Q.17 - Calculus

Question

sketch the parametric curve by plotting points. 

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

Step-by-Step Solution

Answer

The resultant graph is 

Step 1: Given information

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

Step 2: calculation

Consider the parametric curve $x=3 t+1, y=t$ at $t\in [-2,5]$.

The objective is to sketch the parametric curve.

To draw the graph for the parametric equations assume $t=-2,0,2,3,5$.

Substitute different t values in the parametric equations and find the values of x, y.

The point (x, y) When $t=-2$ is,

$(x, y)=(3 t+1, t)$

$$\begin{gathered} (x, y)=\left(3\right(-2)+1,-2) [\sin ce by substituting t=-2] \\ (x, y)=(-6+1,-2) \\ (x, y)=(-5,-2) \{ simplify \} \end{gathered}$$

The point $(x, y)$ When $t=0$ is,

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

The point $(x, y)$ When $t=2$ is,

$$\begin{gathered} (x,y)=(3t+1,t) \\ (x,y)=\left(3\right(2)+1,2) [\text{ since by substituting }t=2] \\ (x,y)=(6+1,2) \\ (x,y)=(7,2)\text{ simplify } \end{gathered}$$

Step 3: Further calculation

The point$(x, y)$ When $t=3$ is,

$$\begin{gathered} (x,y)=(3t+1,t) \\ (x,y)=\left(3\right(3)+1,3) [\text{ since by substituting }t=3] \\ (x,y)=(9+1,3)(x,y)=(10,3)\text{ simplify } \end{gathered}$$

The point $(x, y)$ When $t=5$ is,

$$\begin{gathered} (x,y)=(3t+1,t) \\ (x,y)=\left(3\right(5)+1,5) [\text{ since by substituting }t=5] \\ (x,y)=(15+1,5) \\ (x,y)=(16,5)\text{ simplify } \end{gathered}$$

The tabular representation is as follows:

Step 4: Plot the graph.

The graphical representation is