A Level Maths: Tangents And Normals
Introduction
In this article we are going to look at A Level Maths: Tangents and Normals. You will need to have a very basic knowledge and understanding of differentiation.
We will also consider the second derivative where you are simply differentiating twice and you will see the notation that is used and how it can be used to determine the nature of a turning point.
A Level Maths: Tangents And Normals - Tangents
Consider the general curve y=f(x)
Now a tangent has been drawn at a particular point to the curve. We have seen from Differentiation 1 that from the gradient function of a curve we can determine the gradient value. The gradient value at a particular point is also the gradient of a tangent at that same point.
We can then use the rules for straight line geometry to determine its equation.
A Level Maths: Tangents And Normals - Examples
Example
What is the equation of the tangent to the curve y=x^2+3 x+2 at the point (2 , 12)?
Solution
In order to proceed with a question like this we should be thinking about what is needed. Because we are to find the equation of tangent which is nothing more than a straight line one of the essential requirements that will be needed is the gradient of that line and that is something that we can do quite easily.
\frac{d y}{d x}=2 x+3
At the point x=2, \frac{d y}{d x}=2(2)+3=7 🡨 this is the gradient value and it is also the gradient of the tangent at the point x=2.
Because we know that the tangent makes contact with the curve at (2, 12) we can now find its equation as follows:
Using y=m x+c we have: 12=(7)(2)+c \rightarrow c=-2
So the equation of the tangent is y=7 x-2
A Level Maths: Tangents And Normals – Normals
If we consider the curve y=f(x) then as from above the red line shows the tangent and the blue dashed line is the normal. The normal is perpendicular to a tangent. This then means that the gradient of a normal at a particular point is perpendicular to the tangent at that particular point.
To determine the gradient of the tangent we are simply using the test for perpendicular gradients i.e m_1 \times m_2=-1
The above rule for perpendicular lines you would have met at GCSE and also whilst doing straight line geometry during year 1 A Level Maths course.
Example
What is the equation of the normal to the curve y=x^3-2 x+1 at the point (2,5)?
Solution
First we need to find the gradient function of the curve by finding \frac{d y}{d x}
\frac{d y}{d x}=3 x^2-2At the point x=2 we have \frac{d y}{d x}=10 🡨 this is the gradient value of the tangent.
This then means that the gradient value of the normal will be -\frac{1}{10}.
We can now continue to find the equation of the straight line i.e. the normal.
Using y=m x+c we have: 5=-\frac{1}{10}(2)+c \rightarrow c=\frac{26}{5}
Our equation of the normal is y=-\frac{1}{10} x+\frac{26}{5}
Generally it is good idea not to leave an answer in terms of a fraction in case further work is needed and we make this tidier by multiplying throughout by 10 to give: 10 y+x=52
If you are looking for some additional help just complete the contact form below and we will get back to you within 24 hours.