Question

Use the given velocity vectors $v\left(t\right)=r\text{'}\left(t\right)$and initial positions in exercise to find the position function $r\left(t\right)$.

localid="1650736930792" $v\left(t\right)=e^{t}i+\ln tj,r\left(1\right)=i-6j$

Short answer

$r\left(t\right)=(e^t+1-e)i+(t\ln t-t-5)j$

Step-by-step solution

Given Information

Consider $v\left(t\right)=e^{t}i+\ln tj,r\left(1\right)=i-6j$

The objective is to find the position function $r\left(t\right)$.

since $v\left(t\right)=r\text{'}\left(t\right)$

we have

$$\begin{gathered} r\left(t\right)=\int v\left(t\right)dt \\ =\int (e^{t}i+\ln tj)dt \\ =i\int e^{t}dt+j\int \ln tdt \\ =e^{t}i+(t\ln t-t)j+c \end{gathered}$$

Where c is a vector constant.

$=(e^t+c_1)i+(t\ln t-t+c_2)j$where$c_1$and$c_2$are scalars.

Expression

So

$$\begin{gathered} r\left(t\right)=(e^t+c_1)i+(t\ln t-t+c_2)j \\ r\left(1\right)=(e^1+c_1)i+(1\ln 1-1+c_2)j \\ r\left(1\right)=(e+c_1)i+(-1+c_2)j...\left(1\right) \end{gathered}$$

but the given initial position $r\left(t\right)=i-6j.........\left(2\right)$

Comparing (1) and (2)

$$\begin{gathered} e+c_1=1,-1+c_2=-6 \\ {c}_1=1-e,c_2=-6+1 \\ {c}_1=1-e,c_2=-5 \\ r\left(t\right)=(e^t+1-e)i+(t\ln t-t-5)j \end{gathered}$$

Calculation

Check: Taking the derivative of $r\left(t\right)=(e^t+1-e)i+(t\ln t-t-5)j$,

We obtain $r\text{'}\left(t\right)=e^{t}i+\ln tj$, which is the correct tangent vector function. Next, we evaluate

$$\begin{gathered} r\left(1\right)=(e^t+1-e)t+(1\ln 1-1-5)j \\ =j-6j \end{gathered}$$

Which is true according to the problem.

Thus$r\left(t\right)=(e^t+1-e)i+(t\ln t-t-5)j$