<s>[INST] Steps for Velocity given vector I am asked to find velocity for the space curve
$$r(t) = \langle \cos(3t), 1-\sin(3t), 3\sin(3t)+\cos(3t)\rangle$$
at $t=\pi / 4$
I have differentiated each component and substituted $t$ in said components. Are those the correct steps to find the answers?
I got this as my answer:
$$\langle -3 \sqrt{2},  3 \sqrt{2} , -9 \sqrt{2} - 3 \sqrt{2} \rangle.$$
 [/INST] The steps you stated you did are the correct ones. In particular, differentiating each component leads to
$$\frac{dr(t)}{dt} = \langle -3\sin(3t), -3\cos(3t), 9\cos(3t) - 3\sin(3t) \rangle \tag{1}\label{eq1A}$$
Next, $t = \frac{\pi}{4}$ means $3t = \frac{3\pi}{4}$, with $\sin(3t) = \frac{1}{\sqrt{2}}$ and $\cos(3t) = -\frac{1}{\sqrt{2}}$. Thus, \eqref{eq1A} gives
$$\left\langle -\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}, -\frac{9}{\sqrt{2}} - \frac{3}{\sqrt{2}} \right\rangle \tag{2}\label{eq2A}$$
However, your answer has the result in \eqref{eq2A} multiplied by $2$. Also, note that the $z$ component can be simplified to
$$-\frac{9}{\sqrt{2}} - \frac{3}{\sqrt{2}} = -6\sqrt{2} \tag{3}\label{eq3A}$$
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