Calculus (3rd Edition)

$$\frac{d}{dt}(r(t)\cdot r_1(t))|_2 =62$$
Since $$r_1(t)=\langle t^2,t^3,t \rangle$$ then, we have $$\frac{d}{dt}(r(t)\cdot r_1(t))=\frac{d}{dt}(r(t))\cdot r_1(t) +r(t)\cdot \frac{d}{dt}( r_1(t))\\ =\frac{d}{dt}(r(t))\cdot r_1(t) +r(t)\cdot \langle 2t,3t^2,1\rangle.$$ Now, at $t=2$, we have $$\frac{d}{dt}(r(t)\cdot r_1(t))|_2=\frac{d}{dt}(r(2))\cdot r_1(2) +r(2)\cdot \frac{d}{dt}( r_1(2))\\ =\langle 1,4,3 \rangle\cdot \langle 4,8,2 \rangle+\langle 2,1,0 \rangle\cdot \langle 4,12,1 \rangle\\ =62$$