91Ó°ÊÓ

First, we need to find the cross product of the given vectors. $$\begin{aligned} \left( t^3\mathbf{i}-2t\mathbf{j}-2\mathbf{k}\right) \times\left( t\mathbf{i}-t^2\mathbf{j}-t^3\mathbf{k}\right) \end{aligned}$$ Using the determinant method to find the cross product: $$ \begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ t^3 & -2t & -2 \\ t & -t^2 & -t^3 \end{pmatrix} $$ Calculating the determinant, we get: $$\begin{aligned} C &= (t^3\mathbf{i}-2t\mathbf{j}-2\mathbf{k})\times (t\mathbf{i}-t^2\mathbf{j}-t^3\mathbf{k})\\ &=(-2t)(-t^3) \mathbf{i} -(2t^3)(-t^3) \mathbf{j} +(t)(-t^3)\mathbf{k}-(-2)(-t^2) \mathbf{i} +(-2t)(t) \mathbf{j} -(t^3)(-t^2) \mathbf{k}\\ &=(2t^4+2t^2)\mathbf{i} + (2t^6-2t^2) \mathbf{j} +(t^4+t^3) \mathbf{k} \end{aligned}$$

Now, we compute the derivative of each component of the cross product vector with respect to t: $$\frac{d}{dt}(C)= \frac{d}{dt}((2t^4+2t^2)\mathbf{i}+(2t^6-2t^2)\mathbf{j}+(t^4+t^3)\mathbf{k})$$ Differentiating component by component: $$\begin{aligned} \frac{dC}{dt} &= \left(\frac{d(2t^4+2t^2)}{dt} \mathbf{i} + \frac{d(2t^6-2t^2)}{dt} \mathbf{j} + \frac{d(t^4+t^3)}{dt} \mathbf{k}\right)\\ &= (8t^3 + 4t)\mathbf{i} + (12t^5 - 4t)\mathbf{j} +(4t^3 + 3t^2) \mathbf{k}\end{aligned}$$ So, the derivative of the cross product is: $$\frac{d}{dt}(C) = (8t^3 + 4t)\mathbf{i} + (12t^5 - 4t)\mathbf{j} +(4t^3 + 3t^2) \mathbf{k}$$