Question

Consider the parametric curve described by r(t) =〈2 sin(3t), √4t, 2cos(3t)〉. Find the unit tangent vector of r(t).

Answer 1

The unit tangent vector to the parametric curve $r(t)$ is given by,

$\overrightarrow{T}(t)=\frac{\overrightarrow{r'}(t)}{∥ \overrightarrow{r'}(t)∥}$

Here,

$\overrightarrow{r'}(t)=⟨ 2\frac{d}{dt}(\sin(3t)),\frac{d}{dt}(\sqrt{4}t),2\frac{d}{dt}(\cos(3t))⟩$

$=⟨ 2(3\cos(3t)),\sqrt{4},2(-3\sin(3t))⟩$

$=⟨ 6\cos(3t),2,-6\sin(3t)⟩$

$∥\overrightarrow{r'}(t)∥=\sqrt{(6\cos(3t))^2+(\sqrt{4})^2+(-6\sin(3t))^2}$

$=\sqrt{36\cos^2(3t)+4+36\sin^2(3t)}$

$=\sqrt{36(\cos^2(3t)+\sin^2(3t))+4}$

Use trigonometric identity $\cos^2(x)+sin^2(x)=1$ to simplify the term inside the radical as,

$=\sqrt{36(1)+4}$

$=\sqrt{36+4}$

$=\sqrt{40}$

$=2\sqrt{10}$

Therefore, the unit tangent vector is,

$\overrightarrow{T}(t)=\frac{⟨ 6\cos(3t),2,-6\sin(3t)⟩}{∥ ⟨ 6\cos(3t),\sqrt{4},-6\sin(3t)⟩∥}$

$=\frac{⟨ 6\cos(3t),2,-6\sin(3t)⟩}{2\sqrt{10}}$

$=⟨ \frac{3\cos(3t)}{\sqrt{10}},\frac{1}{\sqrt{10}},-\frac{3\sin(3t)}{\sqrt{10}}⟩$

Therefore, the unit tangent vector to the parametric curve is $\overrightarrow{T}(t)=⟨ \frac{3\cos(3t)}{\sqrt{10}},\frac{1}{\sqrt{10}},-\frac{3\sin(3t)}{\sqrt{10}}⟩$.