Q.20

Sketch the parametric curve by plotting points.

$x = 2 t - 1, y = 3 t + 5, t \in ℝ$

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Answer

The required graph is

1Step:1 Given information

Consider the parametric curves. $x = 2 t - 1, y = 3 t + 5, t \in ℝ$

2Step 2: Calculation

The objective is to sketch the parametric curve.

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

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

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

$(x, y) = (2 t - 1, 3 t + 5) (x, y) = (2 \cdot - 2 - 1, 3 \cdot - 2 + 5) (x, y) = (- 4 - 1, - 6 + 5) (x, y) = (- 5, - 1)$

The point $(x, y)$ When $t = - 1$ is,

$(x, y) = (2 t - 1, 3 t + 5) (x, y) = (2 \cdot - 1 - 1, 3 \cdot - 1 + 5) (x, y) = (- 2 - 1, - 3 + 5) (x, y) = (- 3, 2)$

The point $(x, y)$ When $t = 0$ is,

$(x, y) = (2 t - 1, 3 t + 5) (x, y) = (2 \cdot 0 - 1, 3 \cdot 0 + 5) (x, y) = (- 1, 5)$

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

$(x, y) = (2 t - 1, 3 t + 5) (x, y) = (2 \cdot 1 - 1, 3 \cdot 1 + 5) (x, y) = (2 - 1, 3 + 5) (x, y) = (1, 8)$

3Step 3: Further calculation

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

$(x, y) = (2 t - 1, 3 t + 5) (x, y) = (2 \cdot 2 - 1, 3 \cdot 2 + 5) (x, y) = (4 - 1, 6 + 5) (x, y) = (3, 11)$

The following is a tabular depiction of the point.

Therefore, the graph is