Pre-Calculus Parametrics Worksheet #2 - Free Printable

1. Fill in the table and sketch the parametric equation for t [-2,6]

| t | x | y |
|----|------------|----|
| -2 | √5 ≈ 2.24 | 4 |
| -1 | √2 ≈ 1.41 | 3 |
| 0 | 1 | 2 |
| 1 | √2 ≈ 1.41 | 1 |
| 2 | √5 ≈ 2.24 | 0 |
| 3 | √10 ≈ 3.16 | -1 |
| 4 | √17 ≈ 4.12 | -2 |
| 5 | √26 ≈ 5.10 | -3 |
| 6 | √37 ≈ 6.08 | -4 |

Sketch: Plot the points (x,y) from the table. The curve starts at approximately (2.24, 4) for t=-2, moves down and right to a minimum x-value of 1 at t=0, then moves up and right as t increases to 6. The graph is symmetric about the line t=0 (y-axis in the parameter space) for the x-values, but y decreases linearly with t.

2. x = 6 - t, y = √(3t) - 4
Solve for t from the first equation: t = 6 - x.
Substitute into the second: y = √(3(6 - x)) - 4 = √(18 - 3x) - 4.
Domain: 18 - 3x ≥ 0 → x ≤ 6.
Rectangular equation: y = √(18 - 3x) - 4, for x ≤ 6.

3. x = (1/2)t + 4, y = t³
Solve for t from the first equation: t = 2(x - 4).
Substitute into the second: y = [2(x - 4)]³ = 8(x - 4)³.
Rectangular equation: y = 8(x - 4)³.

4. x = 3 cos t, y = 3 sin t
Use the identity cos²t + sin²t = 1.
(x/3)² + (y/3)² = 1 → x²/9 + y²/9 = 1 → x² + y² = 9.
Rectangular equation: x² + y² = 9.

5. x = cos t, y = 2 sin²t
Use the identity sin²t = 1 - cos²t.
y = 2(1 - cos²t) = 2(1 - x²) = 2 - 2x².
Rectangular equation: y = 2 - 2x², for -1 ≤ x ≤ 1.

6. x = 4 + 2 cos t, y = -1 + 4 sin t
Rearrange: (x - 4)/2 = cos t, (y + 1)/4 = sin t.
Use the identity cos²t + sin²t = 1:
[(x - 4)/2]² + [(y + 1)/4]² = 1 → (x - 4)²/4 + (y + 1)²/16 = 1.
Rectangular equation: (x - 4)²/4 + (y + 1)²/16 = 1.

7. y = (x + 2)³ - 4
Set 1: Let x = t, then y = (t + 2)³ - 4.
Parametric equations: x = t, y = (t + 2)³ - 4.
Set 2: Let x + 2 = t, so x = t - 2, then y = t³ - 4.
Parametric equations: x = t - 2, y = t³ - 4.

8. x = √(y² - 3)
Note: For real x, y² - 3 ≥ 0 → |y| ≥ √3.
Set 1: Let y = t, then x = √(t² - 3), for |t| ≥ √3.
Parametric equations: x = √(t² - 3), y = t, for |t| ≥ √3.
Set 2: Let y = √3 sec t (for t in (-π/2, π/2) excluding 0, or other intervals where sec t ≥ 1 or ≤ -1).
Then x = √(3 sec²t - 3) = √(3(sec²t - 1)) = √(3 tan²t) = √3 |tan t|.
To handle the absolute value, we can define two cases or use a parameter that naturally gives the correct sign. A simpler alternative is to let y = √3 cosh u (hyperbolic cosine, which is always ≥1).
Then x = √(3 cosh²u - 3) = √(3(cosh²u - 1)) = √(3 sinh²u) = √3 |sinh u|.
Since sinh u can be positive or negative, and x must be non-negative (as it's a square root), we take x = √3 |sinh u|. But to avoid the absolute value, we can restrict u ≥ 0, giving x = √3 sinh u, y = √3 cosh u.
Parametric equations (Set 2): x = √3 sinh u, y = √3 cosh u, for u ≥ 0.
(Note: This only gives the upper branch where y ≥ √3. To get the lower branch where y ≤ -√3, we could use y = -√3 cosh u, x = √3 sinh u for u ≥ 0, or use a different parameterization.)