Question #197806

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


1
Expert's answer
2021-05-24T19:16:10-0400
r(t)=(3cost,3sint,4t)\vec{r}(t)=(3\cos t, 3\sin t, 4t)r(t)=(3sint,3cost,4)\vec{r'}(t)=(-3\sin t, 3\cos t, 4)r(t)=(3sint)2+(3cost)2+(4)2=5|\vec{r'}(t)|=\sqrt{(-3\sin t)^2+(3\cos t)^2+(4)^2}=5T(t)=r(t)r(t)=(35sint,35cost,45)\vec{T}(t)=\dfrac{\vec{r'}(t)}{|\vec{r'}(t)|}=(-\dfrac{3}{5}\sin t, \dfrac{3}{5}\cos t, \dfrac{4}{5})T(t)=(35sint)2+(35cost)2+(45)2=1|\vec{T}(t)|=\sqrt{(-\dfrac{3}{5}\sin t)^2+(\dfrac{3}{5}\cos t)^2+(\dfrac{4}{5})^2}=1N(t)=T(t)T(t)\vec{N}(t)=\dfrac{\vec{T'}(t)}{|\vec{T'}(t)|}T(t)=(35cost,35sint,0)\vec{T'}(t)=(-\dfrac{3}{5}\cos t, -\dfrac{3}{5}\sin t, 0)T(t)=(35cost)2+(35sint)2+(0)2=35|\vec{T'}(t)|=\sqrt{(-\dfrac{3}{5}\cos t)^2+(-\dfrac{3}{5}\sin t)^2+(0)^2}=\dfrac{3}{5}N(t)=(cost,sint,0)\vec{N}(t)=(-\cos t, -\sin t, 0)

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