Unit tangent vector in vector analysis. Today it’s all about the unit tangent vector in vector analysis. Have a look!!

**Unit tangent vector**

**Examples of unit tangent vector**

Note: None of these examples are mine. I have chosen these from some 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 (2011)*¹ “Determine the unit tangent vector at the point for the curve with parametric equations ; ; .”

**Solution**

Now here the given curve with parametric equations is: and

Also, I need to determine the equation of the tangent vector at the point

So I’ll start with the differentiation of vectors.

**Step 1**

Now the parametric equations of the curve are: and

Thus the vector equation of the curve will be, say

Next I’ll differentiate this vector with respect to to get

So this means

Now I’ll simplify it further to get

Then comes my second step.

**Step 2**

Next, I will find out the value of at the point This is because I have to get the unit tangent vector at that point.

As before, I already know that the parametric equations of the curve are and

Also -coordinate of the given point is 2. Therefore I can say

Next, I compare -coordinates. Thus it will be .

In the end, I will compare -coordinates to get

Thus I can say that I have got three equations

(1)

(2)

(3)

in .

Now I’ll solve them to get the value of .

From equation (2), I can say that

Also is satisfied by equations (1) and (3) as well.

Next, I’ll get the equation of the unit tangent vector.

**Step 3**

Now at , the equation of the tangent vector will be the value of at .

Now from Step 1, I already know that

So at , it will be

Thus the magnitude of the unit vector will be

Therefore the equation of the unit tangent vector will be

Hence I can conclude that this is the answer to the first example.

Now I’ll go to my next example.

**Example 2**

According to Kreyszig (2005)*² “Given a curve find a tangent vector , a unit tangent vector ,..at .

**Solution**

Now here the given curve is .

Thus the vector equation of the curve will be

So my first step will be to differentiate the curve with respect to .

**Step 1**

Thus I’ll differentiate this vector with respect to to get

So this means

Next I’ll simplify it further to get

Then comes my second step.

**Step 2**

Now I will find out the value of at the point And the reason is that I have to get the unit tangent vector at that point.

As before, I already know that the parametric equations of the curve are and

Also -coordinate of the given point is 3. Therefore I can say

Next, I compare -coordinates. Thus it will be .

In the end, I will compare -coordinates to get

Thus I can say that I have got three equations

(4)

(5)

(6)

in .

Now I’ll solve them to get the value of .

Thus from equation (6), I can say that .

Also is satisfied by equations (4) and (5) as well.

Next, I’ll get the equation of the unit tangent vector. So for that, I will get the equation of the tangent vector first.

**Step 3**

Now at , the equation of the tangent vector will be the value of at .

As I already know from Step 1, the value of is

So at , it will be

Thus the magnitude of the unit vector is

Therefore the equation of the unit tangent vector will be

This is the answer to the last example.

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!!

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