find $\frac{d}{dt}mathbf{r}(t)cdotmathbf{u}(t)$ and $\frac{d}{dt}mathbf{r}(t)\timesmathbf{u}(t)$ in two…

find $\frac{d}{dt}mathbf{r}(t)cdotmathbf{u}(t)$ and $\frac{d}{dt}mathbf{r}(t)\timesmathbf{u}(t)$ in two different ways. (a) (i) find the product first, then differentiate. (ii) apply the properties of the derivative. (b) (i) find the product first, then differentiate. (ii) apply the properties of the derivative. $mathbf{r}(t)=mathbf{i}+9t^{2}mathbf{j}+t^{3}mathbf{k}$, $mathbf{u}(t)=t^{4}mathbf{k}$

find $\frac{d}{dt}mathbf{r}(t)cdotmathbf{u}(t)$ and $\frac{d}{dt}mathbf{r}(t)\timesmathbf{u}(t)$ in two different ways. (a) (i) find the product first, then differentiate. (ii) apply the properties of the derivative. (b) (i) find the product first, then differentiate. (ii) apply the properties of the derivative. $mathbf{r}(t)=mathbf{i}+9t^{2}mathbf{j}+t^{3}mathbf{k}$, $mathbf{u}(t)=t^{4}mathbf{k}$

Answer

Explanation:

Step1: Recall product - rule for dot - product

The product - rule for the derivative of the dot - product of two vector - valued functions $\mathbf{r}(t)$ and $\mathbf{u}(t)$ is $\frac{d}{dt}[\mathbf{r}(t)\cdot\mathbf{u}(t)]=\mathbf{r}'(t)\cdot\mathbf{u}(t)+\mathbf{r}(t)\cdot\mathbf{u}'(t)$. First, find $\mathbf{r}(t)=\mathbf{i}+9t^{2}\mathbf{j}+t^{3}\mathbf{k}$ and $\mathbf{u}(t)=t^{4}\mathbf{k}$. Then $\mathbf{r}'(t)=0\mathbf{i} + 18t\mathbf{j}+3t^{2}\mathbf{k}$ and $\mathbf{u}'(t)=4t^{3}\mathbf{k}$. $\mathbf{r}'(t)\cdot\mathbf{u}(t)=(0\mathbf{i}+18t\mathbf{j}+3t^{2}\mathbf{k})\cdot(t^{4}\mathbf{k}) = 3t^{6}$. $\mathbf{r}(t)\cdot\mathbf{u}'(t)=(\mathbf{i}+9t^{2}\mathbf{j}+t^{3}\mathbf{k})\cdot(4t^{3}\mathbf{k})=4t^{6}$. So, $\frac{d}{dt}[\mathbf{r}(t)\cdot\mathbf{u}(t)]=3t^{6}+4t^{6}=7t^{6}$.

Step2: Recall product - rule for cross - product

The product - rule for the derivative of the cross - product of two vector - valued functions $\mathbf{r}(t)$ and $\mathbf{u}(t)$ is $\frac{d}{dt}[\mathbf{r}(t)\times\mathbf{u}(t)]=\mathbf{r}'(t)\times\mathbf{u}(t)+\mathbf{r}(t)\times\mathbf{u}'(t)$. $\mathbf{r}'(t)\times\mathbf{u}(t)=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\0&18t&3t^{2}\0&0&t^{4}\end{vmatrix}=\mathbf{i}(18t\cdot t^{4}-0)-\mathbf{j}(0 - 0)+\mathbf{k}(0 - 0)=18t^{5}\mathbf{i}$. $\mathbf{r}(t)\times\mathbf{u}'(t)=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\1&9t^{2}&t^{3}\0&0&4t^{3}\end{vmatrix}=\mathbf{i}(36t^{5}-0)-\mathbf{j}(4t^{3}-0)+\mathbf{k}(0 - 0)=36t^{5}\mathbf{i}-4t^{3}\mathbf{j}$. $\frac{d}{dt}[\mathbf{r}(t)\times\mathbf{u}(t)]=18t^{5}\mathbf{i}+36t^{5}\mathbf{i}-4t^{3}\mathbf{j}=54t^{5}\mathbf{i}-4t^{3}\mathbf{j}$.

Answer:

$\frac{d}{dt}[\mathbf{r}(t)\cdot\mathbf{u}(t)] = 7t^{6}$; $\frac{d}{dt}[\mathbf{r}(t)\times\mathbf{u}(t)]=54t^{5}\mathbf{i}-4t^{3}\mathbf{j}$