Question

Find the unit tangent vector at the point 2,0,pi for a curve which is described by the parametric equations

x=2Sina y=3cosa z=2a

Answer 1

$$\vec v(a)=\vec r'(a)=\langle2\cos a, -3\sin a, 2\rangle$$

$$||\vec v(a)||=\sqrt{4\cos^2 a+9\sin^2 a+4}$$

$$=\sqrt{8+5\sin^2 a}$$

Point $(2, 0, \pi)$

$a=\dfrac{\pi}{2}$

$$\vec v(a)=\vec r'(a)=\langle0, -3, 2\rangle$$

$$||\vec v(a)||=\sqrt{8+5(1)^2}=\sqrt{13}$$

$$\vec T(\dfrac{\pi}{2})=\dfrac{-3\vec j+2\vec k}{\sqrt{13}}$$

$$\vec T(\dfrac{\pi}{2})=\langle0, -\dfrac{3\sqrt{13}}{13}, \dfrac{2\sqrt{13}}{13}\rangle$$