Calculate $\frac{d}{dt}[\mathbf{r}_1(t) \cdot \mathbf{r}_2(t)]$ and $\frac{d}{dt}[\mathbf{r}_1(t) \times \mathbf{r}_2(t)]$ first by differentiating the product directly and then by applying the formulas $\frac{d}{dt}[\mathbf{r}_1(t) \cdot \mathbf{r}_2(t)] = \mathbf{r}_1(t) \cdot \frac{d\mathbf{r}_2}{dt} + \frac{d\mathbf{r}_1}{dt} \cdot \mathbf{r}_2(t)$ and $\frac{d}{dt}[\mathbf{r}_1(t) \times \mathbf{r}_2(t)] = \mathbf{r}_1(t) \times \frac{d\mathbf{r}_2}{dt} + \frac{d\mathbf{r}_1}{dt} \times \mathbf{r}_2(t)$. $\mathbf{r}_1(t) = \cos(t)\mathbf{i} + \sin(t)\mathbf{j} + 7t\mathbf{k}$, $\mathbf{r}_2(t) = 6\mathbf{i} + t\mathbf{k}$ $\frac{d}{dt}[\mathbf{r}_1(t) \cdot \mathbf{r}_2(t)] = \text{________}$ $\frac{d}{dt}[\mathbf{r}_1(t) \times \mathbf{r}_2(t)] = \text{________}$
Added by Alexander F.
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[r1r2t] = (costi + sintj + 7tk)(6i + tk) To calculate this, we can use the distributive property and the fact that i, j, and k are orthogonal unit vectors. [r1r2t] = (costi)(6i) + (costi)(tk) + (sintj)(6i) + (sintj)(tk) + (7tk)(6i) + (7tk)(tk) Now, let's Show more…
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