This question aims to find the **unit tangent** and **unit normal vectors** **T(t)** and **N(t)** when **r(t)** is given as

$ < t, 3cost, 3sint > $

The **unit tangent vector** is the unit vector that is directed towards the velocity vector if the differentiable vector-valued function is r(t) and **v(t) = r’(t)** is the velocity vector. The new vector-valued function is tangent to the defined curve.

The vector that is perpendicular to the unit tangent vector T(t) is called the **unit normal vector**. It is represented by **N(t)**.

## Expert Answer

The given equation is:

\[ r ( t ) = < t, 3 cos t, 3 sin t > \]

By taking the first derivative of the given equation **curve-component wise**:

\[ | r’ ( t ) | = \sqrt { 1 ^ 2 + ( – 3 sin t ) ^ 2 + ( 3 cos t ) ^ 2} \]

\[ | r’ ( t ) | = \sqrt { 10 } \]

We will use $ \sqrt { 10 } $ in the form of a fraction and keep it outside the equation to ease the simplification of the unit tangent vector.

The unit tangent vector can be found by:

\[ \tau ( t ) = \frac { r’ ( t ) } { | r’ ( t ) | } = \frac { 1 } { \sqrt {10} } . < t, -3 sin t, 3 cos t > \]

The derivative of this unit tangent vector can be found by:

\[ \tau’ ( t ) = \frac { 1 } { \sqrt {10} } < 0, – 3 cos t, -3 sin t > \]

Taking **3** common:

\[ \tau’ ( t ) = \frac { 3 } { \sqrt {10} } < 0, – cos t, – sin t > \]

The magnitude of $\tau$ can be calculated by:

\[ | \tau’ ( t ) | = \sqrt {(\frac {3}{\sqrt{10}})^2 . (( -cost)^2+ (-sint)^2)}\]

\[ = \frac {3}{\sqrt{10}}. \sqrt{sin^2 t + cos ^ 2 t } \]

\[ = \frac {3}{\sqrt{10}}( 1 )\]

\[ = \frac {3}{\sqrt{10}} \]

By calculating and simplifying the unit normal vector:

\[ N ( t ) = \frac { \tau’ ( t ) } { | \tau’ ( t ) |} \]

\[ = \frac {\frac {3}{\sqrt{10}} . < 0, – cos t, – sin t > } { \frac {3}{\sqrt{10}}} \]

\[ = < 0, – cos t, – sin t > \]

## Numerical Results

**The magnitude of unit tangent vector is $ \frac {3}{\sqrt{10}}$ and the unit normal vector is $< 0, – cos t, – sin t >$.**

## Example

Find the **magnitude of unit tangent vector** when the given equation is $ r ( t ) = < t^2, \frac{2}{3} t^3, t > $ and the point $ < 4, \frac{-16}{3}, -2 > $ occurs at $ t = -2 $.

By finding the derivative:

\[ R’(t) = <2t,2t^2,1> \]

\[ |R’(t)|= \sqrt{ (2t)^2 + (2t^2)^2 + 1^2 }\]

\[ = \sqrt { 4t^2 + 4t^4 + 1 } \]

\[ = \sqrt { ( 2t^2 + 1 )^2 } \]

\[ = 2t^2 + 1 \]

By finding tangent vector:

\[\tau(t)= \frac{R’(t)}{|R’(t)|}\]

\[\tau(t)= \frac{1}{2t^2+1}<2t, 2t^2, 1>\]

\[\tau(-2)= \frac{1}{2(-2)^2+1}<2(-2), 2(-2)^2, 1>\]

\[ = <\frac{-4}{9},\frac{8}{9},\frac{1}{9}> \]

\[|T’(t)| = < \frac{2-4t^2}{(2t^2+1)^2},\frac{4t}{(2t^2+1)^2},\frac{-4t}{2t^2+1)^2}>\]

\[= \sqrt{\frac{(2-4t^2)^2+(4t)^2+(-4t)^2}{(2t^2+1)^4 }}\]

\[ = \frac{1}{2t^2+1)^2} . \sqrt{16t^4+16t^2+4}\]

\[ |T’(t)| = \frac{2}{2t^2+1)}\]

*Image/Mathematical drawings are created in Geogebra**.*

**Previous Question**** < >** **Next Question**

**Previous Question**

**< >**

**Next Question**