Chapter 12 - Vector-Valued Functions - 12.3 Change Of Parameter; Arc Length - Exercises Set 12.3 - Page 866: 13

$\dfrac{d r}{d \tau}=4i+8(4 \tau+1)j$

Here, we have: $r(t)=ti+t^2 j$ and $t=4 \tau+1$ Apply Chain rule: $\dfrac{d r}{d \tau}=\dfrac{d r}{d t} \times \dfrac{d t}{d \tau}$ or, $\dfrac{d r}{d \tau}=(i+2tj) \times 4=\dfrac{d t}{d \tau}=4i+8tj$ By plugging $t=4 \tau+1$ in the above equation, we get: $\dfrac{d r}{d \tau}=4i+8(4 \tau+1)j$