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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 $\hat{i}$?

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  • $\begingroup$ 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 $\endgroup$ – Ng Chung Tak May 17 '16 at 5:48
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The identity for differentiating vector-valued functions of a real variable in Cartesian coordinates is

$$\frac{d}{dt}(\vec v) = \frac d{dt} (v_1\hat i + v_2 \hat j + v_3\hat k) = \frac{dv_1}{dt}\hat i + \frac{dv_2}{dt}\hat j + \frac{dv_3}{dt}\hat k$$

Applying that formula to $\hat i = 1\hat i$, we indeed get $$\frac d{dt} (\hat i) = \frac{d}{dt}(1\hat i) = \left(\frac{d}{dt} 1\right)\hat i = 0\hat i$$

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Yes, you can interprete any variable that is not $t$ as a constant in this case.

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  • $\begingroup$ You have perhaps a typo in your first derivative. The $\vec{j}$ component is $(2t+ 1)\vec{j}$, not $2t\vec{j}$. $\endgroup$ – user247327 May 16 '16 at 16:35
  • $\begingroup$ @user247327: Your comment was meant to be posted under the original question, was it not? $\endgroup$ – Alex M. May 16 '16 at 16:47
  • $\begingroup$ @user247327 thanks edited $\endgroup$ – gbox May 16 '16 at 16:51
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    $\begingroup$ $\hat i$ is not a variable. $\endgroup$ – user137731 May 16 '16 at 16:52

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