Whether it is to complete geometrical work on circles or find gradients of curves, being able to construct and use tangents as well as work out the area under graphs are useful skills in mathematics.

Part of

To construct the tangent to a curve at a certain point A, you draw a line that follows the general direction of the curve at that point. An example of this can be seen below.

Once the tangent is found you can use it to find the gradient of the graph by using the following formula:

\[\text{Gradient to the curve =}~\frac {y_2-y_1} {x_2-x_1}\]

where \(({x_1,~y_1})\) and \(({x_2,~y_2})\) are any two points on the tangent to the curve.

- Question
Estimate the gradient to the curve in the graph below at point A.

First draw the tangent at the point given.

Select any two points on the tangent. The coordinates that we are using are (1, 0) and (2.5, 2000). Then use the formula below:

\[\text{Gradient to the curve =}~\frac {y_2-y_1} {x_2-x_1}\]

\[\frac {2000~-~0} {2.5~-~1}~=~\frac {2000} {1.5}~=~1333.33\]

It is useful to remember that all lines and curves that slope upwards have a positive gradient.

All lines and curves that slope downwards have a negative gradient.

We want to find the gradient of the curve at \(\text{x = -2}\).

First draw the tangent at \(\text{x = -2}\).

Select any two points on the tangent. The coordinates that we are using are (-4, 9) and (0, -3).

\[\text{Gradient to the curve =}~\frac {y_2-y_1} {x_2-x_1}\]

\(\frac {y_2-y_1} {x_2-x_1}\) where \(({x_1,~y_1})~=~({-4},{~9})\) and \(({x_2,~y_2})~=~({0},~{-3})\) are any two points on the tangent to the curve.

Gradient to the curve \(= \frac {-3~-9} {0~-~(-4)}~=~\frac {-12} {4}~=~{-3}\)

If you are finding the gradient to the curve of a distance-time graph then you are calculating the velocity that the object is moving at that particular time.

- Question
The following graph shows the car journey from Chelsea’s house to her mother’s house. Estimate the velocity of the car at \(\text{t = 6.5 s}\). The tangent has been drawn for you.

\[\text{Gradient to the curve =}~\frac {y_2-y_1} {x_2-x_1}\]

\[\frac {140~-~20} {9~-~4}~=~\frac {120} {5}~=~24~ \text{m/s}^{2}\]

Similarly, if you are finding the gradient to the curve of a velocity-time graph, then you are calculating the acceleration of the object at that particular time.