Calculate the unit tangent vector for the vector function \( \vec{r}(t) = t\vhat{i} + (1/t)\vhat{j} \) at the point \(t=1\).

\( \vec{r}'(t) = \vhat{i} - (1/t^2)\vhat{j} \)
\( \vec{r}'(1) = \vhat{i} - \vhat{j} \)
\(\displaystyle{ \vec{T}(1) = \frac{\vec{r}'(1)}{\| \vec{r}'(1) \|} = \frac{\vhat{i}-\vhat{j}}{\sqrt{2}} }\)
Here is a plot of the solution. The black line is the curve \( \vec{r}(t) \) with the black arrow indicating that it is being traced out left to right (or down the curve). (We have plotted only a section of the curve from x=1/4 to x=4.) The red vector is the unit tangent vector. Notice that it is pointing in the direction that the curve is being traced.

\(\displaystyle{ \vec{T}(1) = \frac{\vhat{i} - \vhat{j}}{\sqrt{2}} }\)

Calculate the unit tangent vector for the vector function \( \vec{r}(t) = \cos t \vhat{i} + 3t\vhat{j} + \) \( 2\sin 2t \vhat{k} \) at the point \(t=0\).

\( \vec{T}(0) = (3/5)\vhat{j} + (4/5)\vhat{k} \)

Calculate the unit tangent vector for the vector function \( \vec{r}(t) = \langle 2\sin(t), 4\cos(t), 4\sin^2(t) \rangle \) at the point \(t=\pi/6\).

\(\displaystyle{ \frac{1}{\sqrt{19}} \langle \sqrt{3}, -2, 2\sqrt{3} \rangle }\)

Calculate the unit tangent vector for the vector function \(\vec{r}(t) = \langle t^3,2t^2 \rangle \) at the point \(t=1\).

\( \langle 3/5,4/5 \rangle \)

Find \(\vec{T}(t)\) and \(\vec{T}(0)\) for \(\vec{r}(t)=\langle 5t^2+1, -e^{-3t}, 2\sin(-3t) \rangle \).

\(\displaystyle{ \vec{T}(t) = \frac{\langle 10t, 3e^{-3t}, -6\cos(-3t) \rangle}{\sqrt{100t^2+9e^{-6t}+36\cos^2(-3t)}} }\)
\( \vec{T}(0) = \langle 0,1/\sqrt{5},-2/\sqrt{5} \rangle \)

Calculate the unit tangent vector for the vector function \( \vec{r}(t) = (-t^3+t)\vhat{i} + \) \( (\ln(t^2))\vhat{j} + (\cos(\pi t))\vhat{k} \) at the point \( t=1 \).

\( \langle -\sqrt{2}/2, \sqrt{2}/2, 0 \rangle \)

Calculate the unit tangent vector for the vector function \( \vec{r}(t) = \langle t\sqrt{2}, e^t, e^{-t} \rangle \) at the point \( t=0 \).

\( \vec{T}(0) = \langle \sqrt{2}/2, 1/2, -1/2 \rangle \)

The curves \(\vec{r}_1(t) = \langle 5t+2, \ln t, \sqrt{4-2t-t^2} \rangle\) and \(\vec{r}_2(t) = \langle 7, s^{3/2}\sqrt{2}, e^{s^2-2s} \rangle\) intersect at a point. Find \(\cos\theta\) where \(\theta\) is the angle of intersection (the angle between their respective tangent lines).

Remember the angle between two vectors \(\vec{u}\) and \(\vec{v}\) can be found using \(\displaystyle{ \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{ \norm{\vec{u}} \norm{\vec{v}} } }\)