Chapter 11.2, Problem 38E

(a)

To determine

To calculate: The velocity vector.

Answer to Problem 38E

The required velocity and speed of the particle are $\langle 2\sqrt{2},0\rangle$ and $2\sqrt{2}$ respectively.

Explanation of Solution

Given information:

The position vectors: $\overrightarrow{x}=e^t+e^{-t}$ , $\overrightarrow{y}=e^t-e^{-t}$

Calculation:

Consider the position vector be $\overrightarrow{r}\left(t\right)$ .

So, $\overrightarrow{r}\left(t\right)=\langle e^t+e^{-t},e^t-e^{-t}\rangle$

It is known that for a position vector $\overrightarrow{r}\left(t\right)$ of a particle, the first derivative of $\overrightarrow{r}\left(t\right)$ is the velocity vector of the particle that is, $\overrightarrow{v}\left(t\right)$ .

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

Hence, the required velocity is $\langle e^t-e^{-t},e^t+e^{-t}\rangle$ .

(b)

To determine

To calculate: The value of $\underset{t\to \infty }{\lim }\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ .

Answer to Problem 38E

The required value of $\underset{t\to \infty }{\lim }\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ is 1.

Explanation of Solution

Given information:

The position vectors: $\overrightarrow{x}=e^t+e^{-t}$ , $\overrightarrow{y}=e^t-e^{-t}$

Calculation:

Consider the position vector be $\overrightarrow{r}\left(t\right)$ .

So, $\overrightarrow{r}\left(t\right)=\langle e^t+e^{-t},e^t-e^{-t}\rangle$

Here, $\begin{array}{c}\frac{dy}{dt}=\frac{d}{dt}\left(e^t-e^{-t}\right)\\ =e^t+e^{-t}\end{array}$

Also, $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left(e^t+e^{-t}\right)\\ =e^t-e^{-t}\end{array}$

Now, substitute $e^t+e^{-t}$ for $\frac{dy}{dt}$ and $e^t-e^{-t}$ for $\frac{dx}{dt}$ in the expression $\underset{t\to \infty }{\lim }\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ and simplify.

  $\begin{array}{c}\underset{t\to \infty }{\lim }\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\underset{t\to \infty }{\lim }\left(\frac{e^t+e^{-t}}{e^t-e^{-t}}\right)\\ =\underset{t\to \infty }{\lim }\left(\frac{e^{2t}+1}{e^{2t}-1}\right)\\ =1\end{array}$

Hence, the required value of $\underset{t\to \infty }{\lim }\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ is 1.

(c)

To determine

To prove: Algebraically that the particle moves on the hyperbola $x^2-y^2=4$ .

Explanation of Solution

Given information:

The position vectors: $\overrightarrow{x}=e^t+e^{-t}$ , $\overrightarrow{y}=e^t-e^{-t}$

Calculation:

Consider the position vector be $\overrightarrow{r}\left(t\right)$ .

So, $\overrightarrow{r}\left(t\right)=\langle e^t+e^{-t},e^t-e^{-t}\rangle$

Substitute $e^t+e^{-t}$ for $x$ and $e^t-e^{-t}$ for $y$ in the left hand side of the hyperbola $x^2-y^2=4$ .

  $\begin{array}{c}{x}^2-y^2={\left(e^t+e^{-t}\right)}^2-{\left(e^t-e^{-t}\right)}^2\\ =\left(e^{2t}+2+e^{-2t}\right)-\left(e^{2t}-2+e^{-2t}\right)\\ =\cancel{e^{2t}}+2+\cancel{e^{-2t}}-\cancel{e^{2t}}+2-\cancel{e^{-2t}}\\ =4\end{array}$

Thus, left hand side = right hand side

Hence, it is proved that the particle moves on the hyperbola $x^2-y^2=4$ .

(d)

To determine

To graph: The path of the particle and showing the velocity vector at $t=0$ .

Explanation of Solution

Given information:

The position vectors: $\overrightarrow{x}=e^t+e^{-t}$ , $\overrightarrow{y}=e^t-e^{-t}$

Calculation:

Consider the position vector be $\overrightarrow{r}\left(t\right)$ .

So, $\overrightarrow{r}\left(t\right)=\langle e^t+e^{-t},e^t-e^{-t}\rangle$

It is known that for a position vector $\overrightarrow{r}\left(t\right)$ of a particle, the first derivative of $\overrightarrow{r}\left(t\right)$ is the velocity vector of the particle that is, $\overrightarrow{v}\left(t\right)$ .

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

Substitute 0 for t in the above expression.

  $\begin{array}{c}\overrightarrow{v}\left(t\right)=\langle e^0-e^{-\left(0\right)},e^0+e^{-\left(0\right)}\rangle \\ =\langle 1-1,1+1\rangle \\ =\langle 0,2\rangle \end{array}$

The graph of the path of the particle is shown below: