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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.

Question 1 / 20

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A curve is parameterized by x(t)=2sin⁡(t)+3x(t) = 2\sin(t) + 3x(t)=2sin(t)+3 and y(t)=cos⁡2(t)y(t) = \cos^2(t)y(t)=cos2(t) for all real ttt. What is the range of possible y-values for this curve?

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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

A curve is parameterized by x(t)=2sin⁡(t)+3x(t) = 2\sin(t) + 3x(t)=2sin(t)+3 and y(t)=cos⁡2(t)y(t) = \cos^2(t)y(t)=cos2(t) for all real ttt. What is the range of possible y-values for this curve?

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

Explanation: The y-values of the curve are given by the function y(t)=cos⁡2(t)y(t) = \cos^2(t)y(t)=cos2(t). The range of the function f(t)=cos⁡(t)f(t)=\cos(t)f(t)=cos(t) is [−1,1][-1, 1][−1,1]. When we square the values in this range, the outputs are always non-negative. The smallest possible value is 02=00^2 = 002=0, which occurs when cos⁡(t)=0\cos(t)=0cos(t)=0. The largest possible value is (−1)2=12=1(-1)^2 = 1^2 = 1(−1)2=12=1. Therefore, the range of y(t)=cos⁡2(t)y(t) = \cos^2(t)y(t)=cos2(t) is the closed interval [0,1][0, 1][0,1].

Question 2

Two different parametric representations of a curve are given by C1:x(t)=t,y(t)=t2C_1: x(t) = t, y(t) = t^2C1​:x(t)=t,y(t)=t2 for t∈Rt \in \mathbb{R}t∈R and C2:x(s)=s3,y(s)=s6C_2: x(s) = s^3, y(s) = s^6C2​:x(s)=s3,y(s)=s6 for s∈Rs \in \mathbb{R}s∈R. Which of the following statements correctly compares the two representations?

  1. C1C_1C1​ and C2C_2C2​ trace the same curve, but C2C_2C2​ is traversed more quickly than C1C_1C1​ for ∣s∣>1|s|>1∣s∣>1. (correct answer)
  2. C1C_1C1​ and C2C_2C2​ trace different curves because the equations are different functions of the parameter.
  3. C1C_1C1​ and C2C_2C2​ trace the same curve, but C2C_2C2​ only traces the portion of the curve where x≥0x \ge 0x≥0.
  4. C1C_1C1​ and C2C_2C2​ trace the same curve, but C2C_2C2​ is traversed in the opposite direction of C1C_1C1​.

Explanation: For both representations, eliminating the parameter results in the rectangular equation y=x2y=x^2y=x2. Thus, they trace the same parabola. For C1C_1C1​, as ttt increases, the point moves along the parabola. For C2C_2C2​, as sss increases, the point also moves along the parabola in the same direction. However, the speed of traversal is different. For example, in C1C_1C1​, it takes 2 units of time for ttt to go from 0 to 2, covering x-values from 0 to 2. In C2C_2C2​, it takes 2 units of time for sss to go from 0 to 2, covering x-values from 03=00^3=003=0 to 23=82^3=823=8. Since C2C_2C2​ covers a greater distance along the curve in the same parameter interval, it is traversed more quickly.

Question 3

A curve is defined by the parametric equations x(t)=t2−3tx(t) = t^2 - 3tx(t)=t2−3t and y(t)=2t+1y(t) = 2t + 1y(t)=2t+1. What are the coordinates of the point on the curve when t=2t=2t=2?

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

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

Question 4

The position of a particle is given by x(t)=tx(t) = \sqrt{t}x(t)=t​ and y(t)=t2−1y(t) = t^2 - 1y(t)=t2−1. What is the average rate of change of yyy with respect to xxx as ttt changes from t=1t=1t=1 to t=4t=4t=4?

