1. **State the problem:** Find the equation of the principal normal line and the osculating plane to the curve $$\vec{X}(t) = \cos t \hat{e}_1 + \sin t \hat{e}_2 + t^3 \hat{e}_3$$ at $$t = \frac{\pi}{2}$$. 2. **Recall formulas and definitions:** - The principal normal line passes through the point on the curve and is directed along the principal normal vector $$\vec{N}(t)$$. - The osculating plane is the plane containing the tangent vector $$\vec{T}(t)$$ and the principal normal vector $$\vec{N}(t)$$ at the point. - The tangent vector $$\vec{T}(t) = \frac{\vec{X}'(t)}{\|\vec{X}'(t)\|}$$. - The principal normal vector $$\vec{N}(t) = \frac{\vec{T}'(t)}{\|\vec{T}'(t)\|}$$. - The osculating plane at $$t$$ has normal vector $$\vec{B}(t) = \vec{T}(t) \times \vec{N}(t)$$ (binormal vector). 3. **Calculate derivatives:** $$\vec{X}'(t) = -\sin t \hat{e}_1 + \cos t \hat{e}_2 + 3t^2 \hat{e}_3$$ At $$t = \frac{\pi}{2}$$: $$\vec{X}'\left(\frac{\pi}{2}\right) = -\sin \frac{\pi}{2} \hat{e}_1 + \cos \frac{\pi}{2} \hat{e}_2 + 3\left(\frac{\pi}{2}\right)^2 \hat{e}_3 = -1 \hat{e}_1 + 0 \hat{e}_2 + \frac{3\pi^2}{4} \hat{e}_3$$ 4. **Calculate magnitude of $$\vec{X}'(t)$$ at $$t=\frac{\pi}{2}$$:** $$\|\vec{X}'\left(\frac{\pi}{2}\right)\| = \sqrt{(-1)^2 + 0^2 + \left(\frac{3\pi^2}{4}\right)^2} = \sqrt{1 + \frac{9\pi^4}{16}} = \sqrt{\frac{16 + 9\pi^4}{16}} = \frac{\sqrt{16 + 9\pi^4}}{4}$$ 5. **Find unit tangent vector $$\vec{T}(t)$$ at $$t=\frac{\pi}{2}$$:** $$\vec{T}\left(\frac{\pi}{2}\right) = \frac{\vec{X}'\left(\frac{\pi}{2}\right)}{\|\vec{X}'\left(\frac{\pi}{2}\right)\|} = \frac{-1 \hat{e}_1 + 0 \hat{e}_2 + \frac{3\pi^2}{4} \hat{e}_3}{\frac{\sqrt{16 + 9\pi^4}}{4}} = \frac{4}{\sqrt{16 + 9\pi^4}}\left(-1 \hat{e}_1 + 0 \hat{e}_2 + \frac{3\pi^2}{4} \hat{e}_3\right) = \frac{-4}{\sqrt{16 + 9\pi^4}} \hat{e}_1 + 0 \hat{e}_2 + \frac{3\pi^2}{\sqrt{16 + 9\pi^4}} \hat{e}_3$$ 6. **Calculate second derivative $$\vec{X}''(t)$$:** $$\vec{X}''(t) = -\cos t \hat{e}_1 - \sin t \hat{e}_2 + 6t \hat{e}_3$$ At $$t=\frac{\pi}{2}$$: $$\vec{X}''\left(\frac{\pi}{2}\right) = -\cos \frac{\pi}{2} \hat{e}_1 - \sin \frac{\pi}{2} \hat{e}_2 + 6 \cdot \frac{\pi}{2} \hat{e}_3 = 0 \hat{e}_1 - 1 \hat{e}_2 + 3\pi \hat{e}_3$$ 7. **Calculate $$\vec{T}'(t)$$ using the formula $$\vec{T}'(t) = \frac{d}{dt} \left( \frac{\vec{X}'(t)}{\|\vec{X}'(t)\|} \right)$$ or use the formula for principal normal vector directly:** Alternatively, use the formula for principal normal vector: $$\vec{N}(t) = \frac{\vec{X}''(t) - (\vec{X}''(t) \cdot \vec{T}(t)) \vec{T}(t)}{\|\vec{X}''(t) - (\vec{X}''(t) \cdot \vec{T}(t)) \vec{T}(t)\|}$$ 8. **Calculate dot product $$\vec{X}''\left(\frac{\pi}{2}\right) \cdot \vec{T}\left(\frac{\pi}{2}\right)$$:** $$= (0)(-\frac{4}{\sqrt{16 + 9\pi^4}}) + (-1)(0) + (3\pi)\left(\frac{3\pi^2}{\sqrt{16 + 9\pi^4}}\right) = \frac{9\pi^3}{\sqrt{16 + 9\pi^4}}$$ 9. **Calculate vector $$\vec{X}'' - (\vec{X}'' \cdot \vec{T}) \vec{T}$$ at $$t=\frac{\pi}{2}$$:** $$= \left(0 \hat{e}_1 - 1 \hat{e}_2 + 3\pi \hat{e}_3\right) - \frac{9\pi^3}{\sqrt{16 + 9\pi^4}} \left(-\frac{4}{\sqrt{16 + 9\pi^4}} \hat{e}_1 + 0 \hat{e}_2 + \frac{3\pi^2}{\sqrt{16 + 9\pi^4}} \hat{e}_3\right)$$ Simplify denominator: $$= 0 \hat{e}_1 - 1 \hat{e}_2 + 3\pi \hat{e}_3 + \frac{9\pi^3 \cdot 4}{16 + 9\pi^4} \hat{e}_1 - 0 \hat{e}_2 - \frac{9\pi^3 \cdot 3\pi^2}{16 + 9\pi^4} \hat{e}_3$$ Calculate coefficients: $$= \left(\frac{36\pi^3}{16 + 9\pi^4}\right) \hat{e}_1 - 1 \hat{e}_2 + \left(3\pi - \frac{27\pi^5}{16 + 9\pi^4}\right) \hat{e}_3$$ 10. **Calculate magnitude of this vector:** $$\sqrt{\left(\frac{36\pi^3}{16 + 9\pi^4}\right)^2 + (-1)^2 + \left(3\pi - \frac{27\pi^5}{16 + 9\pi^4}\right)^2}$$ 11. **Find principal normal vector $$\vec{N}\left(\frac{\pi}{2}\right)$$ by dividing the vector in step 9 by its magnitude.** 12. **Find point on curve at $$t=\frac{\pi}{2}$$:** $$\vec{X}\left(\frac{\pi}{2}\right) = \cos \frac{\pi}{2} \hat{e}_1 + \sin \frac{\pi}{2} \hat{e}_2 + \left(\frac{\pi}{2}\right)^3 \hat{e}_3 = 0 \hat{e}_1 + 1 \hat{e}_2 + \frac{\pi^3}{8} \hat{e}_3$$ 13. **Equation of principal normal line:** $$\vec{r}(s) = \vec{X}\left(\frac{\pi}{2}\right) + s \vec{N}\left(\frac{\pi}{2}\right)$$ 14. **Osculating plane normal vector is binormal vector:** $$\vec{B} = \vec{T} \times \vec{N}$$ 15. **Equation of osculating plane:** $$\vec{B} \cdot (\vec{r} - \vec{X}(\frac{\pi}{2})) = 0$$ This completes the derivation of the principal normal line and osculating plane at $$t=\frac{\pi}{2}$$.