Parametric Circular Orbits Quiz: Parametric Equations Of Circular Orbits

1) For the parametric orbit x = 10 cos(2t), y = 10 sin(2t), what is the orbit's radius?

10

2

5

1/20

In a standard circular parametric form x = R cos(ωt), y = R sin(ωt), the radius is R. Here R = 10, so the radius is 10.

Explanation

In a standard circular parametric form x = R cos(ωt), y = R sin(ωt), the radius is R. Here R = 10, so the radius is 10.

2) For x = 7 cos(2t), y = 7 sin(2t), what is the period T of the motion?

π

2π

π/2

7π

The angular speed is ω = 2 rad/s. Period T = 2π/ω = 2π/2 = π seconds.

Explanation

The angular speed is ω = 2 rad/s. Period T = 2π/ω = 2π/2 = π seconds.

3) Let x = 5 cos(t + π/3), y = 5 sin(t + π/3). What is the satellite’s position at t = 0?

(5√3/2,5/2)

(5/2,(5√3)/2)

(0,5)

(−5,0)

At t = 0 the angle is π/3. cos(π/3) = 1/2 and sin(π/3) = √3/2, so (x,y) = (5·1/2, 5·√3/2) = (5/2, (5√3)/2).

Explanation

At t = 0 the angle is π/3. cos(π/3) = 1/2 and sin(π/3) = √3/2, so (x,y) = (5·1/2, 5·√3/2) = (5/2, (5√3)/2).

4) A satellite moves as x = 8 cos(0.25t), y = 8 sin(0.25t). What is its speed magnitude?

0.25

1

4

2

For circular motion with radius R and angular speed ω, speed v = R·|ω|. Here R = 8 and ω = 0.25, so v = 8·0.25 = 2 units/s.

Explanation

For circular motion with radius R and angular speed ω, speed v = R·|ω|. Here R = 8 and ω = 0.25, so v = 8·0.25 = 2 units/s.

5) For x = R cos(ωt + φ), y = R sin(ωt + φ) with ω > 0, the motion is counterclockwise starting at angle φ.

True

False

As t increases, the angle θ = ωt + φ increases when ω > 0, which traces the circle counterclockwise from initial angle φ.

Explanation

As t increases, the angle θ = ωt + φ increases when ω > 0, which traces the circle counterclockwise from initial angle φ.

6) For x = 3 cos t + 4, y = 3 sin t − 2, the center of the orbit is ____.

The translation (h,k) shifts the center from (0,0) to (h,k). Here x = 3 cos t + 4 and y = 3 sin t − 2 give h = 4 and k = −2, so center = (4, −2).

Explanation

The translation (h,k) shifts the center from (0,0) to (h,k). Here x = 3 cos t + 4 and y = 3 sin t − 2 give h = 4 and k = −2, so center = (4, −2).

Submit

7) Given x = 6 cos(0.5t), y = 6 sin(0.5t), what is the period?

4π

2π

π

12π

ω = 0.5 rad/s, so T = 2π/ω = 2π/0.5 = 4π.

Explanation

ω = 0.5 rad/s, so T = 2π/ω = 2π/0.5 = 4π.

8) Which changes increase the size (radius) of the circular orbit for x = R cos(ωt)+h, y = R sin(ωt)+k? Select all that apply.

Increase R

Decrease ω

Change phase φ

Translate center (h,k)

Multiply both cosine and sine by 2 (i.e., double R)

The orbit size is the radius R. Increasing R directly enlarges the orbit. Multiplying both cos and sin by 2 doubles R. Changing ω, φ, or (h,k) does not change the radius.

Explanation

The orbit size is the radius R. Increasing R directly enlarges the orbit. Multiplying both cos and sin by 2 doubles R. Changing ω, φ, or (h,k) does not change the radius.

Submit

9) For x = 4 cos t, y = 4 sin t, what is the position at t = π/2?

(−4,0)

(4,0)

(0,4)

(0,−4)

At t = π/2: cos(π/2) = 0 and sin(π/2) = 1, so (x,y) = (4·0, 4·1) = (0,4).

Explanation

At t = π/2: cos(π/2) = 0 and sin(π/2) = 1, so (x,y) = (4·0, 4·1) = (0,4).

10) For x = 7 cos t, y = 7 sin t (so ω = 1), the acceleration magnitude equals 7.

True

False

For uniform circular motion, acceleration magnitude a = R·ω². With R = 7 and ω = 1, a = 7·1² = 7.

Explanation

For uniform circular motion, acceleration magnitude a = R·ω². With R = 7 and ω = 1, a = 7·1² = 7.

11) Let x = 9 cos(3t), y = 9 sin(3t). What is the smallest positive time t when the satellite is at (9,0) again?