  1. 15 (correct answer)
  2. 5
  3. 7.5
  4. 13\frac{1}{3}31​

Explanation: The average rate of change of yyy with respect to xxx is given by the formula ΔyΔx=y(t2)−y(t1)x(t2)−x(t1)\frac{\Delta y}{\Delta x} = \frac{y(t_2) - y(t_1)}{x(t_2) - x(t_1)}ΔxΔy​=x(t2​)−x(t1​)y(t2​)−y(t1​)​. Here, t1=1t_1=1t1​=1 and t2=4t_2=4t2​=4. First, find the coordinates at these times. At t=1t=1t=1, x(1)=1=1x(1) = \sqrt{1} = 1x(1)=1​=1 and y(1)=12−1=0y(1) = 1^2 - 1 = 0y(1)=12−1=0. At t=4t=4t=4, x(4)=4=2x(4) = \sqrt{4} = 2x(4)=4​=2 and y(4)=42−1=15y(4) = 4^2 - 1 = 15y(4)=42−1=15. Now, calculate the average rate of change: ΔyΔx=15−02−1=151=15\frac{\Delta y}{\Delta x} = \frac{15 - 0}{2 - 1} = \frac{15}{1} = 15ΔxΔy​=2−115−0​=115​=15.

Question 5

A particle's position is given by x(t)=t2−2tx(t) = t^2 - 2tx(t)=t2−2t and y(t)=t3−3ty(t) = t^3 - 3ty(t)=t3−3t. At which positive value of ttt does the particle's vertical motion change direction?

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

Explanation: The vertical motion is described by the function y(t)=t3−3ty(t) = t^3 - 3ty(t)=t3−3t. 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)y(t)y(t), these extrema occur at the critical points. The derivative is y′(t)=3t2−3y'(t) = 3t^2 - 3y′(t)=3t2−3. Setting the derivative to zero gives 3t2−3=03t^2 - 3 = 03t2−3=0, which leads to t2=1t^2 = 1t2=1, so t=1t = 1t=1 or t=−1t = -1t=−1. The question asks for the positive value of ttt, which is t=1t=1t=1.

Question 6

A particle's motion along a circle is described by C1:x(t)=cos⁡(t),y(t)=sin⁡(t)C_1: x(t) = \cos(t), y(t) = \sin(t)C1​:x(t)=cos(t),y(t)=sin(t) for 0≤t≤2π0 \le t \le 2\pi0≤t≤2π. A second particle's motion is described by C2:x(s)=cos⁡(2s),y(s)=sin⁡(2s)C_2: x(s) = \cos(2s), y(s) = \sin(2s)C2​:x(s)=cos(2s),y(s)=sin(2s) for 0≤s≤π0 \le s \le \pi0≤s≤π. Which statement accurately compares the motions?

  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 x2+y2=cos⁡2(θ)+sin⁡2(θ)=1x^2+y^2=\cos^2(\theta)+\sin^2(\theta)=1x2+y2=cos2(θ)+sin2(θ)=1. For C1C_1C1​, as ttt goes from 0 to 2π2\pi2π, the particle makes one full counter-clockwise revolution. For C2C_2C2​, as the parameter sss goes from 0 to π\piπ, the angle 2s2s2s goes from 0 to 2π2\pi2π. Thus, the second particle also makes one full counter-clockwise revolution. However, the first particle takes 2π2\pi2π units of time to complete the circle, while the second takes only π\piπ units of time. This means the second particle traverses the same path at twice the speed.

Question 7

A point on a wheel follows r}(t)=\langle 3\cos(2t),\,3\sin(2t)\rangle (meters); determine the coordinates at t=π/4t=\pi/4t=π/4.

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

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.

Question 8

A rotating arm follows r⃗(t)=⟨3cos⁡(ωt), 3sin⁡(ωt)⟩\vec{r}(t)=\langle 3\cos(\omega t),\,3\sin(\omega t)\rangler(t)=⟨3cos(ωt),3sin(ωt)⟩; what does increasing ω\omegaω change?

  1. It increases the radius of the circle.
  2. It increases the rotation speed. (correct answer)
  3. It moves the center away from the origin.
  4. It changes the path into a parabola.

