AP Precalculus Quiz: Parametrically Defined Circles And Lines

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Parametrically Defined Circles And LinesQuestion 1 of 20

A Ferris wheel has a radius of 40 feet, and its center is 50 feet above the ground. The wheel rotates counterclockwise, making one full revolution every 2 minutes.

A passenger boards the ride at its lowest point. Which of the following parametrizations models the passenger's position $(x(t), y(t))$ where $t$ is time in minutes?

x(t) = 40\cos(\pi t), y(t) = 50 + 40\sin(\pi t)

x(t) = 40\sin(\pi t), y(t) = 50 - 40\cos(\pi t)

x(t) = 40\cos(2\pi t), y(t) = 50 + 40\sin(2\pi t)

x(t) = 40\sin(2t), y(t) = 50 - 40\cos(2t)

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AP Precalculus Quiz

AP Precalculus Quiz: Parametrically Defined Circles And Lines

Practice Parametrically Defined Circles And Lines in AP Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Parametrically Defined Circles And Lines, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Precalculus.

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Question 1

A Ferris wheel has a radius of 40 feet, and its center is 50 feet above the ground. The wheel rotates counterclockwise, making one full revolution every 2 minutes.

A passenger boards the ride at its lowest point. Which of the following parametrizations models the passenger's position $(x(t), y(t))$ where $t$ is time in minutes?

$x(t) = 40\cos(\pi t), y(t) = 50 + 40\sin(\pi t)$
$x(t) = 40\sin(\pi t), y(t) = 50 - 40\cos(\pi t)$ (correct answer)
$x(t) = 40\cos(2\pi t), y(t) = 50 + 40\sin(2\pi t)$
$x(t) = 40\sin(2t), y(t) = 50 - 40\cos(2t)$

Explanation: The center is (0, 50) and radius is 40. The period is 2 minutes, so

P=2 = 2\pi/B

, which gives

B=\pi

. The lowest point is

(0, 50-40) = (0, 10)

. Standard counterclockwise motion from the rightmost point is

x=40\cos(\pi t), y=50+40\sin(\pi t)

. The lowest point is at an angle of

-\pi/2

from the start. A phase shift gives

x(t) = 40\cos(\pi t - \pi/2) = 40\sin(\pi t)

and

y(t) = 50 + 40\sin(\pi t - \pi/2) = 50 - 40\cos(\pi t)

. At

t=0

, this gives

(0, 10)

, the correct starting point.

Question 2

A particle traces a circular path of radius 4 centered at the origin. It completes one full counterclockwise revolution in exactly 10 seconds. Which parametrization models the particle's motion, assuming it starts at (4, 0) at $t=0$ ?

$x(t) = 4\cos(10t), y(t) = 4\sin(10t)$
$x(t) = 4\cos(2\pi t), y(t) = 4\sin(2\pi t)$
$x(t) = 4\cos(\frac{\pi}{5}t), y(t) = 4\sin(\frac{\pi}{5}t)$ (correct answer)
$x(t) = 4\cos(\frac{5}{\pi}t), y(t) = 4\sin(\frac{5}{\pi}t)$

Explanation: The period of the motion is 10 seconds. For a parametrization

x(t) = r\cos(Bt), y(t) = r\sin(Bt)

, the period is

P = 2\pi/|B|

. We are given

P=10

, so

10 = 2\pi/B

, which gives

B = 2\pi/10 = \pi/5

. The radius is 4. Thus, the correct parametrization is

x(t) = 4\cos(\frac{\pi}{5}t), y(t) = 4\sin(\frac{\pi}{5}t)

Question 3

A satellite follows the circle $x=-1+3\cos t,\ y=4+3\sin t$ , while a scan line follows $x=2+k,\ y=4+2k$ . Find the point(s) where the line intersects the circle defined by the given parametric equations.

