Answer to Question #197806 in Differential Geometry | Topology for Fatowore Samson

Question

At any point of the path x=3cos⁡t,y=3sin⁡t,z=4t, what is the Normal vector?

Answer 1

$$\vec{r}(t)=(3\cos t, 3\sin t, 4t)$$ $$\vec{r'}(t)=(-3\sin t, 3\cos t, 4)$$ $$|\vec{r'}(t)|=\sqrt{(-3\sin t)^2+(3\cos t)^2+(4)^2}=5$$ $$\vec{T}(t)=\dfrac{\vec{r'}(t)}{|\vec{r'}(t)|}=(-\dfrac{3}{5}\sin t, \dfrac{3}{5}\cos t, \dfrac{4}{5})$$ $$|\vec{T}(t)|=\sqrt{(-\dfrac{3}{5}\sin t)^2+(\dfrac{3}{5}\cos t)^2+(\dfrac{4}{5})^2}=1$$ $$\vec{N}(t)=\dfrac{\vec{T'}(t)}{|\vec{T'}(t)|}$$ $$\vec{T'}(t)=(-\dfrac{3}{5}\cos t, -\dfrac{3}{5}\sin t, 0)$$ $$|\vec{T'}(t)|=\sqrt{(-\dfrac{3}{5}\cos t)^2+(-\dfrac{3}{5}\sin t)^2+(0)^2}=\dfrac{3}{5}$$ $$\vec{N}(t)=(-\cos t, -\sin t, 0)$$