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

We have given the parametric curve (vector valued) function as

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

Since, we know that tangent vector to $r(t)$ is given by $\frac{dr(t)}{dt}=r'(t)$ .

Thus,

$r'(t)=\big \lt \frac{d(2\sin(3t))}{dt},\frac{d(\sqrt{4t})}{dt},\frac{d(2\cos(3t))}{dt}\big \gt \\ \implies r'(t)=\big \lt 6\cos(3t),\frac{1}{\sqrt{t}},-6\sin(3t)\big \gt$ .

We also know that, unit vector for any vector $v$ is $\hat{v}=\frac{v}{||v||}$ where, ∣∣$v$∣∣ is the norm of $v$. Hence norm of tangent vector of $r(t)$  will be

$||r'(t)||=\sqrt{(6\sin(3t))^2+(\frac{1}{\sqrt{t}})^2+(-6\cos(3t))^2} \\ \implies ||r'(t)||=\sqrt{36+\frac{1}{t}}$

Therefore,

$\widehat{r'(t)}=\frac{r'(t)}{||r'(t)||}\\ \implies \widehat{r'(t)}=\big \lt \frac{6\cos(3t)}{\sqrt{36+\frac{1}{t}}},\frac{1}{\sqrt{t}\sqrt{36+\frac{1}{t}}},\frac{-6\sin(3t)}{\sqrt{36+\frac{1}{t}}}\big \gt \\ \implies \widehat{r'(t)}=\big \lt \frac{6\cos(3t)}{\sqrt{36+\frac{1}{t}}},\frac{1}{\sqrt{36t+1}},\frac{-6\sin(3t)}{\sqrt{36+\frac{1}{t}}}\big \gt$

Hence the required unit vector is $\big \lt \frac{6\cos(3t)}{\sqrt{36+\frac{1}{t}}},\frac{1}{\sqrt{36t+1}},\frac{-6\sin(3t)}{\sqrt{36+\frac{1}{t}}}\big \gt$.