A particle moves in such a way that its position varies with time t as $r=ti+\frac{t^2}{2}j+tk$ Find as a function of time the speed of the particle.
Since it’s speed, I calculated the magnitude 
$$D=\sqrt{t^2+\frac{t^4}{4}+t^2}$$
$$D=\frac{t\sqrt{8+t^2}}{2}$$
Differentiating wrt t, I get $$\frac{t^2+4}{\sqrt{8+t^2}}$$
The answer is given is$\sqrt{t^2+2}$
I feel it’s more of a conceptual problem than a calculation problem. Either ways, please let me what am I doing wrong.
 A: You give that the particle's position at any given time is
$$\vec{r}=t\vec{i}+\frac{t^2}{2}\vec{j}+t\vec{k} \tag{1}\label{eq1}$$
As Vasily Mitch's comment states,

Magnitude of the velocity is not the same as the derivative of magnitude, I.e., $|\vec{r}'| \neq |\vec{r}|'$

Note the distance you calculate is how far the point is at any given time $t$ from the origin, i.e., $(0,0,0)$ (which from \eqref{eq1}, you can see occurs at $t = 0$). However, this doesn't determine how the distance changes between various other points in time, say $t_1$ and $t_2$, which would actually be $\sqrt{(t_1 - t_2)^2 + \frac{(t_1^2 - t_2^2)^2}{4} + (t_1 - t_2)^2}$ instead. Dividing this by $t_1 - t_2$ and taking the limit as $t_2 \to t_1$ would give the speed at $t = t_1$. Your derivative of $D$ gives the rate of change of the component of the velocity along the vector from the origin (e.g., if the movement, i.e., velocity, was on the surface of a sphere a constant distance $D_1$ from the origin, then the component of the velocity relative to the origin vector would be $0$, which matches the derivative of $D = D_1$ being $0$), so it's not the correct expression to use to calculate the total magnitude (i.e., speed) of the velocity at any given time. This is basically why your answer differs from the correct answer.
Instead of differentiating the distance from the origin with respect to time, differentiating \eqref{eq1} with respect to time gives the particle's velocity, i.e.,
$$\frac{d\vec{r}}{dt} = \vec{i} + t\vec{j} + \vec{k} \tag{2}\label{eq2}$$
The speed is the magnitude of the velocity, i.e., the square root of the sum of the $3$ component values squared, which gives
$$s = \sqrt{1 + t^2 + 1} = \sqrt{t^2 + 2} \tag{3}\label{eq3}$$
This matches the answer which was given.
