Q 19. - Calculus

Question

Sketch the parametric curve by plotting points. 

$x = 3t + 1, y = t, t \in [-2, 5]$

Step-by-Step Solution

Answer

The graph  

Step 1: Given information

$x = 3t + 1, y = t, t \in [-2, 5]$

Step 2: Concept

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

Step 3: Calculation

Consider the parametric curves $x=1-2t,y=5-3t,t\in \mathrm{ℝ}$

The goal is to draw the parametric curve.

To draw the graph for the parametric equations assume $t=-2,-1,0,1,2$

Find the values of $x, y$ by substituting different $t$ values in the parametric equations.

The point $(x, y)$ When $t=-2$ is,

$$\begin{gathered} (x,y)=(1-2t,5-3t) \\ (x,y)=(1-2·-2,5-3·-2) [\text{ since by substituting }t=-2] \\ (x,y)=(1+4,5+6) \\ (x,y)=(5,11)\text{ simplify } \end{gathered}$$

The point $(x, y)$ When $t=-1$ is,

$$\begin{gathered} (x,y)=(1-2t,5-3t) \\ (x,y)=(1-2·-1,5-3·-1) \\ (x,y)=(1+2,5+3)[\text{ since by substituting }t=-1] \\ (x,y)=(3,8)\text{ simplify } \end{gathered}$$

The point $(x, y)$ When $t=0$ is,

$$\begin{gathered} (x,y)=(1-2t,5-3t) \\ (x,y)=(1-2·0,5-3·0) \text{ [since by substituting }t=0\left] \\ \right(x,y)=(1,5)\text{ simplify } \end{gathered}$$

The point $(x, y)$ wither $t=1$ is,

$$\begin{gathered} (x,y)=(1-2t,5-3t) \\ (x,y)=(1-2·1,5-3·1) [\text{ since by substituting }t=1] \\ (x,y)=(1-2,5-3) \\ (x,y)=(-1,2)\text{ simplify } \end{gathered}$$

The point $(x, y)$ When $t=2$ is,

$$\begin{gathered} (x,y)=(1-2t,5-3t) \\ (x,y)=(1-2·2,5-3·2) \text{ [since by substituting }t=2\left] \\ \right(x,y)=(1-4,5-6\left) \\ \right(x,y)=(-3,-1)\text{ simplify } \end{gathered}$$

Step 4: Calculation

The tabular representation of the points is as follows,

The graphical representation is shown below,

Therefore, the solution is a required graph.