Question

Compute the following derivatives. $$\frac{d}{dt}((t^3\mathbf{i}-2t\mathbf{j}-2\mathbf{k})\times (t\mathbf{i}-t^2\mathbf{j}-t^3\mathbf{k}))$$

Short answer

Question: Find the derivative with respect to time (t) of the cross product of the vectors $( t^3\mathbf{i}-2t\mathbf{j}-2\mathbf{k})$ and $(t\mathbf{i}-t^2\mathbf{j}-t^3\mathbf{k})$. Answer: $\frac{d}{dt}(C) = (8t^3 + 4t)\mathbf{i} + (12t^5 - 4t)\mathbf{j} +(4t^3 + 3t^2) \mathbf{k}$

Step-by-step solution

Cross product of the vectors

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

Differentiate the cross product with respect to time

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: $$\begin{array}{rl}\frac{dC}{dt}& =(\frac{d(2t^4+2t^2)}{dt}\mathbf{i}+\frac{d(2t^6-2t^2)}{dt}\mathbf{j}+\frac{d(t^4+t^3)}{dt}\mathbf{k})\\ & =(8t^3+4t)\mathbf{i}+(12t^5-4t)\mathbf{j}+(4t^3+3t^2)\mathbf{k}\end{array}$$ So, the derivative of the cross product is: $$\frac{d}{dt}(C)=(8t^3+4t)\mathbf{i}+(12t^5-4t)\mathbf{j}+(4t^3+3t^2)\mathbf{k}$$