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Find the principal unit normal vector to the curve at the specified value of the parameter: R(t) = ti + (4/t)j where t=2

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:

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

\[ At\ t=1 \]

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

\[ At\ t=3 \]

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

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