1. **Problem statement:** Given the vector function $$\vec{r}(t) = (2\sin 3t, t, 2\cos 3t)$$ find at the point $$P(0, \pi, -2)$$: a) The equations of the tangent, normal, and binormal lines. b) The curvature, torsion, and the equations of the Frenet frame. 2. **Find parameter $t$ at point $P$: ** We have $$x=2\sin 3t=0, y=t=\pi, z=2\cos 3t=-2$$ From $$z=2\cos 3t=-2$$, we get $$\cos 3t = -1$$ Thus $$3t=\pi\implies t=\frac{\pi}{3}$$ Check $x$: $$2\sin 3(\frac{\pi}{3}) = 2\sin \pi = 0$$ and $y=\pi$ correct. 3. **Compute derivatives:** $$\vec{r}'(t) = (6\cos 3t, 1, -6\sin 3t)$$ At $$t=\frac{\pi}{3}$$: $$\vec{r}'\left(\frac{\pi}{3}\right) = (6\cos \pi, 1, -6\sin \pi) = (6(-1), 1, 0) = (-6, 1, 0)$$ $$\vec{r}''(t) = (-18\sin 3t, 0, -18\cos 3t)$$ At $$t=\frac{\pi}{3}$$: $$\vec{r}''\left(\frac{\pi}{3}\right) = (-18\sin \pi, 0, -18\cos \pi) = (0, 0, 18)$$ 4. **Tangent vector $$\vec{T}$$:** Unit tangent vector: $$\vec{T} = \frac{\vec{r}'(t)}{|\vec{r}'(t)|} = \frac{(-6,1,0)}{\sqrt{(-6)^2 + 1^2 + 0^2}} = \frac{(-6,1,0)}{\sqrt{36+1}} = \frac{(-6,1,0)}{\sqrt{37}}$$ 5. **Normal vector $$\vec{N}$$: ** $$\vec{N} = \frac{\vec{T}'(t)}{|\vec{T}'(t)|}$$ We calculate $$\vec{T}'(t)$$ by differentiating $$\vec{T}(t) = \frac{\vec{r}'(t)}{|\vec{r}'(t)|}$$ but easier to use formula: $$\vec{N} = \frac{\vec{r}''(t) - (\vec{r}''(t) \cdot \vec{T})\vec{T}}{|\vec{r}''(t) - (\vec{r}''(t) \cdot \vec{T})\vec{T}|}$$ Calculate dot product: $$\vec{r}''(t) \cdot \vec{T} = (0,0,18) \cdot \frac{(-6,1,0)}{\sqrt{37}} = 0$$ So: $$\vec{N} = \frac{(0,0,18)}{|(0,0,18)|} = (0,0,1)$$ 6. **Binormal vector $$\vec{B}$$: ** $$\vec{B} = \vec{T} \times \vec{N} = \frac{(-6,1,0)}{\sqrt{37}} \times (0,0,1) = \frac{(1\cdot1 - 0\cdot0, 0\cdot(-6) - (-6)\cdot1, (-6)\cdot0 - 1\cdot0)}{\sqrt{37}} = \frac{(1, 6, 0)}{\sqrt{37}}$$ 7. **Equations of the tangent, normal, and binormal lines** (lines passing through $P=(0,\pi,-2)$): - Tangent line: $$\vec{l}_T(s) = P + s \vec{T} = \left(0, \pi, -2\right) + s \frac{(-6, 1, 0)}{\sqrt{37}}$$ - Normal line: $$\vec{l}_N(s) = P + s \vec{N} = (0, \pi, -2) + s(0, 0, 1)$$ - Binormal line: $$\vec{l}_B(s) = P + s \vec{B} = (0, \pi, -2) + s \frac{(1, 6, 0)}{\sqrt{37}}$$ 8. **Curvature $$\kappa$$**: $$\kappa = \frac{|\vec{r}'(t) \times \vec{r}''(t)|}{|\vec{r}'(t)|^3}$$ Calculate cross product: $$\vec{r}' \times \vec{r}'' = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 6\cos 3t & 1 & -6\sin 3t \\ -18\sin 3t & 0 & -18\cos 3t \end{vmatrix}$$ At $$t=\frac{\pi}{3}$$: $$\vec{r}' = (-6,1,0), \quad \vec{r}'' = (0,0,18)$$ Cross product: $$(-6,1,0) \times (0,0,18) = (1\cdot18 - 0\cdot0, 0\cdot0 - (-6)\cdot18, -6\cdot0 - 1\cdot0) = (18, 108, 0)$$ Magnitude: $$|\vec{r}' \times \vec{r}''| = \sqrt{18^2 + 108^2 + 0} = \sqrt{324 + 11664} = \sqrt{11988} = 6\sqrt{333}$$ $$|\vec{r}'| = \sqrt{37}$$ Thus $$\kappa = \frac{6\sqrt{333}}{(\sqrt{37})^3} = \frac{6\sqrt{333}}{37\sqrt{37}}$$ 9. **Torsion $$\tau$$:** $$\tau = \frac{(\vec{r}' \times \vec{r}'') \cdot \vec{r}'''}{|\vec{r}' \times \vec{r}''|^2}$$ Compute $$\vec{r}'''(t) = (-54\cos 3t, 0, 54\sin 3t)$$ At $$t=\frac{\pi}{3}$$: $$\vec{r}''' = (-54\cos \pi, 0, 54\sin \pi) = (-54(-1), 0, 0) = (54, 0, 0)$$ Dot product: $$(18, 108, 0) \cdot (54, 0, 0) = 18 \times 54 = 972$$ Magnitude squared: $$(6\sqrt{333})^2 = 36 \times 333 = 11988$$ So, $$\tau = \frac{972}{11988} = \frac{81}{999} = \frac{9}{111} = \frac{1}{12.333...} \approx 0.081$$ 10. **Equations of Frenet frame vectors at $P$:** $$\vec{T} = \frac{(-6,1,0)}{\sqrt{37}}, \quad \vec{N} = (0,0,1), \quad \vec{B} = \frac{(1,6,0)}{\sqrt{37}}$$ These unit vectors define the Frenet frame at point $P$. **Final answers:** a) Tangent line: $$\vec{l}_T = (0, \pi, -2) + s \frac{(-6,1,0)}{\sqrt{37}}$$ Normal line: $$\vec{l}_N = (0, \pi, -2) + s(0,0,1)$$ Binormal line: $$\vec{l}_B = (0, \pi, -2) + s \frac{(1,6,0)}{\sqrt{37}}$$ b) Curvature: $$\kappa = \frac{6\sqrt{333}}{37\sqrt{37}}$$ Torsion: $$\tau \approx 0.081$$ Frenet frame unit vectors: $$\vec{T} = \frac{(-6,1,0)}{\sqrt{37}}, \quad \vec{N} = (0,0,1), \quad \vec{B} = \frac{(1,6,0)}{\sqrt{37}}$$