AP Precalculus Quiz: Parametric Functions

What this quiz covers

This quiz focuses on Parametric Functions, 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. $C_1$ and $C_2$ trace the same curve, but $C_2$ is traversed more quickly than $C_1$ for $|s|>1$. (correct answer)
2. $C_1$ and $C_2$ trace different curves because the equations are different functions of the parameter.
3. $C_1$ and $C_2$ trace the same curve, but $C_2$ only traces the portion of the curve where $x \ge 0$.
4. $C_1$ and $C_2$ trace the same curve, but $C_2$ is traversed in the opposite direction of $C_1$.

Explanation: For both representations, eliminating the parameter results in the rectangular equation $y=x^2$. Thus, they trace the same parabola. For $C_1$, as $t$ increases, the point moves along the parabola. For $C_2$, as $s$ increases, the point also moves along the parabola in the same direction. However, the speed of traversal is different. For example, in $C_1$, it takes 2 units of time for $t$ to go from 0 to 2, covering x-values from 0 to 2. In $C_2$, it takes 2 units of time for $s$ to go from 0 to 2, covering x-values from $0^3=0$ to $2^3=8$. Since $C_2$ covers a greater distance along the curve in the same parameter interval, it is traversed more quickly.

Question 2

1. $t=1$ (correct answer)
2. $t=\sqrt{3}$
3. $t=2$
4. $t=3$

Explanation: The vertical motion is described by the function $y(t) = t^3 - 3t$. The direction of vertical motion changes at a point where the function's rate of change is zero, corresponding to a local maximum or minimum. For the polynomial $y(t)$, these extrema occur at the critical points. The derivative is $y'(t) = 3t^2 - 3$. Setting the derivative to zero gives $3t^2 - 3 = 0$, which leads to $t^2 = 1$, so $t = 1$ or $t = -1$. The question asks for the positive value of $t$, which is $t=1$.

Question 3

1. Both particles trace the same unit circle once, but the second particle moves twice as fast. (correct answer)
2. The second particle traces the unit circle twice, while the first particle traces it once.
3. The first particle moves counter-clockwise, while the second particle moves clockwise.
4. The first particle traces the entire unit circle, while the second only traces a semicircle.

Explanation: Both parameterizations describe a path on the unit circle because $x^2+y^2=\cos^2(\theta)+\sin^2(\theta)=1$. For $C_1$, as $t$ goes from 0 to $2\pi$, the particle makes one full counter-clockwise revolution. For $C_2$, as the parameter $s$ goes from 0 to $\pi$, the angle $2s$ goes from 0 to $2\pi$. Thus, the second particle also makes one full counter-clockwise revolution. However, the first particle takes $2\pi$ units of time to complete the circle, while the second takes only $\pi$ units of time. This means the second particle traverses the same path at twice the speed.

Question 4

1. $\begin{bmatrix}-8\\3\end{bmatrix}$ (correct answer)
2. $\begin{bmatrix}-2\\12\end{bmatrix}$
3. $\begin{bmatrix}-8\\12\end{bmatrix}$
4. $\begin{bmatrix}-8\\-3\end{bmatrix}$

Explanation: This question tests AP Precalculus skills in parametric functions, vectors, and matrices, focusing on diagonal matrix transformations. Parametric functions use parameters to express coordinates, vectors represent direction and magnitude, and matrices perform transformations. In this scenario, a diagonal scaling matrix M=[2,0;0,1/2] transforms vector v⃗=[-4,6]ᵀ by scaling x by 2 and y by 1/2. Choice A is correct because Mv⃗=[2×(-4)+0×6, 0×(-4)+(1/2)×6]ᵀ=[-8+0, 0+3]ᵀ=[-8,3]ᵀ, properly applying the scaling transformation. Choice C is incorrect because it shows [-8,12], which would result from scaling y by 2 instead of 1/2, a common error when misreading diagonal entries. To help students: Emphasize that diagonal matrices scale each component independently by the corresponding diagonal entry. Watch for: Misreading fractional diagonal entries and errors in handling negative components during multiplication.

