t + dt) - h(t) = 10t + 10dt - 5t2 - 10tdt - 5(dt)2 - (10t - 5t2)J= 10t + 10dt - 5t2 - 10tdt - 5(dt)2 - 10t + 5t2 = 10dt - 10tdt - 5(dt)2 As before, we can find the average speed in this very small time interval as the difference in heights between time t and time (t + dt): h(t + dt) - h(t) = JJJdt J 10dt - 10tdt - 5(dt)2 = JJJJJdt 5dt (2 - 2t - dt) = dt 5(2- 2t - dt) = 10 - 10t - 5dt As dt gets smaller (that is as dt tends to zero) the last term in the above equation becomes so small that it is negligible, and the remaining terms give us the instantaneous speed at time t, which is = 10 - 10t, or the derivative of h(t). The symbol dhdt is introduced to stand for the value of dhdt as dt 0 if the gradient tends to a fixed limit.
We can check if this works by substituting into this expression for t = 0.25; this gives us an instantaneous speed of 10 - 10 X 0.25 = 7.5ms-1.
Wendy Fortescue-Hubbard has been awarded a fellowship by the NESTArts (www.nesta.org. uk) to spread maths to the masses. Email her at teacher@ tes.co.uk or write to TES Teacher, Admiral House, 66-68 East Smithfield, London E1W 1BX