Answer to Question #122826 in Differential Geometry | Topology for Ojugbele Daniel

Question

Determine the unit tangent vector at the point (2,4,7) for the circle with parametric equations x=2u, y=u^2 +3 and z=2u^2 +5

Answer 1

Let’s consider $r(u)=\langle2u, \ u^2+3, \ 2u^2+5\rangle$

$r(1)=\langle2,4,7\rangle$

The unit tangent vector is $\frac{r^{\prime}(u)}{|r(u)|}$ .

$r^{\prime}(u)=\langle 2,\ 2u,\ 4u\rangle, \ \ r^\prime (1)=\langle 2,\ 2, \ 4\rangle, \ \ |r^\prime (1)|=\sqrt{2^2+2^2+4^2}=2\sqrt{6}$

$\frac{r^{\prime}(1)}{|r(1)|}=\frac{1}{2\sqrt{6}}\langle 2,\ 2,\ 4\rangle =\frac{1}{\sqrt 6}\langle 1,\ 1,\ 2\rangle$ .

Answer: the unit tangent vector at the point $(2,4,7)$  is $\frac{1}{\sqrt 6}\langle 1,\ 1,\ 2\rangle$ .