# Final Exercises on Chapter 6: Rates of Change

We have now completed the work on Rates of Change. The complete solutions to all problems from exercises 6.1 to 6.4 have now been posted on the Maths Methods Wiki.

You will have noticed by now that Rates of Change involves many of the skills that you have already learned; gradients in particular. When finding an average velocity you really are only finding the gradient between two points. When you are finding the instantaneous velocity, you fit a tangent to the curve and find the gradient of that tangent which, once again, involves finding two points on the tangent line and then using these to calculate the gradient between these two points. Same old skills you have been using all year only in a different context.

You also learned that a rate is a measure of how one variable changes with respect to another variable. In the case of distance time graphs this change of distance with respect to a change in time is called … wait for it … speed (velocity for displacement verses time). So the “new” skills learned in this unit are just previously learned skills within a new context.

This Rates of Change unit we have just completed is actually an introduction to two topics; one from maths and one from physics. The Physics topic is called Kinematics, which is the study of motion. The mathematics topic is Calculus. This is a significant area of study which dominates much of the remainder of your mathematics studies in Year 11 and 12, particularly in Maths Methods and Specialist Maths. More on that in term 3!

Please make sure that you have completed all of Ex 6.1 to 6.4 by the need of the school holidays so you will be ready for the assessment task in week one of term three. See you all then. 🙂

## 2 thoughts on “Final Exercises on Chapter 6: Rates of Change”

1. Haley

I have a quizicle…
In Excercise 6.3 Question 5 c) To find the gradient, did u just pull 2 numbers out of no where?

Haley xD

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2. Hi Haley,
Out of nowhere? How could you say such a thing Haley! This is science not voodoo! Although you are sort of right I have to admit. This is how it worked:

The graph was constructed using the points we calculated in part b of the question. So we ended up with a graph of speed versus time. There were only four points to work with at t= 2, 4, 6 and 8. (Part B was where we calculated the speeds that correspond to those times and we did this by fitting, by eye, a tangent.) Now doing anything by eye means that there will be some error. It should be small but it will still be there. Once the graph was made we noticed that they more or less lie in a straight line. We then drew a line of best fit through the points and had to estimate the rule or formula of that line. So here is where the answer to your question comes in.

How do we calculate the gradient? You know it is basically rise/run, but where do we get the points to calculate this? Well you just pick any two points you like, as long as they are on the line. I pick points that are easy to work with (why use points like (2.6, 17.983) when I could use (1,3) if it was available?). I picked the point (5,2) and (18, 8.5) because they were easy to read from the graph, but I could have used any other points. The gradient this calculates is rise (13) over run (6.5) which is 2. We have to remember that this is approximate though as all the points are estimates based upon a tangent that was drawn BY EYE.

So you see the points were selected from somewhere, rather than “pulled from nowhere”! Hope that helps Haley.

Mr Thwaites

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