Solución

Apartado a

Tenemos r′(t)=6i+(8t+2)j+5k\bold{r'}(t) = 6\bold{i} + (8t + 2)\bold{j} + 5\bold{k} y u′(t)=2ti+2j+(3t2−3)k\bold{u'}(t) = 2t\bold{i} + 2\bold{j} + (3t^2−3)\bold{k}. Por lo tanto, de acuerdo con la propiedad iv.:

ddt[r(t)⋅u(t)]=r′(t)⋅u(t)+r(t)⋅u′(t)=(6i+(8t+2)j+5k)⋅((t2−3)i+(2t+4)j+(t3−3t)k)+((6t+8)i+(4t2+2t−3)j+5tk)⋅(2ti+2j+(3t2−3)k)=6(t2−3)+(8t+2)(2t+4)+5(t3−3t)+2t(6t+8)  +2(4t2+2t−3)+5t(3t2−3)=20t3+42t2+26t−16\begin{aligned} \frac{d}{dt}[\bold{r}(t)\cdot\bold{u}(t)] &= \bold{r'}(t)\cdot\bold{u}(t)+\bold{r}(t)\cdot\bold{u'}(t)\\ &= (6\bold{i}+(8t+2)\bold{j}+5\bold{k})\cdot((t^2−3)\bold{i}+(2t+4)\bold{j}+(t^3−3t)\bold{k})\\ &+ ((6t+8)\bold{i}+(4t^2+2t−3)\bold{j}+5t\bold{k})\cdot(2t\bold{i}+2\bold{j}+(3t^2−3)\bold{k})\\ &= 6(t^2−3)+(8t+2)(2t+4)+5(t^3−3t)+2t(6t+8)\\ \; &+ 2(4t^2+2t−3)+5t(3t^2−3)\\ &= 20t^3+42t^2+26t−16 \end{aligned}

Apartado b

Primero, necesitamos adaptar la propiedad v. para este problema:

ddt[u(t)×u′(t)]=u′(t)×u′(t)+u(t)×u′′(t).\frac{d}{dt}[\bold{u}(t)\times\bold{u'}(t)] = \bold{u'}(t)\times\bold{u'}(t) + \bold{u}(t)\times\bold{u''}(t).

Recuerda que el producto cruz de cualquier vector consigo mismo es cero. Además, u′′(t)\bold{u''}(t) representa la segunda derivada de u(t)\bold{u}(t):

Por lo tanto,

ddt[u(t)×u′(t)]=0+((t2−3)i+(2t+4)j+(t3−3t)k)×(2i+6tk)=∣ijkt2−32t+4t3−3t206t∣=6t(2t+4)i−(6t(t2−3)−2(t3−3t))j−2(2t+4)k=(12t2+24t)i+(12t−4t3)j−(4t+8)k\begin{aligned} \frac{d}{dt}[\bold{u}(t)\times\bold{u'}(t)] &= 0+((t^2−3)\bold{i}+(2t+4)\bold{j}+(t^3−3t)\bold{k})\times(2\bold{i}+6t\bold{k})\\ &= \begin{vmatrix} \bold{i} & \bold{j} & \bold{k}\\ t^2-3 & 2t+4 & t^3-3t\\ 2 & 0 & 6t \end{vmatrix}\\ &= 6t(2t+4)\bold{i}−(6t(t^2−3)−2(t^3−3t))\bold{j}−2(2t+4)\bold{k}\\ &= (12t^2+24t)\bold{i}+(12t−4t^3)\bold{j}−(4t+8)\bold{k} \end{aligned}