2π/3

π/3

π

2π

Being at (R,0) corresponds to angle θ = 2πk. With θ = 3t, the smallest positive solution is 3t = 2π ⇒ t = 2π/3.

Explanation

Being at (R,0) corresponds to angle θ = 2πk. With θ = 3t, the smallest positive solution is 3t = 2π ⇒ t = 2π/3.

12) For x = 5 cos(2t + π/2), y = 5 sin(2t + π/2), the initial position at t = 0 is ____.

At t = 0 the angle is π/2. cos(π/2) = 0 and sin(π/2) = 1, so (x,y) = (0,5).

Explanation

At t = 0 the angle is π/2. cos(π/2) = 0 and sin(π/2) = 1, so (x,y) = (0,5).

Submit

13) Which parameters affect the speed magnitude v of x = R cos(ωt + φ)+h, y = R sin(ωt + φ)+k? Select all that apply.

R

ω

φ

H and k

Neither R nor ω

Speed magnitude v = R·|ω|. It depends on R and the magnitude of ω. It does not depend on φ or on translations (h,k).

Explanation

Speed magnitude v = R·|ω|. It depends on R and the magnitude of ω. It does not depend on φ or on translations (h,k).

Submit

14) For x = 12 cos(πt), y = 12 sin(πt), what is the angular speed ω?

π

12

2π

1/π

The coefficient of t inside the trig functions is ω. Here the angle is πt, so ω = π rad/s.

Explanation

The coefficient of t inside the trig functions is ω. Here the angle is πt, so ω = π rad/s.

15) The curve x = 2 cos t, y = 3 sin t is a circle.

True

False

If the cosine and sine coefficients differ, the locus satisfies (x/2)² + (y/3)² = 1, which is an ellipse, not a circle.

Explanation

If the cosine and sine coefficients differ, the locus satisfies (x/2)² + (y/3)² = 1, which is an ellipse, not a circle.

16) Which equation represents a circular orbit with radius 10, center at (−3, 5), angular speed 0.2 rad/s, and initial angle −π/4 at t = 0?

X = 10 cos(0.2t − π/4) − 3; y = 10 sin(0.2t − π/4) + 5

X = 10 cos(0.2t + π/4) + 3; y = 10 sin(0.2t + π/4) − 5

X = 10 cos(0.2t) − 3; y = 10 sin(0.2t) + 5

X = 5 cos(0.2t − π/4) − 3; y = 5 sin(0.2t − π/4) + 5

General form is x = R cos(ωt + φ) + h, y = R sin(ωt + φ) + k. Here R = 10, ω = 0.2, φ = −π/4, h = −3, k = 5, which matches option A.

Explanation

General form is x = R cos(ωt + φ) + h, y = R sin(ωt + φ) + k. Here R = 10, ω = 0.2, φ = −π/4, h = −3, k = 5, which matches option A.

Submit

18) How far does the satellite travel along the orbit from t = 0 to t = 2 s for x = 6 cos(1.5t), y = 6 sin(1.5t)?

18

1/12

1/9

6

Arc length over time Δt for uniform circular motion is s = v·Δt with v = R·|ω|. Here v = 6·1.5 = 9, so s = 9·2 = 18 units.

Explanation

Arc length over time Δt for uniform circular motion is s = v·Δt with v = R·|ω|. Here v = 6·1.5 = 9, so s = 9·2 = 18 units.

19) Which statements are true for x = R cos(ωt + φ)+h, y = R sin(ωt + φ)+k? Select all that apply.

Speed magnitude is constant

Acceleration always points toward the center

(x − h)² + (y − k)² = R² for all t

Period equals R/(2π)

Changing φ shifts only the starting point around the circle

Speed magnitude v = R|ω| is constant. Acceleration is centripetal, pointing toward the center. Eliminating t gives (x − h)² + (y − k)² = R². The period is T = 2π/|ω|, not R/(2π). Changing φ alters the initial angle but not R, ω, or center.

Explanation

Speed magnitude v = R|ω| is constant. Acceleration is centripetal, pointing toward the center. Eliminating t gives (x − h)² + (y − k)² = R². The period is T = 2π/|ω|, not R/(2π). Changing φ alters the initial angle but not R, ω, or center.

Submit

20) If the phase φ increases by π/2, the starting point rotates 90° counterclockwise on the orbit.

True

False

Initial position at t = 0 is (R cos φ + h, R sin φ + k). Increasing φ by π/2 adds 90° to the angle, which rotates the starting point 90° counterclockwise.

Explanation

Initial position at t = 0 is (R cos φ + h, R sin φ + k). Increasing φ by π/2 adds 90° to the angle, which rotates the starting point 90° counterclockwise.

Parametric Circular Orbits Quiz: Parametric Equations of Circular Orbits