Chapter 11.2, Problem 29E

(a)

To determine

Tocalculate:the velocity vector $v\left(t\right)$ .

Answer to Problem 29E

The velocity vector $v\left(t\right)$ is $\langle e^{-t}-t{e}^{-t},-e^{-t}\rangle$ .

Explanation of Solution

Given information:

Position vector $r\left(t\right)=\langle t{e}^{-t},e^{-t}\rangle$

Formula Used:

Velocity vector of an object with position vector $\langle x\left(t\right),y\left(t\right)\rangle$ at time $t$ is:

  $v\left(t\right)=\langle \frac{d}{dt}x\left(t\right),\frac{d}{dt}y\left(t\right)\rangle$

Calculation:

The given position vector is $r\left(t\right)$ .

Now, calculate the velocity vector. Use the formula $v\left(t\right)=\langle \frac{d}{dt}x\left(t\right),\frac{d}{dt}y\left(t\right)\rangle$ .

  $\begin{array}{c}v\left(t\right)=\frac{d}{dt}r\left(t\right)\\ =\frac{d}{dt}\langle t{e}^{-t},e^{-t}\rangle \\ =\langle \frac{d}{dt}t{e}^{-t},\frac{d}{dt}{e}^{-t}\rangle \\ =\langle e^{-t}-t{e}^{-t},-e^{-t}\rangle \end{array}$

Conclusion:

Hence, the required velocity vector $v\left(t\right)$ is $\langle e^{-t}-t{e}^{-t},-e^{-t}\rangle$ .

(b)

To determine

To calculate:The acceleration vector $a\left(t\right)$ .

Answer to Problem 29E

The acceleration vector $a\left(t\right)$ is $\langle -2e^{-t}+t{e}^{-t},e^{-t}\rangle$ .

Explanation of Solution

Given information:

Position vector $r\left(t\right)=\langle t{e}^{-t},e^{-t}\rangle$

Formula Used:

Acceleration vector of an object with position vector $\langle x\left(t\right),y\left(t\right)\rangle$ at time $t$ is:

  $a\left(t\right)=\langle \frac{d^2}{d{t}^2}x\left(t\right),\frac{d^2}{d{t}^2}y\left(t\right)\rangle$

Calculation:

The given position vector is $r\left(t\right)$ .

Now, calculate the velocity vector. Use the formula $a\left(t\right)=\langle \frac{d^2}{d{t}^2}x\left(t\right),\frac{d^2}{d{t}^2}y\left(t\right)\rangle$ .

  $\begin{array}{c}a\left(t\right)=\frac{d^2}{d{t}^2}r\left(t\right)\\ =\frac{d^2}{d{t}^2}\langle t{e}^{-t},e^{-t}\rangle \\ =\langle \frac{d^2}{d{t}^2}t{e}^{-t},\frac{d^2}{d{t}^2}{e}^{-t}\rangle \\ =\langle -2e^{-t}+t{e}^{-t},e^{-t}\rangle \end{array}$

Conclusion:

Hence, the required acceleration vector $a\left(t\right)$ is $\langle -2e^{-t}+t{e}^{-t},e^{-t}\rangle$ .