Gradients of Real Life Graphs Worksheets | Questions and Revision

# Gradients of Real Life Graphs Worksheets, Questions and Revision

Level 4 Level 5

## What you need to know

Real life graphs are graphs representing real things, these can be straight line graphs and curved graphs.

These graphs can represent anything so getting the basic is important.

Make sure you are happy with the following topics before continuing:

## Gradient Represents a rate of change

For many real life graphs the gradient will represent the rate.
Take a look at the examples shown.

What rate does each graph represent?

A: Gradient = Litres PER Second (Rate of flow)

B: Gradient = Cost PER Day (price per day)

C: Gradient = Metres/ second PER second (Acceleration)

NOTE: C is a curved graph so the gradient changes over time, this means we have to use some other techniques to for finding the gradient.

Calculate the average rate of change by drawing a chord (a straight line between two points on the curve), and then calculating the gradient of this chord. This will correspond to the average gradient of the curve between the chosen two points in time.

## Estimating rate at a given point:

We calculate the instantaneous rate of change by drawing a tangent to the curve (a straight line just touching the curve) at the desired point, and then calculating the gradient of this tangent (which can be worked out using standard straight line methods).

This will correspond to the gradient of the curve at that individual point.

Example: The descent of a high altitude weather balloon from $40$ km above sea level is shown on the graph below.

Work out the rate of descent at after $5$ minutes.

To find the decent rate at $5$ minutes we need to find the gradient of the line at that point.

To find the gradient we can draw a tangent lineto the graph at that point forming a triangle with the hypotenuse being the tangent to the line at $x=5$. (5 minutes in)

Hence,

$\dfrac{\text{Change in } y}{\text{Change in } x}= \dfrac{45-18.5}{9-1} =3.31 \text{ km/m}$ (2dp)

### Example Questions

To find each cost corresponding to 40 litres, first locate 40 litres of the axis and draw a straight line up to both lines. Next draw a line straight across to the y-axis.

$\text{Diesel cost} \approx £46.00$

$\text{Petrol cost} \approx £40.50$

Hence the difference in price for a full 40 litre tank of fuel is,

$£46.00 - £40.50 = £5.50$

To find the cost of a single hire, locate Thursday on the x-axis and draw a straight line upwards to meet the line. Tracing across to the y-axis we find,

$\text{Cost per bike} \approx £37.50$

As the group is looking to hire 4 bikes, multiple the cost by four,

$\text{Total cost } = £37.50 \times 4 = £150$

To find the equivalent value in Dollars, locate the value of £8 on the x-axis and trace upwards until meeting the line. Then drawing a horizontal line across to the y-axis we find that,

$£8 = A\ 11.20$

To find the equivalent to £800, we can multiple this value by 100.

$£800 = A\ 1120$

To find the acceleration at 5 seconds after launch we need to find the gradient of the line at that point. To find the gradient we can draw a large triangle with the hypotenuse being the tangent to the line at x=5.

Hence,

$\dfrac{\text{Change in } y}{\text{Change in } x}= \dfrac{45-5}{7-3} =10 \text{ m/s}$

Level 5-7

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