Question
Let $k$ be a scalar and $\mathbf{r}(t)$ be a differentiable vector function. Prove that $\frac{d}{d t}(k \mathbf{r}(t))=k \mathbf{r}^{\prime}(t) .$ (This is Theorem 11.11 (a).)
Step 1
Step 1: First, we write down the given vector function $\mathbf{r}(t)$ and the scalar multiple $k \mathbf{r}(t)$: \[\mathbf{r}(t)=\langle x(t), y(t)\rangle\] \[k \mathbf{r}(t)=\langle k x(t), k y(t)\rangle\] Show more…
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Let $f(t)$ be a differentiable scalar function and $\mathbf{r}(t)$ be a differentiable vector function. Prove that $$ \frac{d}{d t}(f(t) \mathbf{r}(t))=f^{\prime}(t) \mathbf{r}(t)+f(t) \mathbf{r}^{\prime}(t) $$ (This is Theorem 11.11 (b).)
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Let $\mathbf{r}_{1}(t)$ and $\mathbf{r}_{2}(t)$ be differentiable vector functions with three components each. Prove that $$ \frac{d}{d t}\left(\mathbf{r}_{1}(t) \cdot \mathbf{r}_{2}(t)\right)=\mathbf{r}_{1}^{\prime}(t) \cdot \mathbf{r}_{2}(t)+\mathbf{r}_{1}(t) \cdot \mathbf{r}_{2}^{\prime}(t) $$ (This is Theorem 11.11 (c).)
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