Question

A curve is given by $\vec{r}(t)=t\hat{i}+t^{2}\hat{j}+t^{3}\hat{k}$. The unit vector of the tangent at $t=1$ is
 

• A. $\dfrac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}$
• B. $\dfrac{\hat{i}+2\hat{j}+3\hat{k}}{\sqrt{6}}$
• C. $\dfrac{t\hat{i}+2t^{2}\hat{j}+2t\hat{k}}{3}$
• D. $\dfrac{\hat{i}+2\hat{j}+3\hat{k}}{\sqrt{14}}$

Hint:

Unit tangent vector = derivative of position vector divided by its magnitude.

Answer 1

The Correct Option is D

Step 1: Find the tangent vector.
$\vec{r}'(t) = \frac{d}{dt}(t, t^{2}, t^{3}) = (1, 2t, 3t^{2})$.

Step 2: Evaluate at $t=1$.
$\vec{r}'(1) = (1,2,3)$.

Step 3: Convert to unit vector.
Magnitude $= \sqrt{1^{2}+2^{2}+3^{2}}=\sqrt{14}$.
Unit tangent $= \dfrac{1}{\sqrt{14}}(1,2,3)$.

Step 4: Conclusion.
The required unit tangent vector is $\dfrac{\hat{i}+2\hat{j}+3\hat{k}}{\sqrt{14}}$.