Hello friends, today it’s all about the tangent, normal to any curve. Have a look!!

### Tangent, normal to any curve

Well, it’s an old topic from high school. So I’ll not go into much detail.

Suppose is the equation of any curve.

The equation of the tangent of this curve at the point is

Similarly, the equation of the normal to this curve at the point is

Now I will solve some problems for you.

#### Examples on the tangent, normal to any curve

Disclaimer: None of these examples are mine. I have chosen these from some book or books. I have also given the due reference at the end of the post.

So here is the first example.

##### Example 1

According to Stroud and Booth (2013)* “Find the equations of the tangent and normal to the curve at the point (6, 4).”

##### Solution

Here the given curve is

To find the equations of the tangent and normal to this curve at the point (6, 4), first of all, I’ll find out the value of at this point.

###### Step 1

Now I’ll differentiate both sides of the given equation with respect to to get

Thus the value of at the point (6, 4) is

Now I’ll go to the next step.

###### Step 2

As per the formula, the equation of the tangent to the curve at the point (6, 4) is

Now I will simplify it further to get

This is the equation of the tangent.

Now I’ll get the equation of the normal.

###### Step 3

As per the formula, the equation of the normal to the curve at the point (6, 4) is

Now I will simplify it further to get

This is the equation of the normal.

Thus I can conclude that the equations of the tangent and normal to the curve at the point (6, 4) are and respectively.

This is the answer to the given example.

Now I will go to the second problem.

##### Example 2

According to Stroud and Booth (2013)* “If find the equation of the tangent at ”

##### Solution

Here the given parametric curves are

(1)

and

(2)

To find the equation of the tangent at first of all, I’ll find out the value of at

###### Step 1

Now I’ll differentiate both sides of equation (1) with respect to to get

Next I’ll differentiate both sides of equation (2) with respect to to get

Thus will be

At will be

Now I’ll get the values of and at

###### Step 2

First, I’ll substitute in the equation (1) to get the value of at . Thus it will be

Next, I’ll substitute in the equation (2) to get the value of at . Thus it will be

Therefore I can say that for , the point will be

###### Step 3

Now I’ll get the equation of the tangent at the point Thus it will be

Next, I’ll simplify it to get

Thus I can say that the equation of the tangent at is

This is the answer to this problem.

Dear friends, this is the end of my today’s post. Thank you very much for reading this. Please let me know how you feel about it. Soon I will be back again with a new post. Till then, bye, bye!!

###### *Reference: K. A. Stroud and Dexter J. Booth (2013): Engineering mathematics, Industrial Press, Inc.; 7th Edition (March 8, 2013), Chapter: Tangents, normals and curvature, Test exercise 9, p. 602, Q. No.s 2 (Example 1), 4 (Example 2).

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