The question aims to find the **unit normal vector** to the curve at the specified value of the **parameter.**

The question is based on the concept of **vector geometry, tangent line, and normal vector.** The **tangent line** is defined as a line that only passes through one point of the **curve.** The **normal vector** is the vector that is **perpendicular** to vectors, curves, or planes. The **unit normal vector** is that normal vector that has a **magnitude** of $1$.

## Expert Answer

The **unit normal vector** can be found by finding the **tangent unit vector** of the given equation and then finding the unit vector of its **derivative. **The given equation is given as:

\[ R(t) = ti + \dfrac{4}{t} j, \hspace{0.4in} where\ t = 2 \]

Taking the **derivative** of this equation and finding its unit vector will give us the **tangent vector.** The equation of the tangent vector is the unit vector of the derivative of the given equation, which is given as:

\[ T(t) = \dfrac{R'(t)}{|| R'(t) ||} \hspace{0.5in} (1) \]

Taking the **derivative** of the given equation:

\[ R'(t) = \dfrac{d}{dt} (ti + \dfrac{4}{t} j) \]

\[ R'(t) = i . \frac{d}{dt}t + 4j . \frac{d}{dt} [\frac{1}{t}] \]

\[ R'(t) = i\ -\ 4j . \dfrac{\frac{d}{dt}t}{t^2} \]

\[ R'(t) = i\ -\ \dfrac{4j}{t^2} \]

Finding the **magnitude** of the derivative of the given equation:

\[ || R'(t) || = \sqrt{ (1)^2 + (- \dfrac{4}{t^2})} \]

\[ || R'(t) || = \sqrt{1 + (\dfrac{16}{t^4})} \]

\[ || R'(t) || = \sqrt{\dfrac{t^4 + 16}{t^4}} \]

\[ || R'(t) || = \dfrac{1}{t^2} \sqrt{t^4 + 16} \]

Putting the values in equation $(1)$ will give us:

\[ T(t) = \dfrac{i\ -\ \dfrac{4j}{t^2}}{\dfrac{1}{t^2} \sqrt{t^4 + 16}} \]

\[ T(t) = \dfrac{t^2 (i\ -\ \dfrac{4j}{t^2})}{\sqrt{t^2 + 16}} \]

\[ T(t) = \dfrac{t^2}{\sqrt{t^2 + 16}} i\ -\ \dfrac{4}{\sqrt{t^2 + 16}} j \]

This equation gives us the **tangent vector** of the given equation. To find its unit normal vector, we again take its derivative and find its magnitude to find its unit vector. The equation is given as:

\[ N(t) = \dfrac{T'(t)}{ || T'(t) || } \hspace{0.5in} (2) \]

Taking the **derivative** of the **tangent line** equation:

\[ T'(t) = \dfrac{d}{dt} \bigg{(} \dfrac{t^2}{\sqrt{t^2 + 16}} i\ -\ \dfrac{4}{\sqrt{t^2 + 16}} j \bigg{)} \]

Solving the derivative will give us:

\[ T'(t) = \dfrac{t^3 + 32t}{\sqrt{(t^2 +16)^3}} i + \dfrac{4t}{\sqrt{(t^2 +16)^3}} j \]

Finding its **magnitude** by the **distance formula,** we get:

\[ || T'(t) || = \sqrt{\Big{(} \dfrac{t^3 + 32t}{\sqrt{(t^2 +16)^3}} \Big{)}^2 + \Big{(} \dfrac{4t}{\sqrt{(t^2 +16)^3}} \Big{)}^2} \]

Solving the equation we get:

\[ || T'(t) || = \dfrac{t \sqrt{t^4 + 64t^2 + 1040}}{\sqrt{t^2 + 16}} \]

The equation $(2)$ becomes:

\[ N(t) = \dfrac{(t^3+32t)i + (4t)j}{(t^3+16t)\sqrt{t^4+64t^2+1040}} \]

This is the **unit normal vector** at $t$. For a given value of $t$, we can calculate the vector as:

\[ At\ t = 2 \]

\[ N(2) = \dfrac{((2)^3+32(2))i + (4(2))j}{((2)^3+16(2)\sqrt{(2)^4+64(2)^2+1040}} \]

## Numerical Result

Simplifying the equation, we get the **unit normal vector:**

\[ N(2) = \dfrac{8}{160\sqrt{82}} (9i + j) \]

## Example

Find the **unit normal vector** at $t=1$ and $t=3$. The unit normal vector is given as:

\[ N(t) = \dfrac{(t^3+32t)i + (4t)j}{(t^3+16t)\sqrt{t^4+64t^2+1040}} \]

\[ At\ t=1 \]

\[ N(1) = \dfrac{33}{17\sqrt{1105}}i + \dfrac{4}{17\sqrt{1105}}j \]

\[ At\ t=3 \]

\[ N(3) = \dfrac{1}{33521} (123i + 12j) \]