MYTUTOR SUBJECT ANSWERS

1273 views

How do I find the equation of the normal line given a point on the curve?

The normal line to a curve at a particular point is the line through that point and perpendicular to the tangent.

A simple trick to remembering how to find the normal gradient, n, is that the slope of any line perpendicular to a line that has a gradient, m, is just the negative reciprocal, -1/m.

Example:

Find the normal gradient to the curve y=2x3 +3x+7 at the point (1,1).

So firstly, let’s recap how to calculate the gradient of the tangent line:

By differentiating y=2x3 +3x+7 , we find

dy/dx = 6x2 +3

Then, by substituting in our point, at x=1 we yield dy/dx=9. This is the tangent gradient of the curve (m=9).

Finally we substitute this into our formula for calculating the normal gradient n=-1/m.

Therefore n=-1/9.

 

Now, let’s try another example which demonstrates how we use the normal gradient to find an equation for the normal line.

We will use the formula (y-y0) = n(x-x0), where (x0,y0) is a given point.

Example:

Consider a curve y=x5+3x2 +2. Find the equation of the normal to the curve at the point (-1,2). Leave your answer in the form y=mx+c.

By differentiating the curve, we have dy/dx = 5x4 +6x.

To find the gradient of the tangent line we substitute in x=-1, which yields

dy/dx = 5(-1)4 +6(-1)

                = 5-6

                =-1 = m

Therefore, we know that the normal gradient is n=-1/m

So n=1

 

Finally, we substitute this into our formula for the normal line (y-y0) = n(x-x0):

In our example, (x0, y0) = (-1,2)

 

So   y-2 = 1(x+1)

And my rearranging, we find y = x+3.

Joy M. GCSE Maths tutor, A Level Maths tutor

2 years ago

Answered by Joy, an A Level Maths tutor with MyTutor


Still stuck? Get one-to-one help from a personally interviewed subject specialist

298 SUBJECT SPECIALISTS

£26 /hr

Tom W.

Degree: Mathematics (Masters) - Warwick University

Subjects offered:Maths

Maths

“Third year Maths undergraduate, with experience teaching Maths from GCSE to first year.”

£20 /hr

Felix A.

Degree: Mathematics with Finance (Bachelors) - Leeds University

Subjects offered:Maths

Maths

“About me! I am currently a Maths student at the University of Leeds, and I have always been passionate about Maths. I hope that after lessons with me you will feel the same and will have improved! I am communicative and friendly and ha...”

MyTutor guarantee

£20 /hr

Shaun F.

Degree: Mathematics with International Study (Masters) - Exeter University

Subjects offered:Maths, Further Mathematics

Maths
Further Mathematics

“Hi, I'm Shaun! I'm currently studying for a masters in Mathematics at the University of Exeter..”

MyTutor guarantee

About the author

Joy M.

Currently unavailable: for regular students

Degree: Mathematics (Bachelors) - Bristol University

Subjects offered:Maths

Maths

“Second Year Mathematics Undergraduate at the University of Bristol. Passionate about conveying my enthusiasm for Maths.”

MyTutor guarantee

You may also like...

Other A Level Maths questions

How do you differentiate by first principles?

How do I find the cartesian equation for a curve written in parametric form?

Prove that the d(tan(x))/dx is equal to sec^2(x).

Differentiate y= exp(cos^2(x)+sin^2(x)) by using the chain rule.

View A Level Maths tutors

We use cookies to improve your site experience. By continuing to use this website, we'll assume that you're OK with this. Dismiss

mtw:mercury1:status:ok