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Solution: Given r(t)=\left\langle\sin (3 t), \cos (3 t), 8 t^{7 / 8}\right\rangle at t=1 step 1:\begin{aligned}r^{\prime}(t)= & \frac{d}{d t}\left(\sin (3 t), \cos (3 t), 8 t^{7 / 8}\right\rangle \\= & \left\langle\frac{d}{d t} \sin (3 t), \frac{d}{d t} \cos (3 t), \frac{d}{d t}\left(8 t^{7 / 8}\right)\right\rangle \\= & \left\langle 3 \cos (3 t),-3 \sin (3 t), 8 \times \frac{7}{8} \times t^{7 / 8-1}\right\rangle \\& \sin \theta \frac{d}{d t} t^{n}=n\left(1 t^{n-1}\right) \\r^{\prime}(t)= & \left\langle 3 \cos (3 t),-3 \sin (3 t), 7 t^{-4 / 8}\right\rangle\end{aligned}we lunow that if \gamma\left(t_{0}\right) is a pact and if r\left(t_{0}\right) \neq 0 then the equation of it tangent line at the point to is: \quad 1(t)=r\left(t_{0}\right)+\left(t-t_{0}\right) r^{\prime}\left(t_{0}\right)if t=1, then r(1)=\langle\sin (3), \cos (3), 8\rangle\gamma^{\prime}(1)=\langle 3 \cos (3),-3 \sin (3), 7\rangleSubstitute known values in () we get\begin{array}{l}l(t)=r(1)+(t-1) r^{\prime}(1) \\l(t)=\langle\sin (3)+\cos (3), 8\rangle+(t-1)(3 \cos (3), \\-3 \sin (3), 1\rangle\end{array}\begin{array}{r}1(t)=\langle\sin (3)+3(t-1) \cos (3), \cos (3)-3(t-1) \sin (t), \\8+7(t-1)\rangle \\1(t)=\langle\sin (3)+3(t-1) \cos (3), \cos (3)-3(t-1) \sin (3), \\7 t+1\rangle\end{array}7 t+1> ...