AP Precalculus Quiz: Parametrically Defined Circles And Lines

What this quiz covers

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.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $x(t) = \cos(t), y(t) = \sin(t)$
2. $x(t) = \sin(t), y(t) = \cos(t)$
3. $x(t) = \cos(2t), y(t) = -\sin(2t)$
4. $x(t) = \cos(t), y(t) = 2\sin(t)$ (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 2

1. $x(t) = 5\cos(t), y(t) = 5\sin(t)$ (correct answer)
2. $x(t) = 25\cos(t), y(t) = 25\sin(t)$
3. $x(t) = 5\sin(t), y(t) = 5\cos(t)$
4. $x(t) = 5\cos(t), y(t) = -5\sin(t)$

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 3

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

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

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 5

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

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 6

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

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 7

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

Explanation: A line segment from $(x_1, y_1)$ to $(x_2, y_2)$ 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 8

1. Intersections at $(4,1)$ and $(0,-7)$ (correct answer)
2. Intersections at $(4,1)$ only
3. Intersections at $(10,13)$ and $(4,1)$
4. 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 9

1. Intersections at $(5,6)$ and $(9,2)$ (correct answer)
2. Intersections at $(5,6)$ only
3. Intersections at $(1,2)$ and $(9,2)$
4. 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 10

1. Intersections at $(2,4)$ and $(0,0)$ (correct answer)
2. Intersections at $(2,4)$ only
3. Intersections at $(5,10)$ and $(2,4)$
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 11

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

Question 12

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

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 13

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

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 14

1. Center at (4, -5) and radius 6
2. Center at (-4, 5) and radius 6 (correct answer)
3. Center at (-4, 5) and radius 36
4. 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 15

1. Starts at (1, 4) and ends at (3, -1)
2. Starts at (3, -1) and ends at (1, 4) (correct answer)
3. Starts at (3, -1) and ends at (5, -6)
4. 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 16

1. $\frac{(x-2)^2}{25} - \frac{(y+1)^2}{25} = 1$
2. $(x-2)^2 + (y+1)^2 = 5$
3. $(x+2)^2 + (y-1)^2 = 25$
4. $(x-2)^2 + (y+1)^2 = 25$ (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 17

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

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 18

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

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 $(x_1, y_1)$ to $(x_2, y_2)$, 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 19

1. Both particles trace the full unit circle.
2. The first particle traces the full unit circle, while the second particle traces only the upper half. (correct answer)
3. The second particle traces the full unit circle, while the first particle traces only the upper half.
4. 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 20

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

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