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

Let $\vec k$ be the tangent vector and $\vec e$ the unit tangent vector, then

$$\vec k=\frac {d\vec r}{dt}$$

$$\vec e=\frac {\vec k}{|k|}$$

Find the tangent vector

$$\vec k=\frac {d\vec r}{dt}=〈\frac{d(2 sin(3t))}{dt}, \frac{d\sqrt 4t}{dt}, \frac{d(2cos(3t)}{dt}〉=〈6cos3t, 2, -6sin3t〉$$

Now we find the module of the $\vec k$

$$|\vec k|=\sqrt{(6cos3t)^2+2^2+(-6sin3t)^2}=\sqrt{40}=2\sqrt{10}$$ $$\vec e=\frac{〈6cos3t, 2, -6sin3t〉}{2\sqrt{10}}=〈\frac{3\sqrt{10}}{10} cos3t, \frac{\sqrt{10}}{10}, -\frac{3\sqrt{10}}{10}sin3t〉$$

Answer: $〈\frac{3\sqrt{10}}{10} cos3t, \frac{\sqrt{10}}{10}, -\frac{3\sqrt{10}}{10}sin3t〉$