Intersections at $(2,4)$ and $(0,0)$ (correct answer)
Intersections at $(2,4)$ only
Intersections at $(5,10)$ and $(2,4)$
Intersections at $(-4,4)$ and $(2,4)$

Explanation: This question tests AP Precalculus skills: understanding parametrically defined circles and lines (focus on algebraic and geometric interpretation). Parametric equations represent geometric figures by defining coordinates as functions of a parameter, often time (t) or distance (k). For this question, the circle has center (-1,4) and radius 3, while the line passes through (2,4) with slope 2. Choice A is correct because substituting x=2+k and y=4+2k into (x+1)²+(y-4)²=9 gives k=0 and k=-2, yielding intersection points (2,4) and (0,0). Choice C is incorrect as it finds one correct point but uses the wrong parameter value for the second intersection, resulting in a point outside the circle. To help students: Develop systematic approaches to solving circle-line intersection problems. Practice interpreting parameter values geometrically to understand what they represent on the line.

Question 4

Which of the following parametrizations represents a particle moving counterclockwise on a unit circle, starting at the point (0, -1) at time $t=0$ ?

$x(t) = \cos(t), y(t) = \sin(t)$
$x(t) = -\cos(t), y(t) = -\sin(t)$
$x(t) = \sin(t), y(t) = -\cos(t)$ (correct answer)
$x(t) = -\sin(t), y(t) = \cos(t)$

Explanation: We need the position at

t=0

to be (0, -1). Let's check the options. For C,

x(0) = \sin(0) = 0

and

y(0) = -\cos(0) = -1

. This gives the correct starting point. The motion is counterclockwise. This can be seen as a phase shift of the standard parametrization:

x(t) = \cos(t - \pi/2) = \sin(t)

and

y(t) = \sin(t - \pi/2) = -\cos(t)

represents a shift in starting position.

Question 5

The motion of a particle is described by the parametric equations $x(t) = -4 + 6\sin(t)$ and $y(t) = 5 + 6\cos(t)$ . What are the center and radius of the circular path?

Center at (4, -5) and radius 6
Center at (-4, 5) and radius 6 (correct answer)
Center at (-4, 5) and radius 36
Center at (6, 5) and radius -4

Explanation: The general form for a circle centered at

(h, k)

with radius

r

can be

x(t) = h + r\cos(t)

and

y(t) = k + r\sin(t)

, or with sine and cosine swapped. Here,

x(t) = -4 + 6\sin(t)

and

y(t) = 5 + 6\cos(t)

. The constant terms give the center, so

(h, k) = (-4, 5)

. The coefficient of the trigonometric functions gives the radius, so

r = 6

Question 6

A particle moves along a path defined by $x(t) = 3 - 2t$ and $y(t) = -1 + 5t$ for $0 \le t \le 1$ . What are the starting and ending points of the particle's motion?

Starts at (1, 4) and ends at (3, -1)
Starts at (3, -1) and ends at (1, 4) (correct answer)
Starts at (3, -1) and ends at (5, -6)
Starts at (2, 5) and ends at (-2, 5)

Explanation: The starting point corresponds to

t=0

. Plugging in

t=0

gives

x(0) = 3 - 2(0) = 3

and

y(0) = -1 + 5(0) = -1

. The starting point is (3, -1). The ending point corresponds to

t=1

. Plugging in

t=1

gives

x(1) = 3 - 2(1) = 1

and

y(1) = -1 + 5(1) = 4

. The ending point is (1, 4).

Question 7

Which of the following represents the rectangular equation for the curve defined parametrically by $x(t) = 2 + 5\cos(t)$ and $y(t) = -1 + 5\sin(t)$ ?

$\frac{(x-2)^2}{25} - \frac{(y+1)^2}{25} = 1$
$(x-2)^2 + (y+1)^2 = 5$
$(x+2)^2 + (y-1)^2 = 25$
$(x-2)^2 + (y+1)^2 = 25$ (correct answer)

Explanation: First, isolate the trigonometric functions:

x-2 = 5\cos(t)

and

y+1 = 5\sin(t)

. Then, square both equations:

(x-2)^2 = 25\cos^2(t)

and

(y+1)^2 = 25\sin^2(t)

. Add the two equations together:

(x-2)^2 + (y+1)^2 = 25\cos^2(t) + 25\sin^2(t)

. Using the identity

\cos^2(t) + \sin^2(t) = 1

, we get

(x-2)^2 + (y+1)^2 = 25(1) = 25

Question 8

Which parametrization traces the line segment from $A(-2, 3)$ to $B(0, 7)$ as $t$ increases from 0 to 2?

