How to draw lines of best fit?

In many questions in Physics or often in a practical students are often required to draw a line of best fit.

So what is the line of best fit. The line of best fit is a line that is drawn through a series of points in a graph in order to determine the trend in the points. The line can then be used to determine the gradient and the y-intercept of the line if it is a straight line.

The following are tips that you can use in order to obtain the best line of best fit from a few points. You should read all of them and choose which tips is appropriate.

Tip 1

Sometimes you will obtains points that are as shown below in fig 1. Although I must say that such occurrence is quite rare. In fig 1 below you can see that the points forms an almost perfect straight line. In such a case it would be enough to choose two points far away from each other and then draw a line from one to the the other. Make sure that the line extends on both side to cover all the points.

Fig 1

Tip 2

Most of the time you will be faced with points as in fig 2 below. As such you will have to draw a line such that the points are on both sides of the lines as such. A you see from one side of the line to the other the points alternates up and down. Also make sure that two points that are closest to each other and are on different sides of the line are the same distance from the line.

Fig 2

Tip 3

In case you are unable to make the points alternate on both sides of the line from one end of the line to the other as in fig 3 below. It is acceptable to have them two by two on the same side of the line.

Fig 3

Tip 4

It may happen as in fig 4 below that one point is so far from the others that it would make no sense to include it when drawing the line of best fit. This point is called an outlier and it is often the result of an error in calculation, an error in taking readings such as parallax error or zero error. It is advisable to check the data again.

Fig 4

Tip 5

Although I had said that it is possible to have the points two by two on both side of the line. It is not possible to have them as in fig 5 below. 

Fig 5

 

If you are in the situation that I described as in fig 5 above then you can draw the line as in the fig 6 below.

Fig 6

So if you have any difficulties drawing the line then review the five different tips above you will surely find a tip for you. Otherwise send me a picture of the points and we will see it together.

y = mx + c

We have seen in a previous post about the gradient and how it is calculated. We have also learned about the y-intercept. We are now going to combined the two concept to learn about the equation of a straight line.

All straight line can be represented by a equation that will be in the form of 

y = mx + c

We are now going to use a straight line in order to see how the equation above can be determine and how the x-coordinate or the y-coordinate could be determined if one of the two and the equation is available.

Fig 1

Fig 1 above shows a straight line graph.

The gradient of the graph can be determined using the method described here.

Two coordinates that can be used are (1,6) and  (5,18)

Thus Gradient = (18 –6)/(5-1)

                               = (12)/(4)

                                 = 3

And using the methods described here we can deduce that the y-intercept is 3.

Now the gradient = 3

and y-intercept =3

If we say that gradient = m then m= 3

and if we say that y-intercept = c then c=3

replacing the values for m and c in y= mx +c

the equation of the line becomes

y = 3x + 3

All straight lines can be represented with an equation and the way to obtain the equation is by using the method above.

We are going to use the equation to determine

(i) the value of the y-coordinate when the x-coordinate is 3

(ii) the value of the x coordinate when the y-coordinate is 15

(i) Equation of the line is y = 3x + 3 and x=3

replacing x=3 in the equation we get

y =3(3) +3

=9+3

=12 check on the graph!!!!!!

(ii) y =15

Replacing into the equation we get

15 = 3x +3

15 – 3 = 3x

12 = 3x

x =12/3

x = 4 Check on the graph!!!!!!

Practical skills

The gradient of a line.

The y-intercept of a line. 

y=mx+c

The line of best fit

What is the y-intercept?

We have seen in a previous post the gradient of a straight line graph. Today we are going so have a look at the y-intercept.

Now as you know a graph is composed of two axes. The y-axis and the x-axis as shown in fig 1.

Fig 1

Now in order to find the y-intercept you will need a straight line that passes through the y-axis, i.e it intercept the y-axis,  as seen in fig 2 below.

Fig 2

As you can see in fig 2 above all three lines crosses the y-axis. As a result the three lines would have a y-intercept.

How to obtain the y-intercept of a straight line?

1. If the x-axis starts at 0

As you can see in fig 3 below the point at which the line crosses the y-axis is at the y=2 coordinate. Hence the y-intercept is 2.

We can thus define the y-intercept as being the value of the y-coordinate when the x-coordinate is 0. 

Fig 3

2. If the x-axis does not start at 0

As you can see in fig 4 below the x-axis does not start at 0. Hence as you may have guess wrongly the y-intercept is not 2 since according to the definition the y-intercept is the y-coordinate when the x-coordinate is 0.

So how do you obtain the y coordinate.

Fig 4

You will first have to calculate the gradient of the line using the method described in this post.

The gradient in this case is 1.

You will use the equation y = mx + c  and a coordinate on the line in this case (3,2).

y = 2

x=2

gradient = m =1

Hence the only variable left is c, the y-intercept.

3 = 1*2 + c

c = 3 – 2 =1

The y-intercept of the line is thus 1.

You should thus be very careful to check that the x-axis starts with 0 or does not start with 0 so as to choose which of the two methods to use.

Types of distance-time graph

As we have seen in this post on distance-time graph, it is easier to extract information from a distance-time graph than from a paragraph or from a table of time and distance travelled.

