This method is more accurate as it only involves calculations. On the next pages I used Microsoft Excel to find the gradient of three points on the graph y=x² and y=2x². I will present the coefficient of x² below the actual calculation.
I will write the results of these calculations below with the results of my linear graphs and hope to spot a pattern.
The gradient function for the range of graphs y=ax² would be m=2ax, where x is the x-coordinate of the point at which the gradient should be determined. However, this is not true for linear graphs. So I will have to investigate the range of graphs with formula y=ax³. These calculations will be made for six points on the graph y=x³ and y=2x³.
Now I will list my results and hope to find a pattern for the gradient function for this range of graphs. The following graph shows the results:
When I looked at this, it took a while until I found the relationship, but in the end I noticed that for this range of graphs y=ax³ the gradient function will be m=3ax². I think that this is enough research to find a relationship between these three ranges of graphs and their gradients. I will draw up a table with the results and try to generalise the rule.
The term in the box should determine the gradient of any point on any graph. I suggest that the gradient function for any graph would be m= nax^n-1. But this may not be true for negative or fractional powers, so I will have to test this function on these ranges of graphs. I will start with y=ax^-1. For this purpose I have to calculate some gradients of random points on two graphs with -1 as their power. Near the calculation I will present the result of my formula. If it equals approximately the calculated gradient, my formula is correct. If it is not, then I will have to change my formula. The result of my formula for the first example below was calculated like this: m=nax^n-1 => m=-1*1*1^-2 => m= -1
I typed it in as a formula, so that Excel will automatically calculate the result of the formula for me. Here the gradient according to my formula is correct, but I will have to test it more than one time to be sure. The gradient calculated by my formula should be the exact number, and the formula calculated by using the chord method is only an approximation of the gradient at this point. So both numbers must be very close in order for the formula to be correct. But the number calculated by using the chord method in Excel is only an approximation itself, so it could be that both numbers are equal, which also proves my formula to be correct. I will also work out the gradient for the calculation on the next page: m= nax^n-1 => m=-1*1*2.5^-2 => m= -0.16.
The function proves to be correct for negative powers. This function may be the gradient function for any graph y=ax^n at any point, but I do not know if the function will determine the correct gradient for graphs with fractional powers. To ensure this, I will test the function on the range of graphs with formula y=ax^0.5. For the example below I will show how the gradient is worked out by using the formula: m= nax^n-1 => m=0.5*1*1^0.5 => m=0.5
The function appears to be correct so far, but I will have to use many more examples in order to show that the gradient function I developed is generally correct.
I will show working out the gradient for the calculation below by using the formula: m=nax^n-1 => m=0.5*3.31*1^-0.5 => m=1.655
I think that this is enough evidence to show that the gradient function is m= nax^n-1.
Now I will use this formula to predict the gradient of two complex functions by combining simpler ones. My first graph to consider will be y=x²+3x+5. To work out the gradient function for this graph I will need to look at each summand separately and use the rule for it. Then I will need to add the gradients together:
Graph: y= x² + 3x + 5
Gradient: 2x + 3 + 0
So the gradient for the graph y=x²+3x+5 will be 2x+3. It is as simple as that. The following two calculations will prove that this is correct. Both calculations are based on the graph y=x²+3x+5:
The is the gradient according to my formula is this: m=2x+3 => m=2*1+3 => m=5
And this is indeed the gradient of this point. The x-coordinate of my next point of this graph will be 2.7.
The calculation of the gradient using my formula looks like this: m=2x+3 => m=2*2.7+3 => m=8.4.
So you can see that it is possible to calculate the gradient of more complex functions by combining simpler ones. Another example could be y=3x³-5x²+5.5.
Graph: y= 3x³ - 5x² + 5.5
Gradient: 3*3x² - 5*2x + 0
So the gradient for the graph y=3x³-5x²+5.5 will be at any point m=9x²-10x.
More about the Gradient Function and Calculus
I used an A-level-book to get additional information on this topic.
We already know that the chord method is only an approximation for the actual gradient of a point. The closer together the points are, the more accurate our result will be, but we will only get 100% accuracy when the distance between the two points is zero. Here by using calculus, you can obtain a simple formula that tells you the gradient of the tangent lines at any point on the curve. This formula, which I found out to be m=nax^n-1 is called the derivative or, more helpfully, the gradient function. The process of finding the derivative is what is known as differentiation and belongs under the topic of calculus.
But let us go back to the chord method. This method will never be accurate, as we could never calculate the gradient of the chord which joins two points, which have the distance zero. The formula m=δy / δx would not work for this case (δ is a Greek letter and just means 'difference', so the notation literally means 'the difference in y over the difference in x equals the gradient). The result would be 0/0 which is undefined and would not give a sensible answer. We call this the limit, and we cannot calculate this. What we can do though is to make the difference between the two points approach zero as much as possible without it becoming zero. This will make the calculated gradient more and more accurate.
