Find the curve’s unit tangent vector. Also, find the length of the indicated portion of the curve.

$r(t) = (2cost)i + (2sint)j + \sqrt{5} k 0 \leq t \geq \pi$

This problem aims to familiarize us with differential curves and their unit tangent vectors. The problem holds the background of calculus and it’s important to recall the concepts of arc length parameter and tangent vector.

If we look at arc length, it is the absolute distance between two points along a portion of a curve. Another term that is most commonly used is the rectification of curve, which is the length of an uneven arc segment defined by approximating the arc segment as small interconnected line segments.

The unit tangent vector is the derivative of a vector-valued function that provides a unique vector-valued function that is tangent to the specified curve. In order to obtain the unit tangent vector, we require the absolute length of the tangent vector whereas the analog to the slope of the tangent line is the direction of the tangent line.

The formula to find the curve’s unit tangent vector is:

$T = \dfrac{v}{|v|}$

And the formula to find the length of the indicated portion of the curve can be written as:

$L = \int_a^b |v| dt$

So both the formulas requires $v$, and the formula to find $v$ is thus:

$v = \dfrac{dr}{dt}$

Therefore, putting the value of &r& and differentiating with respect to &dt& to find $v$:

$v = \dfrac{d}{dt} ((2cost)i + (2sint)j + \sqrt{5} k)$

$v$ comes out to be:

$v = (-2sint)i + (2cost)j + \sqrt{5} k$

Taking the magnitude $|v|$:

$|v| = \sqrt { (-2sint)^2 + (2cost)^2 + (\sqrt {5})^2 }$

$= \sqrt { 4sin^2 t + 4cos^2 t + 5 }$

$= \sqrt { 4(sin^2 t + cos^2 t) + 5 }$

Using the property $sin^2 t + cos^2 t = 1$:

$= \sqrt { 4(1) + 5 }$

$|v|$ comes out to be:

$|v| = 3$

Inserting the values of $v$ and $|v|$ to the tangent vector’s formula:

$T = \dfrac{v}{|v|} = \dfrac{(-2sint)i + (2cost)j + \sqrt{5} k} {3}$

$T = \dfrac{-2sint}{3}i + \frac{2cost}{3}j + \dfrac{\sqrt{5}} {3} k$

Now solving for $L$:

$L = \int_a^b |v| dt = \int_0^\pi 3dt$

$= [3t]_0^\pi = 3(\pi) – 3(0)$

$L = 3\pi$

Numerical Result

$T = \dfrac{-2sint}{3}i + \frac{2cost}{3}j + \dfrac{\sqrt{5}} {3} k$

$L = 3\pi$

Example

Find the unit tangent vector of the curve. Also, find the indicated portion of the curve’s length.

$r(t) = ti + \dfrac{2}{3}t^{3/2} 0 \leq t \geq 8$

$v = \dfrac{d}{dt} ( ti + \dfrac{2}{3}t^{3/2} )$

$v = i + t^{1/2}k$

$|v| = \sqrt { (1)^2 + (t^{1/2})^2 } = \sqrt{1+t}$

$T = \dfrac{v}{|v|} = \dfrac{i + (t^{1/2}) k } { \sqrt{1+t} }$

$T = \dfrac{1} { \sqrt{1+t} }i + \dfrac{ t^{1/2}} {\sqrt{1+t}} k$

Now solving for $L$:

$L = \int_a^b |v| dt = \int_0^8 \sqrt{1+t} dt$

$= \left( \dfrac{2}{3} (1+t)^{3/2} \right)_0^8$

$L = \dfrac{52}{3}$