Explanation: This question tests AP Precalculus skills in parametric functions, vectors, and matrices, focusing on parametric equations for circular motion with variable angular velocity. Parametric functions use parameters to express coordinates, vectors represent direction and magnitude, and matrices perform transformations. In this scenario, r(t)=⟨3cos(ωt), 3sin(ωt)⟩ describes a rotating arm of fixed length 3, where ω controls the angular velocity. Choice B is correct because increasing ω increases the coefficient of t inside the trigonometric functions, causing the angle to change more rapidly with time, thus increasing the rotation speed. Choice A is incorrect because ω doesn't affect the radius—the coefficient 3 outside the trig functions determines the radius and remains constant. To help students: Emphasize distinguishing between parameters that affect size (amplitude/radius) versus those that affect rate (frequency/angular velocity), practice analyzing how each parameter influences motion, and use animations to visualize parameter effects. Watch for: Confusion between radius and angular velocity parameters, misunderstanding the role of coefficients inside vs outside trig functions, and incorrect geometric interpretations.

Question 9

A projectile is r⃗(t)=⟨18t, 2+12t−4.9t2⟩\vec{r}(t)=\langle 18t,\,2+12t-4.9t^2\rangler(t)=⟨18t,2+12t−4.9t2⟩; determine r⃗(1)\vec{r}(1)r(1).

  1. ⟨18, 9.1⟩\langle 18,\,9.1\rangle⟨18,9.1⟩ (correct answer)
  2. ⟨18, 7.1⟩\langle 18,\,7.1\rangle⟨18,7.1⟩
  3. ⟨36, 9.1⟩\langle 36,\,9.1\rangle⟨36,9.1⟩
  4. ⟨18, 18.9⟩\langle 18,\,18.9\rangle⟨18,18.9⟩

Explanation: This question tests AP Precalculus skills in parametric functions, vectors, and matrices, focusing on projectile motion with quadratic parametric equations. Parametric functions use parameters to express coordinates, vectors represent direction and magnitude, and matrices perform transformations. In this scenario, r(t)=⟨18t, 2+12t-4.9t²⟩ describes projectile motion with constant horizontal velocity 18 m/s and vertical motion under gravity. Choice A is correct because substituting t=1 gives x(1)=18×1=18 and y(1)=2+12×1-4.9×1²=2+12-4.9=14-4.9=9.1, resulting in ⟨18, 9.1⟩. Choice B is incorrect because it miscalculates the y-component as 7.1 instead of 9.1, possibly from an arithmetic error in combining the three terms. To help students: Emphasize careful evaluation of each term in multi-term expressions, practice substituting values systematically, and understand the physical meaning of each term (initial height, initial velocity, gravity). Watch for: Order of operations errors, sign mistakes with the gravity term, and computational errors when combining multiple terms.

Question 10

A rotating arm has r}(t)=\langle R\cos(\omega t),R\sin(\omega t)\rangle with R=2R=2R=2; how does increasing \omega affect the motion?

  1. It increases the radius of the circle.
  2. It decreases the radius of the circle.
  3. It increases the angular speed around the circle. (correct answer)
  4. It changes the circle into an ellipse.

Explanation: This question tests AP Precalculus skills in parametric functions, vectors, and matrices, focusing on understanding parameters in circular motion equations. Parametric functions use parameters to express coordinates, vectors represent direction and magnitude, and matrices perform transformations. In this scenario, a rotating arm follows r(t) = ⟨Rcos(ωt), Rsin(ωt)⟩ where R is the radius and ω is the angular frequency, describing uniform circular motion. Choice C is correct because increasing ω increases the coefficient of t inside the trigonometric functions, causing the angle ωt to grow faster with time, thus increasing the angular speed around the circle. Choice A is incorrect because ω appears inside the trig functions, not as a coefficient of the radius R, so it doesn't affect the circle's size. To help students: Emphasize distinguishing between parameters that affect size (amplitude) versus speed (frequency), practice interpreting coefficients in parametric equations, and use animations to visualize how parameters affect motion. Watch for: Confusing the roles of different parameters, thinking all coefficients affect size, and not recognizing frequency's effect on speed.

Question 11

A point moves on a circle: r⃗(t)=⟨5cos⁡(2t), 5sin⁡(2t)⟩\vec{r}(t)=\langle 5\cos(2t),\,5\sin(2t)\rangler(t)=⟨5cos(2t),5sin(2t)⟩; find r⃗(π/2)\vec{r}(\pi/2)r(π/2).

