Question

For each of the vector-valued functions, find the unit tangent vector .

$r\left(t\right)=(t,t^2)$

Short answer

$\frac{1}{\sqrt{1+4t^2}}⟨1,2t⟩$

Step-by-step solution

Step1. Given Information

Step2. Continue

$\begin{array}{l}∥r^{\mathrm{\prime }}\left(t\right)∥=\parallel ⟨1,2t⟩\parallel \\ =\sqrt{1^2+(2t{)}^2}\\ =\sqrt{1+4t^2}\\ \text{The unit tangent vector to}r\left(t\right)\text{is}\\ \begin{array}{l}T\left(t\right)=\frac{r^{\mathrm{\prime }}\left(t\right)}{∥r^{\prime \prime }\left(t\right)∥}\\ =\frac{⟨1,2t⟩}{\sqrt{1+4t^2}}\\ \text{Thus the unit tangent vector to the given vector function is}\\ \frac{1}{\sqrt{1+4t^2}}⟨1,2t⟩\end{array}\end{array}$