The main purpose of this question is to find the derivative of a given vector-valued function.
A vector function accepts one or perhaps many variables and yields a vector. Computer graphics, computer vision, and machine learning algorithms frequently use vector-valued functions. They are especially helpful for determining space curve parametric equations. It is a function possessing two characteristics such as having a domain as a set of real numbers and its range comprising of a set of vectors. Commonly, these functions are the extended form of the scalar functions.
The vector-valued function can take a scalar or a vector as an input. Moreover, the dimensions of range and domain of such a function are not related to each other. This function typically depends upon one parameter, that is, $t$ often regarded as time, and results in a vector $\textbf{v}(t)$. And in terms of $\textbf{i}$, $\textbf{j}$ and $\textbf{k}$, i.e., the unit vectors, the vector-valued function has a specific form such as: $\textbf{r}(t)=x(t)\textbf{i}+y(t)\textbf{j}+z(t)\textbf{k}$.
Expert Answer
Let $\dfrac{d}{dt}[\textbf{r}(t)]=\textbf{r}'(t)$, then:
$\textbf{r}'(t)=\dfrac{d}{dt}[e^{t^2}\textbf{i}-\textbf{j}+\ln(1+3t)\textbf{k}]$
Using the chain rule on the first and third term, and the power rule on the second term as:
$\textbf{r}'(t)=e^{t^2}\cdot \dfrac{d}{dt}[t^2]\textbf{i}-0\cdot\textbf{j}+\dfrac{1}{1+3t}\dfrac{d}{dt}[1+3t]\textbf{k}$
$\textbf{r}'(t)=e^{t^2}(2t)+\dfrac{1}{1+3t}(3)\textbf{k}$
$\textbf{r}'(t)=2te^{t^2}+\dfrac{3}{1+3t}\textbf{k}$
Example 1
Find the derivative of the following vector-valued function:
$\textbf{r}(t)=\cos t\textbf{i}+\sin t\textbf{j}+\tan t\textbf{k}$
Solution
The graph of the vector-valued function given in Example 1.
$\textbf{r}'(t)=-\sin t\textbf{i}+\cos t\textbf{j}+\sec^2 t\textbf{k}$
Example 2
Find the derivative of the following vector-valued function:
$\textbf{r}(t)=t^2\ln 2t\textbf{i}+3e^{2t}\textbf{j}+(t^3+\cos t)\textbf{k}$
Solution
Using the product rule on the first term, the chain rule on the second term, and the sum rule on the last term as:
$\textbf{r}'(t)=\left[t^2\dfrac{d}{dt}(\ln 2t)+\ln 2t\dfrac{d}{dt}(t^2)\right]\textbf{i}+3\dfrac{d}{dt}(e^{2t})\textbf{j}+\dfrac{d}{dt}[t^3+\cos t]\textbf{k}$
$\textbf{r}'(t)=\left(t^2\cdot\left(\dfrac{1}{2t}\cdot 2\right)+\ln 2t\cdot 2t\right)\textbf{i}+3\cdot 2 e^{2t}\textbf{j}+(3t^2-\sin t)\textbf{k}$
$\textbf{r}'(t)=(t+2t\ln 2t)\textbf{i}+6e^{2t}\textbf{j}+(3t^2-\sin t)\textbf{k}$
Example 3
Let the two vectors be given by:
$\textbf{r}(t)=(t+1)\textbf{i}-3t\textbf{j}+(t^2+4)\textbf{k}$ and $\textbf{v}(t)=(2t+6)\textbf{i}+t\textbf{j}+(t^3-3)\textbf{k}$
Find $\dfrac{d}{dt}[\textbf{r}(t)\cdot \textbf{v}(t)]$.
Solution
Since $\dfrac{d}{dt}[\textbf{r}(t)\cdot \textbf{v}(t)]=\textbf{r}'(t)\cdot \textbf{v}(t)+\textbf{r}(t)\cdot \textbf{v}'(t)$
Now, $\textbf{r}'(t)=\textbf{i}-3\textbf{j}+2t\textbf{k}$
and $\textbf{v}'(t)=2\textbf{i}+\textbf{j}+3t^2\textbf{k}$
Also, $\textbf{r}'(t)\cdot \textbf{v}(t)=(\textbf{i}-3\textbf{j}+2t\textbf{k})\cdot((2t+6)\textbf{i}+t\textbf{j}+(t^3-3)\textbf{k})$
$=(2t+6)-3t+2t(t^3-3)$
$=2t+6-3t+2t^4-6t$
$=2t^4-7t+6$
And $\textbf{r}(t)\cdot \textbf{v}'(t)=((t+1)\textbf{i}-3t\textbf{j}+(t^2+4)\textbf{k})\cdot(2\textbf{i}+\textbf{j}+3t^2\textbf{k})$
$=2(t+1)-3t+3t^2(t^2+4)$
$=2t+2-3t+3t^4+12t^2$
$=3t^4+12t^2-t+2$
Finally, we have:
$\dfrac{d}{dt}[\textbf{r}(t)\cdot \textbf{v}(t)]=2t^4-7t+6+3t^4+12t^2-t+2$
$=5t^4+12t^2-8t+8$
Example 4
Consider the same functions as in example 3. Find $\dfrac{d}{dt}[\textbf{r}(t)-\textbf{v}(t)]$.
Solution
Since $\dfrac{d}{dt}[\textbf{r}(t)-\textbf{v}(t)]=\dfrac{d}{dt}[\textbf{r}(t)]-\dfrac{d}{dt}[\textbf{v}(t)]$
or $\dfrac{d}{dt}[\textbf{r}(t)-\textbf{v}(t)]=\textbf{r}'(t)-\textbf{v}'(t)$
Therefore, $\dfrac{d}{dt}[\textbf{r}(t)]=\textbf{r}'(t)=\textbf{i}-3\textbf{j}+2t\textbf{k}$
and $\dfrac{d}{dt}[\textbf{v}(t)]=\textbf{v}'(t)=2\textbf{i}+\textbf{j}+3t^2\textbf{k}$
So that, $\dfrac{d}{dt}[\textbf{r}(t)-\textbf{v}(t)]=(\textbf{i}-3\textbf{j}+2t\textbf{k})-(2\textbf{i}+\textbf{j}+3t^2\textbf{k})$
$=[(1-2)\textbf{i}+(-3-1)\textbf{j}+(2t-3t^2)\textbf{k}]$
$=-\textbf{i}-4\textbf{j}+(2t-3t^2)\textbf{k}$
or $\dfrac{d}{dt}[\textbf{r}(t)-\textbf{v}(t)]=-\textbf{i}-4\textbf{j}+t(2-3t)\textbf{k}$
Images/mathematical drawings are created with GeoGebra.