Question 5

1. At $(0, -3)$ and $(0, 21)$ (correct answer)
2. At $(0, 0)$ and $(0, 32)$
3. At $(0, -3)$ only
4. At $(0, 21)$ only

Explanation: The curve intersects the y-axis when the x-coordinate is 0. We set $x(t) = 0$ to find the values of $t$ at the intersections: $t^2 - 9 = 0 \implies t^2 = 9 \implies t = 3$ or $t = -3$. To find the points of intersection, we substitute these $t$ values into the equation for $y(t)$. For $t=3$, $y(3) = (3)^2 - 4(3) = 9 - 12 = -3$. For $t=-3$, $y(-3) = (-3)^2 - 4(-3) = 9 + 12 = 21$. Thus, the points of intersection with the y-axis are $(0, -3)$ and $(0, 21)$.

Question 6

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

Explanation: A parameterization for a line segment from $(x_1, y_1)$ to $(x_2, y_2)$ for $0 \le t \le 1$ is given by $x(t) = x_1 + (x_2 - x_1)t$ and $y(t) = y_1 + (y_2 - y_1)t$. With P(2, -1) as the starting point and Q(5, 3) as the ending point, we have $x(t) = 2 + (5 - 2)t = 2 + 3t$ and $y(t) = -1 + (3 - (-1))t = -1 + 4t$. Checking the endpoints: at $t=0$, the point is $(2, -1)$; at $t=1$, the point is $(5, 3)$.

Question 7

1. A circle with center $(3, -2)$ and radius 5. (correct answer)
2. A circle with center $(-3, 2)$ and radius 5.
3. A circle with center $(3, -2)$ and radius 25.
4. An ellipse with center $(3, -2)$ and semi-axes of length 3 and 5.

Explanation: The standard parametric equations for a circle with center $(h, k)$ and radius $r$ are $x(t) = h + r\cos(t)$ and $y(t) = k + r\sin(t)$. By comparing the given equations to the standard form, we can identify $h=3$, $k=-2$, and $r=5$. Thus, the path is a circle centered at $(3, -2)$ with a radius of 5.

Question 8

1. $(x-1)^2 + (y+3)^2 = 16$ (correct answer)
2. $(x+1)^2 + (y-3)^2 = 16$
3. $(x-1)^2 + (y+3)^2 = 4$
4. $(x-4)^2 + (y+4)^2 = 1$

Explanation: To convert to a rectangular equation, we isolate the trigonometric terms and use the Pythagorean identity $\cos^2(t) + \sin^2(t) = 1$. From the given equations, we have $x-1 = 4\cos(t) \implies \cos(t) = \frac{x-1}{4}$ and $y+3 = 4\sin(t) \implies \sin(t) = \frac{y+3}{4}$. Substituting into the identity gives $\left(\frac{x-1}{4}\right)^2 + \left(\frac{y+3}{4}\right)^2 = 1$. This simplifies to $\frac{(x-1)^2}{16} + \frac{(y+3)^2}{16} = 1$, which is equivalent to $(x-1)^2 + (y+3)^2 = 16$.

Question 9

1. $t=3$ (correct answer)
2. $t=2$
3. $t=5$
4. No such positive value of $t$ exists.

Explanation: We need to find a single value of $t$ that satisfies both $x(t)=5$ and $y(t)=12$. First, solve $x(t)=5$: $2t-1 = 5 \implies 2t = 6 \implies t=3$. Then, check if this value of $t$ works in the second equation: $y(3) = (3)^2+3 = 9+3 = 12$. Since $t=3$ satisfies both equations and is positive, it is the correct value.

