# Consider the curve r(t) = t^2i + 2tj + tk. Find the unit tangent, the principal normal, and write...

## Question:

Consider the curve {eq}r(t) = t^2i + 2tj + tk. {/eq} Find the unit tangent, the principal normal, and write an equation in {eq}x, \ y, \ z {/eq} for the osculating plane at point where {eq}t = 1. {/eq}

## Finding Unit Vectors and Equation of Osculating Plane:

The higher order derivatives are required to find the unit vectors and the equation of the osculating plane at the point.

The derivatives are obtained by differentiating the vector function.

Let us consider the curve {eq}\displaystyle r(t) = t^2 \vec{i} + 2t \vec{j} + t \vec{k} {/eq} at {eq}\displaystyle t = 1 {/eq}.

Finding the unit tangent vector:

{eq}\begin{align*} \displaystyle {r}'(t) &= 2t \vec{i} + 2 \vec{j} + 1 \vec{k} \\ \displaystyle \left \| {r}'(t) \right \| &=\sqrt{(2t)^{2}+(2)^{2}+(1)^{2}} \\ \displaystyle \left \| {r}'(t) \right \| &=\sqrt{4t^2+5} \\ \displaystyle \vec{T}(t) &=\frac{{\vec{r}}'(t)}{\left \| {\vec{r}}'(t) \right \|} \\ \displaystyle \vec{T}(t) &=\frac{2t}{\sqrt{4t^2+5}} \vec{i} + \frac{2}{\sqrt{4t^2+5}} \vec{j} + \frac{1}{\sqrt{4t^2+5}} \vec{k} \\ \displaystyle \vec{T}(1) &=\frac{2(1)}{\sqrt{4(1)^2+5}} \vec{i} + \frac{2}{\sqrt{4(1)^2+5}} \vec{j} + \frac{1}{\sqrt{4(1)^2+5}} \vec{k} \\ \displaystyle \vec{T}(1) &=\frac{2}{3} \vec{i} + \frac{2}{3} \vec{j} + \frac{1}{3} \vec{k} \end{align*} {/eq}

The unit tangent vector at point is {eq}\ \displaystyle \mathbf{\color{blue}{ \vec{T}(1) =\frac{2}{3} \vec{i} + \frac{2}{3} \vec{j} + \frac{1}{3} \vec{k} }} {/eq}.

Finding the principal unit normal vector:

{eq}\begin{align*} \displaystyle \vec{T}'(t) &=\frac{10}{\left(4t^2+5\right)^{\frac{3}{2}}} \vec{i} -\frac{8t}{\left(4t^2+5\right)^{\frac{3}{2}}} \vec{j} -\frac{4t}{\left(4t^2+5\right)^{\frac{3}{2}}} \vec{k} \\ \displaystyle \vec{T}'(1) &=\frac{10}{\left(4(1)^2+5\right)^{\frac{3}{2}}} \vec{i} -\frac{8(1)}{\left(4(1)^2+5\right)^{\frac{3}{2}}} \vec{j} -\frac{4(1)}{\left(4(1)^2+5\right)^{\frac{3}{2}}} \vec{k} \\ \displaystyle \vec{T}'(1) &=\frac{10}{27} \vec{i} -\frac{8}{27} \vec{j} -\frac{4}{27} \vec{k} \\ \displaystyle \left \| \vec{T}'(1) \right \| &=\sqrt{\left(\frac{10}{27}\right)^2+\left(-\frac{8}{27}\right)^2+\left(-\frac{4}{27}\right)^2} \\ \displaystyle \left \| \vec{T}'(1) \right \| &=\frac{2\sqrt{5}}{9} \\ \displaystyle \vec{N}(t) &=\frac{{\vec{T}}'(t)}{ \left \| {\vec{T}}'(t) \right \|} \\ \displaystyle \vec{N}(1) &=\frac{\frac{10}{27}}{\frac{2\sqrt{5}}{9}} \vec{i} -\frac{\frac{8}{27}}{\frac{2\sqrt{5}}{9}} \vec{j} -\frac{\frac{4}{27}}{\frac{2\sqrt{5}}{9}} \vec{k} \\ \displaystyle \vec{N}(1) &=\frac{\sqrt{5}}{3} \vec{i} -\frac{4}{3\sqrt{5}} \vec{j} -\frac{2}{3\sqrt{5}} \vec{k} \end{align*} {/eq}

The principal unit normal vector is {eq}\ \displaystyle \mathbf{\color{blue}{ \vec{N}(1)=\frac{\sqrt{5}}{3} \vec{i} -\frac{4}{3\sqrt{5}} \vec{j} -\frac{2}{3\sqrt{5}} \vec{k} }} {/eq}.

Finding the cross product of first and second order derivatives:

{eq}\begin{align*} \displaystyle r(t) &= t^2 \vec{i} + 2t \vec{j} + t \vec{k} \\ \displaystyle r(1) &= (1)^2 \vec{i} + 2(1) \vec{j} + 1 \vec{k} \\ \displaystyle r(1) &= 1 \vec{i} + 2 \vec{j} + 1 \vec{k} \\ \displaystyle {r}'(1) &= 2 \vec{i} + 2 \vec{j} + 1 \vec{k} \\ \displaystyle {r}''(1) &= 2 \vec{i} + 0 \vec{j} + 0 \vec{k} \\ \displaystyle {r}'(1) \times {r}''(1) &=\begin{vmatrix} i & j & k\\ 2 & 2 & 1\\ 2 & 0 & 0 \end{vmatrix} \\ \displaystyle &=((0)(2)-(0)(1))\vec{i}-((0)(2)-(2)(1))\vec{j}+((0)(2)-(2)(2))\vec{k} \\ \displaystyle {r}'(1) \times {r}''(1) &= 0 \vec{i} +2 \vec{j} -4 \vec{k} \end{align*} {/eq}

Finding an equation of the osculating plane at point:

{eq}\begin{align*} \displaystyle \vec{n} \cdot (x-x_{0}, y-y_{0}, z-z_{0}) &=0 \\ \displaystyle (0, 2, -4) \cdot (x-1, y-2, z-1) &=0 \\ \displaystyle (0)(x-1)+(2)(y-2)+(-4)(z-1) &=0 \\ \displaystyle 2y-4-4z+4 &=0 \\ \displaystyle 2y-4z-4+4 &=0 \\ \displaystyle 2y-4z &=0 \end{align*} {/eq}

An equation of the osculating plane is {eq}\ \displaystyle \mathbf{\color{blue}{ 2y-4z =0 }} {/eq}.