Parametrically Defined Circles and Lines - AP Precalculus

1. 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?
   Answer: $x(t) = 1 + 4t, y(t) = 2 + 6t$ 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 $y(t) = 2 + (8 - 2)t = 2 + 6t$.
2. Which of the following parametrizations does NOT describe the unit circle defined by $x^2 + y^2 = 1$?
   Answer: $x(t) = \cos(t), y(t) = 2\sin(t)$ 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.
3. 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)?
   Answer: $x(t) = 5\cos(t), y(t) = 5\sin(t)$ 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.
4. Which of the following is a parametrization for the right half of the circle $x^2 + y^2 = 9$, traced in a counterclockwise direction?
   Answer: $x(t) = 3\cos(t), y(t) = 3\sin(t)$ for $-\frac{\pi}{2} \le t \le \frac{\pi}{2}$ The circle has radius 3, so a standard parametrization is $x(t) = 3\cos(t), y(t) = 3\sin(t)$. The right half of the circle corresponds to values where $x \ge 0$. For the standard parametrization, $x(t) = 3\cos(t) \ge 0$ when $-\frac{\pi}{2} \le t \le \frac{\pi}{2}$. This interval traces the curve from (0, -3) up to (0, 3) in a counterclockwise direction.
5. Which of the following represents the rectangular equation for the curve defined parametrically by $x(t) = 2 + 5\cos(t)$ and $y(t) = -1 + 5\sin(t)$?
   Answer: $(x-2)^2 + (y+1)^2 = 25$ 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 $(x-2)^2 + (y+1)^2 = 25(1) = 25$.
6. Which of the following parametrizations represents a particle moving counterclockwise on a unit circle, starting at the point (0, -1) at time $t=0$?
   Answer: $x(t) = \sin(t), y(t) = -\cos(t)$ 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.
7. 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?
   Answer: Starts at (3, -1) and ends at (1, 4) 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).
8. 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.
   Answer: Intersections at $(4,1)$ and $(0,-7)$ 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.
9. 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?
   Answer: The first particle traces the full unit circle, while the second particle traces only the upper half. 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).
10. 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?
    Answer: $x(t) = 2t, y(t) = 3t$, for $0 \le t \le 2$ 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 $0 \le t \le 2$.
11. 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?
    Answer: $x(t) = 40\sin(\pi t), y(t) = 50 - 40\cos(\pi t)$ 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.
12. What is the rectangular equation for the line containing the segment parametrized by $x(t) = 3t - 1$ and $y(t) = 6t + 2$?
    Answer: $y = 2x + 4$ To find the rectangular equation, we eliminate the parameter $t$. From the equation for $x$, we solve for $t$: $x = 3t - 1 \implies x + 1 = 3t \implies t = \frac{x+1}{3}$. Now, substitute this expression for $t$ into the equation for $y$: $y = 6(\frac{x+1}{3}) + 2 = 2(x+1) + 2 = 2x + 2 + 2 = 2x + 4$.
13. 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.
    Answer: Intersections at $(2,4)$ and $(0,0)$ 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.
14. Which parametrization traces the line segment from $A(-2, 3)$ to $B(0, 7)$ as $t$ increases from 0 to 2?
    Answer: $x(t) = -2 + t, y(t) = 3 + 2t$ 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 $(-2+2, 3+4) = (0, 7)$.
15. 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.
    Answer: Intersections at $(5,2)$ and $(0,-3)$ 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.
16. 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 $0 \le t \le 1$?
    Answer: $x(t) = 3 - 2t, y(t) = 3 + t$ 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 $0 \le t \le 1$.
17. Which of the following parametrizations describes the line segment from $(-3, 5)$ to $(1, -1)$, traced over the time interval $0 \le t \le 2$?
    Answer: $x(t) = -3 + 2t, y(t) = 5 - 3t$ The path starts at $(-3, 5)$. The total change is $(1 - (-3), -1 - 5) = (4, -6)$. This change must occur over the interval $0 \le t \le 2$. The rate of change is $(4/2, -6/2) = (2, -3)$. So, the equations are $x(t) = -3 + 2t$ and $y(t) = 5 - 3t$. At $t=2$, the position is $(-3+4, 5-6) = (1, -1)$, which is the correct endpoint.
18. A circle is defined by $x(t) = 1 + 2\sin(t)$ and $y(t) = 3 + 2\cos(t)$. What is the starting point at $t=0$ and what is the direction of motion as $t$ increases?
    Answer: Starts at (1, 5), clockwise At $t=0$, the position is $x(0) = 1 + 2\sin(0) = 1$ and $y(0) = 3 + 2\cos(0) = 5$. The starting point is (1, 5). To find the direction, check a slightly later time, like $t=\pi/2$. The position is $x(\pi/2) = 1 + 2\sin(\pi/2) = 3$ and $y(\pi/2) = 3 + 2\cos(\pi/2) = 3$. The particle moves from (1, 5) to (3, 3), which is a clockwise motion on the circle centered at (1, 3).
19. 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?
    Answer: $x(t) = 2 + 3\cos(t), y(t) = -1 + 3\sin(t)$ 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 $y(t) = -1 + 3\sin(t)$.
20. 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 $t=0$?
    Answer: $x(t) = 4\cos(\frac{\pi}{5}t), y(t) = 4\sin(\frac{\pi}{5}t)$ 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 $x(t) = 4\cos(\frac{\pi}{5}t), y(t) = 4\sin(\frac{\pi}{5}t)$.

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