Question 10

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

Explanation: One common method to parametrize an equation where $y$ is a function of $x$ is to set $x=t$. Then, we substitute $t$ for $x$ in the original equation to find $y$ in terms of $t$. If $x(t) = t$, then substituting into $y = (x+2)^2 - 1$ yields $y(t) = (t+2)^2 - 1$. This is a valid parameterization. Other parameterizations exist, but this is the most direct one among the choices.

Question 11

1. A parabola that opens to the right. (correct answer)
2. A parabola that opens upward.
3. A line with a slope of 2.
4. A hyperbola.

Explanation: To identify the curve, we eliminate the parameter $t$. From the equation for y, we can solve for t: $t = \frac{y}{8}$. Now we substitute this expression for $t$ into the equation for x: $x = 4\left(\frac{y}{8}\right)^2 = 4\left(\frac{y^2}{64}\right) = \frac{y^2}{16}$. The equation $x = \frac{1}{16}y^2$ is the equation of a parabola. Since $x$ is expressed as a function of $y^2$ and the coefficient is positive, the parabola opens in the positive x-direction, which is to the right.

Question 12

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

Explanation: To find the rectangular equation, we eliminate the parameter $t$. First, solve the equation for $x$ in terms of $t$: $x = 3t - 1 \implies x+1 = 3t \implies t = \frac{x+1}{3}$. Next, substitute this expression for $t$ into the equation for $y$: $y = 6\left(\frac{x+1}{3}\right) + 2$. Simplifying gives $y = 2(x+1) + 2 = 2x + 2 + 2 = 2x + 4$.

Question 13

1. The particle always moves to the left. (correct answer)
2. The particle always moves to the right.
3. The particle moves to the right for $t > 0$.
4. The particle moves to the left for $t < 0$ and then to the right for $t > 0$.

Explanation: The horizontal motion is determined by the function $x(t) = 5 - t^2$. As $t$ increases from 0, $t^2$ increases. Since $t^2$ is being subtracted from 5, the value of $x(t)$ will decrease. A decrease in the x-coordinate corresponds to motion to the left. Since this is true for all $t > 0$, the particle always moves to the left for the given domain $t \ge 0$.

Question 14

1. $s \in [-2, 2]$ (correct answer)
2. $s \in [0, 4]$
3. $s \in [-1, 1]$
4. $s \in [1, 5]$

Explanation: First, determine the starting and ending points and the path of the original segment. At $t=-1$, the point is $x(-1)=-2, y(-1)=4$. At $t=1$, the point is $x(1)=2, y(1)=4$. The rectangular equation is found by solving for $t$: $t=x/2$, so $y=4(x/2)^2 = 4(x^2/4) = x^2$. The segment is on the parabola $y=x^2$ from $(-2,4)$ to $(2,4)$. Now consider the new parameterization $x(s)=s, y(s)=s^2$. This also traces the parabola $y=x^2$. To trace the same segment, the x-values, which are equal to $s$, must range from -2 to 2. Therefore, the required domain is $s \in [-2, 2]$.

Question 15

1. 144 feet (correct answer)
2. 80 feet
3. 64 feet
4. 2 seconds

Explanation: The maximum height is determined by the maximum value of the vertical position function, $y(t) = -16t^2 + 64t + 80$. This is a quadratic function representing a downward-opening parabola. The maximum value occurs at the vertex. The t-coordinate of the vertex is $t = -\frac{b}{2a} = -\frac{64}{2(-16)} = -\frac{64}{-32} = 2$. To find the maximum height, we evaluate $y(t)$ at $t=2$: $y(2) = -16(2)^2 + 64(2) + 80 = -16(4) + 128 + 80 = -64 + 128 + 80 = 144$. So, the maximum height is 144 feet.

