Precalculus: Parametric Equations Worksheet

Question 1

For the parametric equations x = t + 1 and y = t^2, find the ordered pairs for t = -2, -1, 0, 1, and 2.

Accepted Answer

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.

Question 2

Eliminate the parameter for x = 3t - 2 and y = 6t + 1. Write the rectangular equation.

Accepted Answer

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.

Question 3

Eliminate the parameter for x = t^2 and y = t + 4. State any restriction on x.

Accepted Answer

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.

Question 4

The equations x = 4 cos t and y = 4 sin t are graphed for 0 <= t <= 2π. Describe the graph and its direction.

Accepted Answer

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.

Question 5

Eliminate the parameter for x = 2 + 3 cos t and y = -1 + 5 sin t. Identify the shape.

Accepted Answer

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).

Question 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.

Accepted Answer

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.

Question 7

Determine whether x = 2t, y = 4t^2 traces the same rectangular curve as x = s, y = s^2. Explain your answer.

Accepted 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.

Question 8

Write one possible parametric representation for the rectangular equation y = 3x^2 - 2.

Accepted Answer

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.

Question 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.

Accepted Answer

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.

Question 10

Determine whether the point (5, 12) lies on the parametric curve x = 2t - 1 and y = t^2 + 3.

Accepted Answer

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.

Question 11

Find the intersection points of the two parametric curves x = t, y = t + 2 and x = s^2, y = 4 - s.

Accepted Answer

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).

Question 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.

Accepted Answer

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.

Question 13

For x = t^2 + 1 and y = t^3, find the slope dy/dx at t = 2.

Accepted Answer

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.

Question 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.

Accepted Answer

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.

Question 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)?

Accepted Answer

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).

Precalculus: Parametric Equations

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