Discussion
A curve is given by \(\vec{r}(t) = t\hat{\imath} + t^{2}\hat{\jmath} + t^{3}\hat{k}\). The unit vector of the tangent to the curve at \(t = 1\) is
Answer: D
The unit tangent is \(\frac{\hat{\imath} + 2\hat{\jmath} + 3\hat{k}}{\sqrt{14}}\).
The tangent is the derivative of the position vector, divided by its length:
\[ \begin{aligned} \frac{d\vec{r}}{dt} &= \hat{\imath} + 2t\hat{\jmath} + 3t^{2}\hat{k} \\ \text{at } t = 1: &\ \hat{\imath} + 2\hat{\jmath} + 3\hat{k} \\ \left|\,\cdot\,\right| &= \sqrt{1 + 4 + 9} = \sqrt{14} \end{aligned} \]
The first option is the position vector itself normalised, not its derivative. The others use wrong coefficients for the \(t^{2}\) and \(t^{3}\) terms; for example \(\hat{\imath} + 2\hat{\jmath} + 2\hat{k}\) forgets that \(d(t^{3})/dt = 3t^{2}\).
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