Question 16

1. $[0, 1]$ (correct answer)
2. $[-1, 1]$
3. $[1, 5]$
4. $[0, 4]$

Explanation: The y-values of the curve are given by the function $y(t) = \cos^2(t)$. The range of the function $f(t)=\cos(t)$ is $[-1, 1]$. When we square the values in this range, the outputs are always non-negative. The smallest possible value is $0^2 = 0$, which occurs when $\cos(t)=0$. The largest possible value is $(-1)^2 = 1^2 = 1$. Therefore, the range of $y(t) = \cos^2(t)$ is the closed interval $[0, 1]$.

Question 17

1. $(-2, 5)$ (correct answer)
2. $(10, 5)$
3. $(2, 3)$
4. $(-2, 3)$

Explanation: To find the coordinates of the point at a specific value of the parameter $t$, substitute $t=2$ into both parametric equations. For the x-coordinate: $x(2) = (2)^2 - 3(2) = 4 - 6 = -2$. For the y-coordinate: $y(2) = 2(2) + 1 = 4 + 1 = 5$. Therefore, the coordinates of the point on the curve when $t=2$ are $(-2, 5)$.

Question 18

1. $120$ (correct answer)
2. $0$
3. $-120$
4. $60$

Explanation: This question tests AP Precalculus skills in parametric functions, vectors, and matrices, focusing on evaluating sinusoidal functions in AC circuit analysis. Parametric functions use parameters to express coordinates, vectors represent direction and magnitude, and matrices perform transformations. In this scenario, voltage follows v(t)=120sin(120πt) and we need to evaluate at t=1/240. Choice A is correct because v(1/240)=120sin(120π×1/240)=120sin(π/2)=120×1=120. Choice C is incorrect because it represents -120, which would occur at t=3/240 when sin(3π/2)=-1, showing a phase error. To help students: Practice substituting specific time values into sinusoidal functions and simplifying the arguments. Watch for: Common errors in simplifying fractions within trigonometric arguments and sign errors when evaluating standard angles.

Question 19

1. 15 (correct answer)
2. 5
3. 7.5
4. $\frac{1}{3}$

Explanation: The average rate of change of $y$ with respect to $x$ is given by the formula $\frac{\Delta y}{\Delta x} = \frac{y(t_2) - y(t_1)}{x(t_2) - x(t_1)}$. Here, $t_1=1$ and $t_2=4$. First, find the coordinates at these times. At $t=1$, $x(1) = \sqrt{1} = 1$ and $y(1) = 1^2 - 1 = 0$. At $t=4$, $x(4) = \sqrt{4} = 2$ and $y(4) = 4^2 - 1 = 15$. Now, calculate the average rate of change: $\frac{\Delta y}{\Delta x} = \frac{15 - 0}{2 - 1} = \frac{15}{1} = 15$.

Question 20

1. $\langle 0,\,3\rangle$ (correct answer)
2. $\langle 3,\,0\rangle$
3. $\langle 0,\,-3\rangle$
4. $\langle 2.12,\,2.12\rangle$

Explanation: This question tests AP Precalculus skills in parametric functions, vectors, and matrices, focusing on circular motion described by parametric equations. Parametric functions use parameters to express coordinates, vectors represent direction and magnitude, and matrices perform transformations. In this scenario, a point on a wheel follows r(t) = ⟨3cos(2t), 3sin(2t)⟩, describing circular motion with radius 3 meters and angular frequency 2 rad/s. Choice A is correct because at t=π/4, we have x = 3cos(2·π/4) = 3cos(π/2) = 0 and y = 3sin(2·π/4) = 3sin(π/2) = 3, giving coordinates ⟨0, 3⟩. Choice D is incorrect because it appears to use t=π/4 directly in the trig functions without the factor of 2, resulting in cos(π/4) = sin(π/4) = √2/2, giving approximately ⟨2.12, 2.12⟩. To help students: Emphasize the role of the coefficient of t in parametric equations as angular frequency, practice evaluating trigonometric functions at key angles, and visualize how the parameter affects position on the circle. Watch for: Forgetting to multiply t by the coefficient inside trig functions, confusion between radians and degrees, and misremembering special angle values.