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

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.

## Expert Answer

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 $L$:

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

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

\[L = \dfrac{52}{3} \]