Precalculus: Parametric Equations
Graphing, interpreting, and converting parametric relationships
Precalculus: Parametric Equations
Graphing, interpreting, and converting parametric relationships
Math - Grade 9-12
- 1
For the parametric equations x = t + 1 and y = t^2, find the ordered pairs for t = -2, -1, 0, 1, and 2.
Substitute each t-value into both equations.
The ordered pairs are (-1, 4), (0, 1), (1, 0), (2, 1), and (3, 4). These points form a parabola opening upward as t increases from -2 to 2. - 2
Eliminate the parameter for x = 3t - 2 and y = 6t + 1. Write the rectangular equation.
Solving x = 3t - 2 gives t = (x + 2)/3. Substituting into y = 6t + 1 gives y = 2(x + 2) + 1, so the rectangular equation is y = 2x + 5. - 3
Eliminate the parameter for x = t^2 and y = t + 4. State any restriction on x.
Solve the linear equation for t first.
Since y = t + 4, t = y - 4. Substituting into x = t^2 gives x = (y - 4)^2. The restriction is x >= 0 because t^2 cannot be negative. - 4
The equations x = 4 cos t and y = 4 sin t are graphed for 0 <= t <= 2π. Describe the graph and its direction.
Use the identity cos^2 t + sin^2 t = 1.
The graph is a circle centered at the origin with radius 4, so its rectangular equation is x^2 + y^2 = 16. It starts at (4, 0) when t = 0 and moves counterclockwise. - 5
Eliminate the parameter for x = 2 + 3 cos t and y = -1 + 5 sin t. Identify the shape.
The equations give cos t = (x - 2)/3 and sin t = (y + 1)/5. Using cos^2 t + sin^2 t = 1 gives (x - 2)^2/9 + (y + 1)^2/25 = 1. The shape is an ellipse centered at (2, -1). - 6
For x = 1 + 2t and y = 5 - t with 0 <= t <= 3, find the endpoints of the graph and describe the direction of motion.
Evaluate the equations at the smallest and largest t-values.
At t = 0, the point is (1, 5). At t = 3, the point is (7, 2). The graph is a line segment traced from (1, 5) to (7, 2) as t increases. - 7
Determine whether x = 2t, y = 4t^2 traces the same rectangular curve as x = s, y = s^2. Explain your answer.
Yes, both trace the rectangular curve y = x^2. For the first pair, x = 2t, so t = x/2 and y = 4(x/2)^2 = x^2. The second pair already gives y = x^2 when x = s. - 8
Write one possible parametric representation for the rectangular equation y = 3x^2 - 2.
A common choice is to let x equal the parameter.
One possible representation is x = t and y = 3t^2 - 2. This works because substituting x = t into y = 3x^2 - 2 gives y = 3t^2 - 2. - 9
A projectile is modeled by x = 40t and y = 5 + 30t - 16t^2, where t is time in seconds and x and y are measured in feet. Find the projectile's position at t = 1.5 seconds.
Substitute 1.5 for t in both equations.
At t = 1.5, x = 40(1.5) = 60 and y = 5 + 30(1.5) - 16(1.5)^2 = 14. The projectile is at (60, 14), meaning it is 60 feet horizontally from the start and 14 feet above the ground. - 10
Determine whether the point (5, 12) lies on the parametric curve x = 2t - 1 and y = t^2 + 3.
Using x = 2t - 1, the equation 5 = 2t - 1 gives t = 3. Substituting t = 3 into y = t^2 + 3 gives y = 12, so the point (5, 12) does lie on the curve. - 11
Find the intersection points of the two parametric curves x = t, y = t + 2 and x = s^2, y = 4 - s.
Set the x-values equal and the y-values equal, then use substitution.
At an intersection, t = s^2 and t + 2 = 4 - s. Substituting gives s^2 + 2 = 4 - s, so s^2 + s - 2 = 0. This factors as (s + 2)(s - 1) = 0, so s = -2 or s = 1. The intersection points are (4, 6) and (1, 3). - 12
A particle moves according to x = 3t^2 and y = 4t for 0 <= t <= 3. Find its starting point, ending point, and displacement distance from start to end.
At t = 0, the particle starts at (0, 0). At t = 3, it ends at (27, 12). The displacement distance is sqrt((27 - 0)^2 + (12 - 0)^2) = sqrt(873) = 3sqrt(97), which is about 29.55 units. - 13
For x = t^2 + 1 and y = t^3, find the slope dy/dx at t = 2.
Use dy/dx = (dy/dt)/(dx/dt), as long as dx/dt is not zero.
The derivatives are dx/dt = 2t and dy/dt = 3t^2. Therefore, dy/dx = (dy/dt)/(dx/dt) = 3t^2/(2t). At t = 2, the slope is 12/4 = 3. - 14
For the parametric curve x = t - sin t and y = 1 - cos t, find the points when t = 0, t = π, and t = 2π.
At t = 0, the point is (0, 0). At t = π, the point is (π, 2). At t = 2π, the point is (2π, 0). These points are key points on one arch of the curve. - 15
The parametric equations x = 5 cos t and y = 5 sin t trace a circle. What interval of t traces only the upper semicircle from (5, 0) to (-5, 0)?
The upper semicircle has y >= 0, which matches sin t >= 0 on this interval.
The interval 0 <= t <= π traces the upper semicircle. At t = 0 the point is (5, 0), at t = π/2 the point is (0, 5), and at t = π the point is (-5, 0).