Question

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

Answer 1

$$\vec r(t)=⟨3\cos t,3\sin t, 4t ⟩$$

$$\vec r\ '(t)=⟨-3\sin t,3\cos t, 4 ⟩$$

$$||\vec r\ '(t)||=\sqrt{(-3\sin t)^2+(-3\cos t)^2+(4)^2}=5$$

$$\vec T(t)=\dfrac{\vec r\ '(t)}{||\vec r\ '(t)||}=⟨-\dfrac{3}{5}\sin t,\dfrac{3}{5}\cos t, \dfrac{4}{5} ⟩$$

$$\vec T\ '(t)=⟨-\dfrac{3}{5}\cos t,-\dfrac{3}{5}\sin t, 0 ⟩$$

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

$$\vec N(t)=\dfrac{\vec T\ '(t)}{||\vec T\ '(t)||}=⟨-\cos t,-\sin t, 0 ⟩$$