$x(t) = -2 + 2t, y(t) = 3 + 4t$
$x(t) = -2 + t, y(t) = 3 + 2t$ (correct answer)
$x(t) = -2 - t, y(t) = 3 - 2t$
$x(t) = -2 + 0.5t, y(t) = 3 + t$

Explanation: The total change in coordinates from A to B is

(0 - (-2), 7 - 3) = (2, 4)

. This change must occur over a time interval of length 2. The rate of change is

(2/2, 4/2) = (1, 2)

. Using the starting point A(-2, 3) and this rate, the parametrization is

x(t) = -2 + 1t

and

y(t) = 3 + 2t

. Checking the endpoints: at

t=0

, we have (-2, 3). At

t=2

, we have

(-2+2, 3+4) = (0, 7)

Question 9

A circle has a radius of 3 and is centered at the point (2, -1). Which of the following equations parametrize this circle with a counterclockwise orientation?

$x(t) = 2 + 3\cos(t), y(t) = -1 + 3\sin(t)$ (correct answer)
$x(t) = -2 + 3\cos(t), y(t) = 1 + 3\sin(t)$
$x(t) = 2 - 3\cos(t), y(t) = -1 - 3\sin(t)$
$x(t) = 3 + 2\cos(t), y(t) = -1 + 2\sin(t)$

Explanation: For a circle centered at

(h, k)

with radius

r

, the counterclockwise parametrization is

x(t) = h + r\cos(t)

and

y(t) = k + r\sin(t)

. Given the center is (2, -1) and radius is 3, the equations are

x(t) = 2 + 3\cos(t)

and

y(t) = -1 + 3\sin(t)

Question 10

The path of a particle is given by $x(t) = 1 + 2t, y(t) = 4 - t$ for $0 \le t \le 1$ . Which of the following parametrizations traces the same line segment but in the opposite direction for $0 \le t \le 1$ ?

$x(t) = 1 + 2t, y(t) = 4 - t$
$x(t) = 3 - 2t, y(t) = 3 + t$ (correct answer)
$x(t) = 1 - 2t, y(t) = 4 + t$
$x(t) = 3 + 2t, y(t) = 3 - t$

Explanation: The original path starts at

t=0

at (1, 4) and ends at

t=1

at (3, 3). The opposite direction means starting at (3, 3) and ending at (1, 4). Using the standard parametrization for a line segment from

(x_1, y_1)

(x_2, y_2)

, we have

x(t) = 3 + (1 - 3)t = 3 - 2t

and

y(t) = 3 + (4 - 3)t = 3 + t

for

0 \le t \le 1

Question 11

The motion of a particle is described by $x(t) = \cos(2t)$ and $y(t) = \sin(2t)$ for $0 \le t \le \pi$ . Another particle's motion is described by $x(t) = \cos(t)$ and $y(t) = \sin(t)$ for $0 \le t \le \pi$ . Which statement correctly compares their paths?

Both particles trace the full unit circle.
The first particle traces the full unit circle, while the second particle traces only the upper half. (correct answer)
The second particle traces the full unit circle, while the first particle traces only the upper half.
The first particle moves twice as slowly as the second particle.

Explanation: The first particle's parametrization has a period of

P = 2\pi/2 = \pi

. Over the interval

0 \le t \le \pi

, it completes one full revolution. The second particle's parametrization has a period of

P = 2\pi/1 = 2\pi

. Over the interval

0 \le t \le \pi

, it completes half a revolution, which corresponds to the upper semicircle from (1,0) to (-1,0).

Question 12

An object moves in a straight line from point A(1, 4) to point B(7, 0) in 3 seconds at a constant speed.

Which of the following parametric equations models the object's path for the time interval $0 \le t \le 3$ ?

