This is really a homework problem for which I'd like a fresh set of eyes to look the solution over. The math isn't working out right. This means I've done something incorrect. Without further ado.
The problem is the find the angle between the velocity vector and the acceleration vector of the given vector function at time $t=0$. The formula for getting the angle between two vectors is $\theta = cos^{-1}(\frac{\mathbf{u} \cdot \mathbf{v}}{\left| \mathbf{u}\right|\cdot \left|\mathbf{v}\right|})$.
Ok, not a big deal. So, I differentiate $\mathbf{r}$ twice to get $\mathbf{v}$ and $\mathbf{a}$. Here's what I've got:
$$
\begin{array}{rcll}
\mathbf{r}(t) & = & \frac{4}{9}(1+t)^{3/2}\mathbf{i} + \frac{4}{9}(1-t)^{3/2}\mathbf{j}+\frac{1}{3}t\mathbf{k} \\
\frac{d\mathbf{r}}{dt} & = & \frac{2}{3}(1+t)^{1/2}\mathbf{i}+\frac{2}{3}(1-t)^{1/2}\mathbf{j}+\frac{1}{3}\mathbf{k} = \mathbf{v}(t)\\
\frac{d\mathbf{v}}{dt} & = & \frac{1}{3}(1+t)^{-1/2}\mathbf{i}+\frac{1}{3}(1-t)^{-1/2}\mathbf{j} = \mathbf{a}(t)\\
\end{array}
$$
Ok, so if that's correct, then the following should be reasonable:
$$
\begin{array} {rcl}
\mathbf{v}(0) & = & \frac{2}{3}(1+0)^{1/2} \mathbf{i} + \frac{2}{3}(1-0)^{1/2}\mathbf{j} + \frac{1}{3}\mathbf{k} \\
& = & \frac{2}{3}\mathbf{i} + \frac{2}{3}\mathbf{j} + \frac{1}{3}\mathbf{k} \\
\mathbf{a}(0) & = & \frac{1}{3}(\frac{1}{(1+0)^{1/2}})\mathbf{i} + \frac{1}{3}(\frac{1}{(1-0)^{1/2}})\mathbf{j} \\
& = & \frac{1}{3}\mathbf{i} + \frac{1}{3}\mathbf{j}
\end{array}
$$
Great, now I have my vectors. So, I now need to know the magnitude of the vectors:
$$
\begin{array} {rcl}
\left|\mathbf{v}(0)\right| & = & \sqrt{(\frac{2}{3})^{2} + (\frac{2}{3})^{2} +(\frac{1}{3})^{2}} \\
& = & \sqrt{\frac{9}{9}} \\
& = & 1 \\
\left|\mathbf{a}(0)\right| & = & \sqrt{(\frac{1}{3})^{2} + (\frac{1}{3})^{2}} \\
& = & \sqrt{\frac{2}{9}} \\
\end{array}
$$
Which, if anything I've done is correct, then the angle I'm looking for is $\theta = cos^{-1}\left(\frac{\frac{2}{3} + \frac{2}{3}}{\sqrt{\frac{2}{9}}}\right)=cos^{-1}\left(\frac{4}{3\sqrt{\frac{2}{9}}}\right)$.
However, as you can see, this yields a value larger than 1 and so the arccos function complains. So, please, would some kind soul please explain where I went wrong? I suspect I missed something in differentiation. It has been nearly two years since I've done derivatives.
Best Answer
Two words: chain rule :)
(More precisely you forgot to take the derivative of "inside the parantheses" of the $j$th component of $r$. That mistake only affects $u\cdot v$ by the way, so it's easy to fix it I believe.)