AP Precalculus Quiz: Parametric Functions

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Parametric FunctionsQuestion 1 of 20

Two different parametric representations of a curve are given by $C_1: x(t) = t, y(t) = t^2$ for $t \in \mathbb{R}$ and $C_2: x(s) = s^3, y(s) = s^6$ for $s \in \mathbb{R}$ . Which of the following statements correctly compares the two representations?

C_1

and

C_2

trace the same curve, but

C_2

is traversed more quickly than

C_1

for

|s|>1

C_1

and

C_2

trace different curves because the equations are different functions of the parameter.

C_1

and

C_2

trace the same curve, but

C_2

only traces the portion of the curve where

x \ge 0

C_1

and

C_2

trace the same curve, but

C_2

is traversed in the opposite direction of

C_1

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AP Precalculus Quiz

AP Precalculus Quiz: Parametric Functions

Practice Parametric Functions in AP Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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.

All questions

Question 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)
$C_1$ and $C_2$ trace different curves because the equations are different functions of the parameter.
$C_1$ and $C_2$ trace the same curve, but $C_2$ only traces the portion of the curve where $x \ge 0$ .
$C_1$ 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

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

A particle's position is given by $x(t) = t^2 - 2t$ and $y(t) = t^3 - 3t$ . At which positive value of $t$ does the particle's vertical motion change direction?

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

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

t = -1

. The question asks for the positive value of

t

, which is

t=1

Question 3

A particle's motion along a circle is described by $C_1: x(t) = \cos(t), y(t) = \sin(t)$ for $0 \le t \le 2\pi$ . A second particle's motion is described by $C_2: x(s) = \cos(2s), y(s) = \sin(2s)$ for $0 \le s \le \pi$ . Which statement accurately compares the motions?

Both particles trace the same unit circle once, but the second particle moves twice as fast. (correct answer)
The second particle traces the unit circle twice, while the first particle traces it once.
The first particle moves counter-clockwise, while the second particle moves clockwise.
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

A linear map uses $M=\begin{bmatrix}2&0\\0&\tfrac12\end{bmatrix}$ on $\vec{v}=\begin{bmatrix}-4\\6\end{bmatrix}$ . Find the transformed vector $M\vec{v}$ .

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

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

An image vector $\vec{v}=\begin{bmatrix}1\\2\end{bmatrix}$ is scaled by $M=\begin{bmatrix}2&0\\0&3\end{bmatrix}$ ; find $M\vec{v}$ .

$\begin{bmatrix}2\\6\end{bmatrix}$ (correct answer)
$\begin{bmatrix}5\\5\end{bmatrix}$
$\begin{bmatrix}6\\2\end{bmatrix}$
$\begin{bmatrix}2\\3\end{bmatrix}$

Explanation: This question tests AP Precalculus skills in parametric functions, vectors, and matrices, focusing on matrix transformations for scaling operations. Parametric functions use parameters to express coordinates, vectors represent direction and magnitude, and matrices perform transformations. In this scenario, a diagonal matrix M = [[2,0],[0,3]] scales an image vector v = [[1],[2]], with different scaling factors for each component. Choice A is correct because Mv = [[2,0],[0,3]][[1],[2]] = [[2(1)+0(2)],[0(1)+3(2)]] = [[2],[6]], representing horizontal scaling by 2 and vertical scaling by 3. Choice C is incorrect because it reverses the result, giving [[6],[2]] instead of [[2],[6]], suggesting confusion about which component gets which scaling factor. To help students: Emphasize that diagonal matrices scale each component independently, practice identifying transformation types from matrix structure, and visualize scaling effects geometrically. Watch for: Mixing up which diagonal element affects which component, arithmetic errors in simple multiplication, and confusion about matrix-vector multiplication order.

Question 6

A drone's displacement vectors are $\vec{a}=\langle 4,-2\rangle$ km and $\vec{b}=\langle -1,5\rangle$ km; compute $\vec{a}+\vec{b}$ .