I will do calculations that deal with differentiation and explain some expressions.
Suggest that the second point is called Q and has the coordinates (x+δx; y+δy) and that the first point is called P and has the coordinates (x, y). The small y-difference of these two points would thus be δy and the tiny x-difference would be δx. δx or δy are both a single symbol and are often called the increment in x or y. The gradient of the chord PQ would thus be:
(y+ δx) –y δy
(x+ δx) +x δx
Because the equation of the curve is y=f(x), the coordinates of P can also be written as [x, f(x)] and the coordinates of Q as [(x+δx), f(x+δx)]. The value of δx can be made as small as you like. If you decrease the value of δx, the value of δy will also decrease, and the point Q will become closer to the point Q and approaches the definite limit I talked about earlier. The definite limit here is δx being equal to zero. It would be the gradient of the tangent of the curve at the point P. It is called the rate of change of y with respect to x at the point. This is denoted by dy / dx. This symbol is called the derivative or the differential coefficient of y with respect to x and is also the limit of δy / δx as δx tends to zero, where ‘tends to’ just means ‘approaches’. The following equation denotes this:
dy δy
dx δx
By finding dy / dx from y I used a procedure called differentiating y with respect to x.
The gradient function I developed is indeed a generally used function in calculus, and this backs up the findings of this assignment. For any function y=ax^n the differential coefficient or derivative equals nax^n-1. The mathematical function is:
dy
dx
What is Calculus and who invented it?
Calculus is the study of how things change. It provides a way for us to construct quantitative models of change like I did to find the gradient function. Calculus is most commonly used in physics and engineering.
Calculus was first invented by Newton, who discovered that acceleration, which means change of speed of objects could be modelled by his ‘laws of motion’.
The set of positions and times that we use to describe motion is what we call a function.
Calculus often deals with the following topics:
1. How to find the instantaneous change (called the "derivative") of various functions. (The process of doing so is called "differentiation".)
2. How to use derivatives to solve various kinds of problems.
3. How to go back from the derivative of a function to the function itself. (This process is called "integration".)
4. Study of detailed methods for integrating functions of certain kinds.
5. How to use integration to solve various geometric problems, such as computations of areas and volumes of certain regions.
As I have mentioned earlier calculus was first created by Newton and also Leibniz, although some of the ideas were already used by Fermat and even Archimedes.
Calculus is divided into two parts. The part of calculus which I dealt with on the previous two pages is called differential calculus and the other part is called integral calculus.
Integral calculus is concerned with area and volume.
The basic idea of integral calculus is this: the simplest shape whose area we can compute is the rectangle. The area is the length of the rectangle multiplied by its width. For instance, a "square mile" is a piece of land with as much area as a square plot of land with sides measuring one mile each. To compute the area of a more complicated region, we chop up the region into lots and lots of little rectangles. When we do this, we will not be able to succeed completely because there will always be pieces with curved sides, generally. But the key idea is that the sum of the areas of the rectangular pieces will be a very close approximation of the actual area, and the more pieces we cut, the closer our approximation will be. At one point the approximation will be so close that the difference with the real volume would not matter anymore.
Differential calculus would be used for the following problem: Imagine you go for a car ride. Suppose you know your position at all times. Just from the knowledge of your position at all times, you can reconstruct what your speedometer showed at any time by using differential calculus.
The basic idea of differential calculus is this: The simplest situation where you can compute what the speedometer read is when you drove at the same speed over the entire distance. Then, you can use the formula: speed equals distance divided by time. For instance, if you drive 50 miles in one hour all at the same speed, then your speedometer read 50 miles per hour the whole trip. In the situation where you didn't drive at the same speed, the idea is to imagine your trip as lots and lots of short trips, say, one trip involving pulling the car out of the garage, another trip getting the car onto the road, and so on. Over each of these trips, your speed doesn't change much and you can pretend that your speed didn't change at all. This puts you in the situation where you know how to calculate the speed for each partial trip, and gives you a good idea of what your speedometer. However, the assumption that the speed didn't change over each tiny trip is generally wrong, and so you only get an approximation to the correct answer. But the key idea is that the smaller you make the tiny trips used in your computation, the more accurate you will be able to compute the actual speedometer reading.
Conclusion
In this assignment I wrote about a function that I wanted to develop by using the chord and tangent method. I finally found this function to be m=nax^n-1, which proved to be correct for both fractional and negative powers and also of course positive powers. I also showed how you can work out the gradient of complex functions by combining the results of my formula on simpler parts of the function. In the last part I researched into calculus and in particular differentiation, as this relates to my assignment. It showed that the gradient function I developed was actually used in calculus, which proves my findings to be correct. Overall I created a significant piece of coursework, which gave me an initial view of calculus and differentiation, and what I can expect when I attend an A-level Mathematics course in 6th form.