Question

parametrizations are provided for portions of the same function. For each problem do the following:

(i) Eliminate the parameter to show that the curves are portions of the same function.

(ii) Describe the portion of the graph that each parametrization describes.

(iii) Discuss the direction of motion along with the graph for each parametrization.

$$\begin{gathered} \left(a\right)x=t,y=t^2-1,t\ge 0 \\ \left(b\right)x=-t,y=t^2-1,t\ge 0 \end{gathered}$$

Short answer

Part (a) (i) The equation after eliminating the parameter is $y=x^2-1$

(ii) The derivative, $\frac{dy}{dx}=2x>0$the curve advances down the right-hand side of the parabola.

(iii) $y=x^2-1$the right half of the parabola, As t grows, the motion shifts to the right.

Part (b) (i) The required equation after eliminating the parameter is $y=x^2-1$

(ii) The derivative $x^1=-1<0$of the curve advances down the left-hand side of the parabola.

(iii) $y=x^2-1$left half of the parabola, the motion is to the left as $t$ increases.

Step-by-step solution

Part (a) Step 1: Given information

$x=t,y=t^2-1,t\ge 0$

Part (a) Step 2: Concept

The formula used: Parabola$y=x^2-1$

Part (a) Step 3: Calculation

(i) Consider the parametric equations,

$x=t,y=t^2-1,t\ge 0$

The goal is to remove the parameter and describe the motion in conjunction with the graph. Take the parametric curves for example $x=t,y=t^2-1$

Now substitute $x=t$in the equation $y=t^2-1$

$y=x^2-1$[since by substituting $\left.t=x\right]$

After removing the parameter $t$is the needed equation is $y=x^2-1$

(ii) The equation $y=x^2-1$represents a parabola.

The derivative of the function $y=x^2-1$is as follows,

Differentiate with respect to $t$then,

$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\left(x^2-1\right) \\ \frac{dy}{dx}=\frac{d}{dx}{x}^2-\frac{d1}{dx} \\ \frac{dy}{dx}=2x-0 \end{gathered}$$

Thus the derivative, $\frac{dy}{dx}=2x>0$

As a result, the curve advances down the right-hand side of the parabola.

iii) On the right side of the parabola, the parabola moves.

As the $t$value increases the curve moves on the right side.

Therefore, for the parametric curves $x=t,y=t^2-1$the answer is

$y=x^2-1$the right half of the parabola, and the motion is to the right as $t$increases.

Part (b) Step 1: Given information

$x=-t,y=t^2-1,t\ge 0$

Part (b) Step 2: Calculation

consider the parametric equations,

$x=-t,y=t^2-1,t\ge 0$

The goal is to remove the parameter and describe the motion in conjunction with the graph.

Take the parametric curves $x=-t,y=t^2-1$

Now substitute $x=-t$in the equation $y=t^2-1$

$y=x^2-1[$since by substituting $t=-x]$

Thus, the required equation after eliminating the parameter $t$is $y=x^2-1$

Consider the parametric equations $x=-t,y=t^2-1$

The function's derivative is as follows:

Differentiate with respect to $t$then,

$x^1=-\frac{d}{dt}t$

Thus, the derivative $x^1=-1<0$

As a result, the curve advances along the parabola's left half.

On the left side of the parabola, the parabola moves..

The curve shifts to the left as the $t$value rises.

As a result, the answer for the parametric curves $x=t,y=t^2-1$is $y=x^2-1$the left half of the parabola, with motion to the left as$t$ grows.