$\langle 3,3\rangle$ (correct answer)
$\langle 5,-7\rangle$
$\langle -3,7\rangle$
$\langle 3,-3\rangle$

Explanation: This question tests AP Precalculus skills in parametric functions, vectors, and matrices, focusing on vector addition for displacement calculations. Parametric functions use parameters to express coordinates, vectors represent direction and magnitude, and matrices perform transformations. In this scenario, a drone's total displacement is found by adding two displacement vectors a = ⟨4, -2⟩ km and b = ⟨-1, 5⟩ km, representing sequential movements. Choice A is correct because vector addition is performed component-wise: a + b = ⟨4+(-1), -2+5⟩ = ⟨3, 3⟩, giving the net displacement. Choice B is incorrect because it appears to subtract the vectors instead of adding them, computing ⟨4-(-1), -2-5⟩ = ⟨5, -7⟩, a common sign error. To help students: Emphasize that vector addition represents combining displacements tip-to-tail, practice component-wise operations carefully, and use graphical representations to visualize vector sums. Watch for: Sign errors when adding negative components, confusing addition with subtraction, and misaligning components during calculation.

Question 7

A curve is defined by the parametric equations $x(t) = t^2 - 9$ and $y(t) = t^2 - 4t$ . At what points does the curve intersect the y-axis?

At $(0, -3)$ and $(0, 21)$ (correct answer)
At $(0, 0)$ and $(0, 32)$
At $(0, -3)$ only
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

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 8

Which of the following sets of parametric equations describes the line segment from point P(2, -1) to point Q(5, 3) for $0 \le t \le 1$ ?

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

Explanation: A parameterization for a line segment from

(x_1, y_1)

(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 9

The path of a particle is given by the parametric equations $x(t) = 3 + 5\cos(t)$ and $y(t) = -2 + 5\sin(t)$ . Which of the following describes the path?

A circle with center $(3, -2)$ and radius 5. (correct answer)
A circle with center $(-3, 2)$ and radius 5.
A circle with center $(3, -2)$ and radius 25.
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 10

A curve is defined by the parametric equations $x(t) = 4\cos(t)+1$ and $y(t) = 4\sin(t)-3$ . Which of the following is the corresponding rectangular equation?

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

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 11

A particle's path is described by $x(t) = 2t-1$ and $y(t) = t^2+3$ . For which positive value of the parameter $t$ is the particle at the point $(5, 12)$ ?

$t=3$ (correct answer)
$t=2$
$t=5$
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 12

Which of the following is a possible parametric representation of the parabola given by the equation $y = (x+2)^2 - 1$ ?

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

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 13

A curve is defined by the parametric equations $x(t) = 4t^2$ and $y(t) = 8t$ . Which of the following describes the curve?

A parabola that opens to the right. (correct answer)
A parabola that opens upward.
A line with a slope of 2.
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 14

A parametric curve is given by the equations $x(t) = 3t - 1$ and $y(t) = 6t + 2$ . Which of the following is the rectangular equation for this curve?

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

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 15

The motion of a particle in the xy-plane is described by the parametric equations $x(t) = 5 - t^2$ and $y(t) = t^3 - 4t$ for $t \ge 0$ . Which of the following describes the particle's horizontal motion as $t$ increases?

The particle always moves to the left. (correct answer)
The particle always moves to the right.
The particle moves to the right for $t > 0$ .
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 16

The parametric equations $x(t) = 2t$ and $y(t) = 4t^2$ for $t \in [-1, 1]$ describe a segment of a parabola. Which of the following domains for the parameterization $x(s) = s$ and $y(s) = s^2$ would trace the exact same segment?

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

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)

(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 17

The vertical position of a projectile is given by $y(t) = -16t^2 + 64t + 80$ , where $t$ is time in seconds. The horizontal position is given by $x(t) = 20t$ .

What is the maximum height reached by the projectile?

144 feet (correct answer)
80 feet
64 feet
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)

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 18

A curve is parameterized by $x(t) = 2\sin(t) + 3$ and $y(t) = \cos^2(t)$ for all real $t$ . What is the range of possible y-values for this curve?

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

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)

[-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 19

A curve is defined by the parametric equations $x(t) = t^2 - 3t$ and $y(t) = 2t + 1$ . What are the coordinates of the point on the curve when $t=2$ ?

$(-2, 5)$ (correct answer)
$(10, 5)$
$(2, 3)$
$(-2, 3)$

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 20

An AC circuit uses $v(t)=120\sin(120\pi t)$ and $i(t)=6\sin(120\pi t-\pi/3)$ . Find $v\!\left(\tfrac{1}{240}\right)$ .

$120$ (correct answer)
$0$
$-120$
$60$

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.