Chapter 10.2, Problem 47E

$\text{(a)}$

To determine

To find: The velocity.

Answer to Problem 47E

The velocity: $\frac{-4t}{{\left(1+t^2\right)}^2},\frac{2-2t^2}{{\left(1+t^2\right)}^2}$

Explanation of Solution

Given information:

The particle moves on the circle $x^2+y^2=1$ .

Any time $t\ge 0$ is $\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}$

Calculation:

  $x(t)=\frac{1-t^2}{1+t^2}$

Use quotient rule,

  $\begin{array}{c}\frac{dx(t)}{dt}=\frac{-2t\left(1+t^2\right)-2t\left(1-t^2\right)}{{\left(1+t^2\right)}^2}\\ \frac{dx(t)}{dt}=\frac{-2t-2t^3-2t+2t^3}{{\left(1+t^2\right)}^2}\\ \frac{dx(t)}{dt}=\frac{-4t}{{\left(1+t^2\right)}^2}\end{array}$

And,

  $y(t)=\frac{2t}{1+t^2}$

Use quotient rule,

  $\begin{array}{c}\frac{dy(t)}{dt}=\frac{2\left(1+t^2\right)-2t(2t)}{{\left(1+t^2\right)}^2}\\ \frac{dy(t)}{dt}=\frac{2+2t^2-4t^2}{{\left(1+t^2\right)}^2}\\ \frac{dy(t)}{dt}=\frac{2-2t^2}{{\left(1+t^2\right)}^2}\end{array}$

Then,

  $\begin{array}{c}\overrightarrow{v}(t)=\frac{dx(t)}{dt},\frac{dy(t)}{dt}\\ =\frac{-4t}{{\left(1+t^2\right)}^2},\frac{2-2t^2}{{\left(1+t^2\right)}^2}\end{array}$

Therefore, the required velocity is $\frac{-4t}{{\left(1+t^2\right)}^2},\frac{2-2t^2}{{\left(1+t^2\right)}^2}$ .

$\text{(b)}$

To determine

To find: The particle is or never rest.

Answer to Problem 47E

Because the velocity components are never both $0$ at the same.

Explanation of Solution

Given information:

The particle moves on the circle $x^2+y^2=1$ .

Any time $t\ge 0$ is $\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}$

Calculation:

For the particle to be at rest both $\frac{dx}{dt}$ and $\frac{dy}{dt}$ have to be $0$ at the same.

Then,

  $\begin{array}{c}\frac{dx}{dt}=\frac{-4t}{{\left(1+t^2\right)}^2}\\ =0\\ \Rightarrow -4t=0\\ \Rightarrow t=0\\ \frac{dy}{dt}=\frac{2-2t^2}{{\left(1+t^2\right)}^2}\\ =0\\ \Rightarrow 2-2t^2=0\\ \Rightarrow 2t^2=2\\ \Rightarrow t^2=1\\ \Rightarrow t=1\end{array}$

Therefore, the required the velocity components are never both $0$ at the same.

$(c)$

To determine

To find: The coordinates of the point.

Answer to Problem 47E

The answer: $(-1,0)$

Explanation of Solution

Given information:

The particle moves on the circle $x^2+y^2=1$ .

Any time $t\ge 0$ is $\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}$

Calculation:

Find the limit of each coordinate as $t$ approaches infinity.

  

Indeterminate form - apply L' Hospital

  

Indeterminate form - apply L' Hospital

  

Therefore, the required particle approaches $(-1,0)$ as $t$ increases without bound.