Points, Vectors, and Functions Exercises

Answer

• Since the equation x(t) = t³ can have only one solution while the equation y(t) = t⁴ can have two, we'll start with the x equation.

• When x = 8 we must have t = 2. Then y = (2)⁴ = 16.The point (8, 16) is on the graph and occurs when t = 2.
• When x = 8 we have t = 2. Then y = 16 not -16. Since the only value of t that makes x = 8 doesn't make y = -16, the point(8, -16) isn't on the graph.
• When x = -27 we have t = -3. Theny = (-3)⁴ = 81.The point (-27, 81) is on the graph when t = -3.

Answer

We'll use the y-equation first since we'll find only one t-value that way. When y = -6 we must have t = -2. Then x = (-2)² + 1 = 5. The point (5, -6) occurs on the graph when t = -2.

Answer

• Using the y-equation first, if y = 9 then t = 3. Then x = (3)² + 1 = 10. Since x ≠ 11 at the only t-value that makes y = 9, the point (11, 9) is not on the graph.
• It's also possible to determine the points at which parametric equations intersect. This is how we'd know if Mario ran into a red koopa flying loftily over the canyon during his jump.