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Find the unit tangent and unit normal vectors T(t) and N(t).

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.

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