# derivative of a vector

find $\frac{d^2\vec{S}}{dt}$ where $\vec{S}=(t+1)\hat{i}+(t^2+t+1)\hat{j}+(t^3+t^2+t)\hat{k}$

So $\frac{d\vec{S}}{dt}=\hat{i}+(2t+1)\hat{j}+(3t^2+2t+1)\hat{k}$

now when I take the derivative again with respect to $t$ the $\hat{i}$ component is $0$ because I look at it as the derivative of $\hat{i}$?

• In Cartesian coordinates, $\mathbf{i,j,k}$ are invariants all the time derivates are zero. For curvillinear coordinates, the bases are spatial dependent and hence their time derivatives are no longer zero. See the case of spherical polar coordinates: mathworld.wolfram.com/SphericalCoordinates.html – Ng Chung Tak May 17 '16 at 5:48

$$\frac{d}{dt}(\vec v) = \frac d{dt} (v_1\hat i + v_2 \hat j + v_3\hat k) = \frac{dv_1}{dt}\hat i + \frac{dv_2}{dt}\hat j + \frac{dv_3}{dt}\hat k$$
Applying that formula to $\hat i = 1\hat i$, we indeed get $$\frac d{dt} (\hat i) = \frac{d}{dt}(1\hat i) = \left(\frac{d}{dt} 1\right)\hat i = 0\hat i$$
Yes, you can interprete any variable that is not $t$ as a constant in this case.
• You have perhaps a typo in your first derivative. The $\vec{j}$ component is $(2t+ 1)\vec{j}$, not $2t\vec{j}$. – user247327 May 16 '16 at 16:35
• $\hat i$ is not a variable. – user137731 May 16 '16 at 16:52