We have seen in this post the gradient of the distance-time graph is the speed. Hence you must be able to deduce how the speed of an object varies according to the shape of its distance time graph.

Let us now look at some distance-time graph and see how the speed varies with time. We will then look at the corresponding speed-time graph for the object.

1. Fig 1 below shows a distance-time graph shows that the distance travelled by the object from time  0 s to t s  is 0 m. This means that the object has not moved at all and as a result it is motionless at the starting point.

The gradient of the graph is 0, hence it means that form time o s to time t s, the speed is 0 m/s.

The speed-time graph is thus as shown in fig 2.

Fig 1

Fig 2

2. Fig 3 below is a distance-time graph that shows an object whose distance travelled is constant form 0 s to time t s. We can thus assume that the object is stationary. The gradient of the graph is thus 0 which means that the speed is also  0 m/s from 0 s to t s. The speed-time graph would thus be as shown in fig 4 below.

Fig 3

 

Fig 4

3. Fig 5 below is a distance-time graph that shows an objects moving and the distance travelled is increasing.

The gradient of the graph is the speed. From the graph the gradient is a constant value but is not zero. Hence the speed-time graph is as shown in fig 6.

Fig 5

Fig 6

4. In fig 7 below we have a distance-time graph that shows an object that is moving as a result the distance travelled is increasing.

As we know the gradient of the graph is the speed and since the gradient is increasing then is it means that the speed is increasing. Hence the speed-time graph of the object is as shown in fig 8.

Fig  7

Fig 8

Distance-time graph

The ability to draw graphs and to obtain information from them is one of the most important skills that a physicist needs to develop. You are also able to extract information more easily using graphs.

Example 1

A boy starts to walk at t = 0 s and walks from point A to point B a distance of 100 m for 10 s . He then stops and walk back towards his starting point in a time of 10 s.  The motion is as shown in fig 1.

 

Fig 1

If you use a table to present this information then it would be as follows:

time /s	Distance travelled /m
0	0
10	100
20	200

Table 1

After 10 s the boy has walked a distance of 100 m. And after 20 s the boy has walked a distance of 200 m( 100 m from A to B and another 100 m from point B to A).

Hence we can plot this on a graph as shown in fig 2 below.

Fig 2

As you can see from this graph the different coordinates will the give you the distance that the boy has walked after a particular time.

Now that you can plot the motion of an object on a graph. Let us see what you can do with a distance-time graph.

You would remember that the speed of an object is the rate of change of distance with time and that it can be calculated using the following equation

speed = distance travelled / time taken

With the distance-time graph the speed of an object at a particular time is the gradient of the line at that particular point.

Now what is the speed of he object at 5 s?

You will have to determine the gradient of the line at 5 s.

Hence the two coordinates that can be used are

(0,0)  and (10,100)

The gradient is thus

gradient = (y1 –y2)/(x1-x2)

                   = (0-100)/(0-10)

                   = –100/-10

                   10

Hence since the gradient a distance-time graph is the speed

speed at 5 s = 10 m/s

What is the gradient and how to determine it?

The gradient of a line or the slope of a line is an indication of how steep a line is at a particular point. The gradient is particularly important in Physics and as a result it is particularly important that you understand how to calculate it perfectly.

Gradient of a straight line

We have in fig 1 below two straight lines that have different slopes. We are going to see how to determine the gradient of both straight lines.

As you can see, there are two straight lines: a blue one and a red one.

It is clear that the red one is steeper than the blue one and as a result its slope would be greater. Consequently the gradient of the red line is greater than that of the blue line.

But do you know how to determine the gradient of the two straight lines mathematically?

Let us look at the red line first and then you are going to look at the blue line.

Loot at the line and the take two coordinated off the line.

I have taken (1,2) and (4,8)

These two coordinates would be considered as (x1,y1) and (x2,y2) such that

From the first coordinate you have x1 =1 and y1=2

from the second coordinate you have x2 =4 and y2 = 8

In order to calculate the gradient of the red line we are going to use the equation below

Hence using the equation above the gradient of the red line is calculated as shown

gradient of red line = (8-2)/(4-1) =6/3 = 2

If you want to check if you can calculate the gradient of the blue line you can do so by repeating the process above.

What you would also find is that any pair of coordinates that you take will give you the same gradient. Which mean that at any point along the straight line the gradient is the same.

Gradient of a curve

The method to calculate the gradient at a point along a curve is slightly different. In that you first have to draw a tangent to the curve at the point that you want to calculate the gradient.

Fig 2

As you can see in fig 2 above, I want to calculate the gradient of the curve at the point (3, 27) as a result I have drawn a tangent to the curve at that point. If you do not know how to draw a gradient I can show you later on in a later post.

Now that you have drawn a tangent at the point that we want (3,27) you will need to choose any two coordinates on the tangent line.

I have chosen (2.5 , 4) and (4 , 60). You can choose any coordinate on the tangent.

Hence using the coordinated below the gradient of the curve at the point (3,27) is

Gradient of curve at point (3,27) = (60 – 4)/(4 - 2.5)= 56/1.5 = 37.3 =37 (2 sf)

As you can see it is easy to calculate the gradient of a straight line or a curve at a particular point.