$x(t) = 1 + 6t, y(t) = 4 - 4t$
$x(t) = 1 + 2t, y(t) = 4 - \frac{4}{3}t$ (correct answer)
$x(t) = 7 - 2t, y(t) = 3t$
$x(t) = 1 + 3t, y(t) = 4 + 3t$

Explanation: The starting point is A(1, 4). The total displacement is B - A =

(7-1, 0-4) = (6, -4)

. This displacement occurs over 3 seconds. The constant velocity vector (rate of change) is

(6/3, -4/3) = (2, -4/3)

. Therefore, the position at time

t

is given by the initial position plus

t

times the velocity:

x(t) = 1 + 2t

and

y(t) = 4 - \frac{4}{3}t

Question 13

Which of the following parametrically defines the line segment starting at P(1, 2) and ending at Q(5, 8) as $t$ increases from 0 to 1?

$x(t) = 1 + 4t, y(t) = 2 + 6t$ (correct answer)
$x(t) = 5 - 4t, y(t) = 8 - 6t$
$x(t) = 1 + 5t, y(t) = 2 + 8t$
$x(t) = 1 - 4t, y(t) = 2 - 6t$

Explanation: A line segment from

(x_1, y_1)

(x_2, y_2)

for

0 \le t \le 1

can be parametrized as

x(t) = x_1 + (x_2 - x_1)t

and

y(t) = y_1 + (y_2 - y_1)t

. Here,

(x_1, y_1) = (1, 2)

and

(x_2, y_2) = (5, 8)

. So,

x(t) = 1 + (5 - 1)t = 1 + 4t

and

y(t) = 2 + (8 - 2)t = 2 + 6t

Question 14

Which of the following parametrizations does NOT describe the unit circle defined by $x^2 + y^2 = 1$ ?

$x(t) = \cos(t), y(t) = \sin(t)$
$x(t) = \sin(t), y(t) = \cos(t)$
$x(t) = \cos(2t), y(t) = -\sin(2t)$
$x(t) = \cos(t), y(t) = 2\sin(t)$ (correct answer)

Explanation: A parametrization describes the unit circle if

x(t)^2 + y(t)^2 = 1

. For option D,

x(t)^2 + y(t)^2 = (\cos(t))^2 + (2\sin(t))^2 = \cos^2(t) + 4\sin^2(t)

, which does not equal 1 for all

t

. This parametrization actually describes an ellipse.

Question 15

Which of the following is a parametrization for a circle centered at the origin with a radius of 5, traced counterclockwise starting from the point (5, 0)?

$x(t) = 5\cos(t), y(t) = 5\sin(t)$ (correct answer)
$x(t) = 25\cos(t), y(t) = 25\sin(t)$
$x(t) = 5\sin(t), y(t) = 5\cos(t)$
$x(t) = 5\cos(t), y(t) = -5\sin(t)$

Explanation: The standard counterclockwise parametrization for a circle with radius

r

centered at the origin is

x(t) = r\cos(t), y(t) = r\sin(t)

. With a radius of 5, this becomes

x(t) = 5\cos(t), y(t) = 5\sin(t)

. At

t=0

, the position is

(5\cos(0), 5\sin(0)) = (5,0)

, which is the correct starting point.

Question 16

A line segment from (0,0) to (4,6) is traced by the parametrization $x(t)=4t, y(t)=6t$ for $0 \le t \le 1$ . Which parametrization traces the same segment but takes 2 seconds to complete?

$x(t) = 8t, y(t) = 12t$ , for $0 \le t \le 2$
$x(t) = 4t, y(t) = 6t$ , for $0 \le t \le 2$
$x(t) = 2t, y(t) = 3t$ , for $0 \le t \le 2$ (correct answer)
$x(t) = 2t, y(t) = 3t$ , for $0 \le t \le 1$

Explanation: The path starts at (0,0) and needs to end at (4,6) when

t=2

. The parametrization will be of the form

x(t)=at, y(t)=bt

. At

t=2

, we need

x(2)=4

and

y(2)=6

. So,

a(2)=4 \implies a=2

, and

b(2)=6 \implies b=3

. The correct parametrization is

x(t)=2t, y(t)=3t

for the interval

0 \le t \le 2

Question 17

Which of the following parametrizations describes a unit circle centered at the origin being traced clockwise, starting from the point (1, 0)?

