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

$$r(t)=\big<2\sin(3t),\sqrt{4t},2\cos(3t)\big>$$

Since, we know that tangent vector to $r(t)$ is given

$$\frac{dr(t)}{dt}=r'(t)$$

Thus,

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

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,

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

Hence the required unit vector is

$$\big<\frac{6\cos(3t)}{\sqrt{36+\frac{1}{t}}},\frac{1}{\sqrt{36t+1}},\frac{-6\sin(3t)}{\sqrt{36+\frac{1}{t}}}\big>$$

We are done.