1. Statement of the problem: Given the time-dependent vectors $A(t)=2t^3\mathbf{i}+t^4\mathbf{j}+2t\mathbf{k}$ and $B(t)=4t^6\mathbf{i}-1\mathbf{j}-4t\mathbf{k}$, compute $\dfrac{d}{dt}(A\times B)$, $\dfrac{d^2}{dt^2}(A\times B)$, and $\dfrac{d}{dt}(A\cdot B)$.\n 2. Write components explicitly so we can compute componentwise: $A_x=2t^3$, $A_y=t^4$, $A_z=2t$, and $B_x=4t^6$, $B_y=-1$, $B_z=-4t$.\n 3. Use the cross-product component formulas $ (A\times B)_x=A_yB_z-A_zB_y$, $ (A\times B)_y=A_zB_x-A_xB_z$, and $ (A\times B)_z=A_xB_y-A_yB_x$ to compute each component.\n 4. Compute the $x$-component: $(A\times B)_x=A_yB_z-A_zB_y=t^4\cdot(-4t)-(2t)\cdot(-1)=-4t^5+2t$.\n 5. Compute the $y$-component: $(A\times B)_y=A_zB_x-A_xB_z=(2t)\cdot(4t^6)-(2t^3)\cdot(-4t)=8t^7+8t^4$.\n 6. Compute the $z$-component: $(A\times B)_z=A_xB_y-A_yB_x=(2t^3)\cdot(-1)-t^4\cdot(4t^6)=-2t^3-4t^{10}$.\n 7. Collecting the components gives the cross product as the vector below.\n $$A\times B=\langle -4t^5+2t,\;8t^7+8t^4,\;-4t^{10}-2t^3\rangle$$\n 8. Differentiate componentwise to obtain $\dfrac{d}{dt}(A\times B)$.\n 9. Differentiate the $x$-component: $\dfrac{d}{dt}(-4t^5+2t)=-20t^4+2$.\n 10. Differentiate the $y$-component: $\dfrac{d}{dt}(8t^7+8t^4)=56t^6+32t^3$.\n 11. Differentiate the $z$-component: $\dfrac{d}{dt}(-4t^{10}-2t^3)=-40t^9-6t^2$.\n 12. Assemble the first derivative as the vector below.\n $$\dfrac{d}{dt}(A\times B)=\langle -20t^4+2,\;56t^6+32t^3,\;-40t^9-6t^2\rangle$$\n 13. Differentiate once more componentwise to get the second derivative $\dfrac{d^2}{dt^2}(A\times B)$.\n 14. Differentiate the $x$-component again: $\dfrac{d}{dt}(-20t^4+2)=-80t^3$.\n 15. Differentiate the $y$-component again: $\dfrac{d}{dt}(56t^6+32t^3)=336t^5+96t^2$.\n 16. Differentiate the $z$-component again: $\dfrac{d}{dt}(-40t^9-6t^2)=-360t^8-12t$.\n 17. Assemble the second derivative as the vector below.\n $$\dfrac{d^2}{dt^2}(A\times B)=\langle -80t^3,\;336t^5+96t^2,\;-360t^8-12t\rangle$$\n 18. Now compute the dot product $A\cdot B$ using $A\cdot B=A_xB_x+A_yB_y+A_zB_z$ and simplify.\n 19. Substitute the components: $A\cdot B=(2t^3)(4t^6)+t^4(-1)+(2t)(-4t)=8t^9-t^4-8t^2$.\n 20. Differentiate the dot product to obtain $\dfrac{d}{dt}(A\cdot B)=\dfrac{d}{dt}(8t^9-t^4-8t^2)=72t^8-4t^3-16t$.\n 21. Final answers collected for clarity are below.\n $$\dfrac{d}{dt}(A\times B)=\langle -20t^4+2,\;56t^6+32t^3,\;-40t^9-6t^2\rangle$$\n $$\dfrac{d^2}{dt^2}(A\times B)=\langle -80t^3,\;336t^5+96t^2,\;-360t^8-12t\rangle$$\n $$\dfrac{d}{dt}(A\cdot B)=72t^8-4t^3-16t$$\n