$x(t) = \cos(t), y(t) = \sin(t)$
$x(t) = \cos(t), y(t) = -\sin(t)$ (correct answer)
$x(t) = -\cos(t), y(t) = \sin(t)$
$x(t) = \sin(t), y(t) = \cos(t)$

Explanation: To reverse the direction of a counterclockwise parametrization to clockwise, one common method is to replace

t

with

-t

. Starting with

x(t) = \cos(t)

and

y(t) = \sin(t)

, this gives

x(-t) = \cos(-t) = \cos(t)

and

y(-t) = \sin(-t) = -\sin(t)

. This traces the circle clockwise and starts at

(1,0)

when

t=0

. Another method is to negate the sine component.

Question 18

A sensor sweeps the circle $x=-2+6\cos t,\ y=1+6\sin t$ , and a signal line follows $x=4+r,\ y=1+2r$ . Find the point(s) where the line intersects the circle defined by the given parametric equations.

Intersections at $(4,1)$ and $(0,-7)$ (correct answer)
Intersections at $(4,1)$ only
Intersections at $(10,13)$ and $(4,1)$
No intersection points occur

Explanation: This question tests AP Precalculus skills: understanding parametrically defined circles and lines (focus on algebraic and geometric interpretation). Parametric equations represent geometric figures by defining coordinates as functions of a parameter, often time (t) or distance (r). For this question, the circle has center (-2,1) and radius 6, while the line passes through (4,1) with slope 2. Choice A is correct because substituting x=4+r and y=1+2r into (x+2)²+(y-1)²=36 gives r=0 and r=-4, yielding intersection points (4,1) and (0,-7). Choice C is incorrect as it finds one correct point but miscalculates the second intersection by using the wrong parameter value. To help students: Practice solving quadratic equations that arise from circle-line intersections. Verify solutions by checking that both points satisfy the original circle equation.

Question 19

A skier traces the circle $x=1+4\cos t,\ y=2+4\sin t$ , and a tow cable follows $x=5+u,\ y=6-u$ . Find the point(s) where the line intersects the circle defined by the given parametric equations.

Intersections at $(5,6)$ and $(9,2)$ (correct answer)
Intersections at $(5,6)$ only
Intersections at $(1,2)$ and $(9,2)$
Intersections at $(3,4)$ and $(7,0)$

Explanation: This question tests AP Precalculus skills: understanding parametrically defined circles and lines (focus on algebraic and geometric interpretation). Parametric equations represent geometric figures by defining coordinates as functions of a parameter, often time (t) or distance (u). For this question, the circle has center (1,2) and radius 4, while the line passes through (5,6) with slope -1. Choice A is correct because substituting x=5+u and y=6-u into the circle equation (x-1)²+(y-2)²=16 gives u=0 and u=4, yielding intersection points (5,6) and (9,2). Choice C is incorrect due to an algebraic error in solving the quadratic equation, resulting in wrong parameter values. To help students: Emphasize careful algebraic manipulation when substituting parametric equations. Practice converting between parametric and Cartesian forms, and verify solutions by substituting back into both original equations.

Question 20

A beacon rotates on the circle $x=0+5\cos t,\ y=2+5\sin t$ , and a survey line follows $x=5+p,\ y=2+p$ . Find the point(s) where the line intersects the circle defined by the given parametric equations.

Intersections at $(5,2)$ and $(0,-3)$ (correct answer)
Intersections at $(5,7)$ and $(0,-3)$
Intersections at $(5,2)$ only
Intersections at $(3,4)$ and $(2,5)$

Explanation: This question tests AP Precalculus skills: understanding parametrically defined circles and lines (focus on algebraic and geometric interpretation). Parametric equations represent geometric figures by defining coordinates as functions of a parameter, often time (t) or distance (p). For this question, the circle has center (0,2) and radius 5, while the line passes through (5,2) with slope 1. Choice A is correct because substituting x=5+p and y=2+p into x²+(y-2)²=25 gives p=0 and p=-5, yielding intersection points (5,2) and (0,-3). Choice B is incorrect as it miscalculates the y-coordinate of the first intersection point, likely due to a sign error when evaluating the parametric equation. To help students: Carefully track signs when substituting parameter values back into parametric equations. Verify that calculated points satisfy both the circle and line equations.