# Find the unit tangent vector T(t) and find a set of parametric equations for the line of the...

## Question:

Find the unit tangent vector {eq}T(t){/eq} and find a set of parametric equations for the line of the tangent to the curve at point, {eq}P{/eq}.

(a) {eq}r(t) = 4 cos(t)\vec{i} + 4 sin(t)\vec{j}, P(2, \sqrt{2}, 2 \sqrt{2}){/eq}

(b) {eq}r(t) = t^2\vec{i}+t\vec{j}+\frac{4}{3}\vec{k}, P(1,1,\frac{4}{3}){/eq}

## Parametric Curves

The tangent line to a parametric curve at a point, {eq}\displaystyle t_0 \text{ is } \mathbf{r}(t_0)+t\ \mathbf{r}'(t_0), t\in \mathbf{R}. {/eq}

The unit tangent vector to a parametric curve is {eq}\displaystyle \mathbf{T}(t)=\frac{\mathbf{r}'(t_0)}{|\mathbf{r}'(t_0)|}. {/eq}

(a) To find the unit tangent vector for the curve {eq}\displaystyle \mathbf{r}(t)=\langle 4\cos t, 4\sin t\langle, {/eq}

we need the derivative vector, first.

{eq}\displaystyle \begin{align}\mathbf{r}'(t)&=\langle -4\sin t, 4\cos t\langle \implies |\mathbf{r}'(t)|=\sqrt{16\sin^2 t+16\cos^2 t}=4, &\left[\text{ using the identity }\sin^2 t+\cos^2 t=1\right]\\ \mathbf{T}(t)&=\frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}=\boxed{\langle -\sin t, \cos t\rangle}. \end{align} {/eq}

The tangent line to the curve at the point {eq}\displaystyle P(2\sqrt{2}, 2\sqrt{2}) {/eq}, which corresponds to the parameter {eq}\displaystyle t=\frac{\pi}{4}, {/eq}

is given by {eq}\displaystyle \mathbf{r}(t)=\langle 2\sqrt{2}, 2\sqrt{2}\rangle +t\mathbf{T}\left(\frac{\pi}{4}\right)=\langle 2\sqrt{2}, 2\sqrt{2}\rangle +t\left\langle -\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2} \right\rangle \text{ or } \boxed{\langle 2\sqrt{2}, 2\sqrt{2}\rangle +\frac{t\sqrt{2}}{2}\left\langle -1,1 \right\rangle, t\in\mathbf{R}}. {/eq}

(b) To find the unit tangent vector for the curve {eq}\displaystyle \mathbf{r}(t)=\left\langle t^2, t, \frac{4}{3}\right\rangle, {/eq}

we will need the derivative vector.

{eq}\displaystyle \begin{align}\mathbf{r}'(t)&=\langle 2t, 1, 0\langle \implies |\mathbf{r}'(t)|=\sqrt{4t^2+1}\\ \mathbf{T}(t)&=\frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}=\boxed{\frac{1}{4t^2+1}\langle 2t,1,0\rangle}. \end{align} {/eq}

The tangent line to the curve at the point {eq}\displaystyle P\left(1, 1, \frac{4}{3}\right) {/eq}, which corresponds to the parameter {eq}\displaystyle t=1, {/eq}

is given by {eq}\displaystyle \mathbf{r}(t)=\left\langle 1, 1, \frac{4}{3}\right\rangle +t\mathbf{T}\left(1\right)=\boxed{\left\langle 1, 1, \frac{4}{3}\right\rangle +\frac{t}{\sqrt{5}}\left\langle 2,1,0 \right\rangle , t\in\mathbf{R}}. {/eq} 