Chapter 8.4, Problem 2E

1. (a) True or False? The same curve can be described by parametric equations in many different ways.
2. (b) The parametric equations x = 2t, y = (2t)² model the position of a moving object at time t. When t = 0, the object is at $(\boxed{},\boxed{})$, and when t = 1, the object is at $(\boxed{},\boxed{})$.
3. (c) If we eliminate the parameter, we get the equation y = __________, which is the same equation as in Exercise 1(c). So the objects in Exercises 1(b) and 2(b) move along the same __________ but traverse the path differently. Indicate the position of each object when t = 0 and when t = 1 on the following graph.

To determine

To fill: Whether the same curve can be described by different parametric equations.

Answer to Problem 2E

The statement is true that the same curve. can be described by different parametric equations.

Explanation of Solution

The parabolic equation $y=x^2$ have two sets of parametric equations.

The first set of parametric equations is,

$x=t$ and $y={\left(t\right)}^2$.

The other set of parametric equations are,

$x=2t$ and $y={\left(2t\right)}^2$

Thus, the statement is true that the same curve can be described by different parametric equations.

To determine

To fill: The position coordinates $\left(x,y\right)$ of a moving object at time $t=0$ and time $t=1$.

Answer to Problem 2E

The position coordinate $\left(x,y\right)$ of a moving object with parametric equations $x=t$ and $y=t^2$ at time $t=0$ is $\left(0,0\right)$ and at time $t=1$ is $\left(1,1\right)$.

Explanation of Solution

The two parametric equations are $x=2t$ and $y={\left(2t\right)}^2$.

Substitute $0$ for $t$ in the above equations.

The value of $x$ is $0$ and the value of $y$ is $0$.

Clearly, the position coordinate $\left(x,y\right)$ of a moving object at time $t=0$ is $\left(0,0\right)$.

Substitute $1$ for $t$ in the above parametric equations.

The value of $x$ is $1$ and the value of $y$ is $1$.

Clearly, the position coordinate $\left(x,y\right)$ of a moving object at time $t=1$ is $\left(2,4\right)$.

Thus, the position coordinate $\left(x,y\right)$ of a moving object with parametric equations $x=2t$ and $y={\left(2t\right)}^2$ at time $t=0$ is $\left(0,0\right)$ and at time $t=1$ is $\left(2,4\right)$.

To determine

To fill: The equation after elimination of the parameter and the nature of the path of parametric equations $x=t$, $y={\left(t\right)}^2$ and $x=2t$, $y={\left(2t\right)}^2$ is same.

Answer to Problem 2E

The equation after elimination of parameter is $y=x^2$ and the path of the parametric equations $x=t$, $y={\left(t\right)}^2$ and $x=2t$, $y={\left(2t\right)}^2$ along the curve is same.

Explanation of Solution

From part (b) the two parametric equations are $x=2t$ and $y={\left(2t\right)}^2$.

Eliminate the parameter in the above parametric equations.

Substitute x for $2t$ in the equation $y={\left(2t\right)}^2$.

$y=x^2$

Thus, the path of the parametric equations $x=t$, $y={\left(t\right)}^2$ and $x=2t$, $y={\left(2t\right)}^2$ along the curve is same.

The below graph is of the equation $y=x^2$

Figure (1)

The above graph represent the position of object at time $t=0$ and $t=1$.