Q 11. - Calculus

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.

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

Step-by-Step Solution

Answer

Part (a) (i) The equation after eliminating the parameter $t$ is $y=\sin x$

(ii) The derivative $y^{\text{'}}=\cos t$ So, the curve moves on the right of the $y$ axis.

(iii) In the parametric curves $x=t,y=\sin t$ the answer is $y=\sin x$ right of the $y-$axis, In the direction of motion is to the right as $t$ increases.

Part (b) The required equation after eliminating the parameter $t$ is $y=\sin x$

(ii) The derivative $y^1=\cos (t-1)$ So, the curve moves on the right of the $y-$ axis.

(iii) In The parametric curves $x=t-1,y=\sin (t-1)$ the answer is $y=\sin x$ right of the $y-$axis, the direction of motion is to the right as $t-$ increases.

Part (a) Step 1: Given information

$x = t, y = \sin t, t \ge 0$

Part (a) Step 2: Concept

A collection of quantities is defined as a function of one or more independent variables called parameters in a parametric equation.

Part (a) Step 3: Calculation

(i) consider the parametric equations,

$x=t,y=\sin t,t\ge 0$

The goal is to remove the parameter and characterize the graph's motion in terms of direction. Take the parametric curves $x=t,y=\sin t$

Take $x=t$

Now substitute $t=x$ in the equation $y=\sin t$

$y=\sin x [$ since by substituting $t=x]$

Thus, the required equation after eliminating the parameter $t$ is $y=\sin x$

(ii) Consider the parametric equations $x=t,y=\sin t$

The derivative of the function $x=t$ is as follows,

Differentiate with respect to t then,

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

Thus, the derivative $x^1=1>0$

Take $y=\sin t$

The derivative of the function $y=\sin t$ is as follows,

Differentiate with respect to $t$ then,

$$\begin{gathered} y^{\text{'}}=\frac{d}{dt}\sin t \\ {y}^{\text{'}}=\cos t \end{gathered}$$

Thus, the derivative $y^{\text{'}}=\cos t$

So, the curve moves on the right of the $y$ axis

(iii) The curve moves on the right side of the $y$ axis.

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

Therefore, for the parametric curves, $x=t,y=\sin t$ the answer is

$y=\sin x$ right of the $y-$axis, the direction of motion is to the right as $t$ increases.

Part (b) Step 1: Given information

$x=t-1,y=\sin (t-1),t>1$

Part (b) Step 2: Calculation

(i) Consider the parametric equations,

$x=t-1,y=\sin (t-1),t>1$

The goal is to remove the parameter and describe the motion in conjunction with the graph. Take the parametric curves$x=t-1,y=\sin (t-1)$

Take $x=t-1$

On all sides of the equation, add one.

$$\begin{gathered} x+1=t-1+1 \\ x+1=t \\ t=x+1 \end{gathered}$$

Now substitute $t=x+1$ in the equation $y=\sin (t-1)$

$$\begin{gathered} y=\sin (x+1-1)\text{ [since by substituting }t=x+1] \\ y=\sin (x+1-1)\text{ ) } \\ y=\sin x \end{gathered}$$

Thus, the required equation after eliminating the parameter $t$ is $y=\sin x$

(ii) The derivative of the function $x=t-1$ is as follows,

Differentiate with respect to $t$ then,

$\frac{dx}{dt}=1$

Thus, the derivative $x^1=1>0$

Take $y=\sin (t-1)$

The derivative of the function $y=\sin (t-1)$ is as follows.

Differentiate with respect to $t$ then,

$$\begin{gathered} y^{\text{'}}=\frac{d}{dt}\sin (t-1) \\ {y}^{\text{'}}=\cos (t-1) \end{gathered}$$

Thus, the derivative $y^1=\cos (t-1)$

So, the curve moves on the right of the $y$ axis.

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

Therefore, for the parametric curves $x=t-1,y=\sin (t-1)$ the answer is

$y=\sin x$ right of the $y-$axis, the direction of motion is to the right as $t$ increases.