  1. ⟨0, 5⟩\langle 0,\,5\rangle⟨0,5⟩
  2. ⟨−5, 0⟩\langle -5,\,0\rangle⟨−5,0⟩ (correct answer)
  3. ⟨5, 0⟩\langle 5,\,0\rangle⟨5,0⟩
  4. ⟨0, −5⟩\langle 0,\,-5\rangle⟨0,−5⟩

Explanation: This question tests AP Precalculus skills in parametric functions, vectors, and matrices, focusing on circular motion using trigonometric parametric equations. Parametric functions use parameters to express coordinates, vectors represent direction and magnitude, and matrices perform transformations. In this scenario, r(t) = ⟨5cos(2t), 5sin(2t)⟩ describes a circle of radius 5 with angular velocity 2, where the parameter 2t controls the rotation speed. Choice B is correct because substituting t=π/2 gives x(π/2)=5cos(2×π/2)=5cos(π)=5×(-1)=-5 and y(π/2)=5sin(2×π/2)=5sin(π)=5×0=0, resulting in ⟨-5,0⟩. Choice C is incorrect because it has the wrong sign for the x-component, likely from evaluating cos(π) as 1 instead of -1. To help students: Emphasize understanding how the coefficient of t affects the angle, practice evaluating trigonometric functions at key angles, and visualize the circular path. Watch for: Confusion about radians vs degrees, incorrect evaluation of trig functions at multiples of π, and forgetting to multiply the parameter by its coefficient.

Question 12

A circular path is r}(t)=\langle 5\cos t,\,5\sin t\rangle; determine the coordinates at t=π/2t=\pi/2t=π/2.

  1. ⟨5, 0⟩\langle 5,\,0\rangle⟨5,0⟩
  2. ⟨0, 5⟩\langle 0,\,5\rangle⟨0,5⟩ (correct answer)
  3. ⟨0, −5⟩\langle 0,\,-5\rangle⟨0,−5⟩
  4. ⟨−5, 0⟩\langle -5,\,0\rangle⟨−5,0⟩

Explanation: This question tests AP Precalculus skills in parametric functions, vectors, and matrices, focusing on evaluating trigonometric parametric equations at special angles. Parametric functions use parameters to express coordinates, vectors represent direction and magnitude, and matrices perform transformations. In this scenario, a circular path with radius 5 is described by r(t) = ⟨5cos t, 5sin t⟩, representing counterclockwise motion starting from (5,0). Choice B is correct because at t=π/2, x(π/2) = 5cos(π/2) = 5(0) = 0 and y(π/2) = 5sin(π/2) = 5(1) = 5, giving coordinates ⟨0, 5⟩. Choice C is incorrect because it has the wrong sign for the y-coordinate, suggesting confusion about the quadrant or the value of sin(π/2). To help students: Emphasize memorizing unit circle values at key angles, practice visualizing motion around circles, and connect parameter values to positions. Watch for: Confusion between cosine and sine values at special angles, sign errors based on quadrants, and mixing up coordinates.

Question 13

A point is transformed by M=[20012]M=\begin{bmatrix}2&0\\0&\tfrac12\end{bmatrix}M=[20​021​​]; compute M[−46]M\begin{bmatrix}-4\\6\end{bmatrix}M[−46​].

  1. [−83]\begin{bmatrix}-8\\3\end{bmatrix}[−83​] (correct answer)
  2. [−212]\begin{bmatrix}-2\\12\end{bmatrix}[−212​]
  3. [83]\begin{bmatrix}8\\3\end{bmatrix}[83​]
  4. [−812]\begin{bmatrix}-8\\12\end{bmatrix}[−812​]

Explanation: This question tests AP Precalculus skills in parametric functions, vectors, and matrices, focusing on matrix transformations with scaling. Parametric functions use parameters to express coordinates, vectors represent direction and magnitude, and matrices perform transformations. In this scenario, the diagonal matrix M=[2 0; 0 1/2] represents a scaling transformation that stretches horizontally by 2 and compresses vertically by 1/2. Choice A is correct because M[-4; 6] = [2×(-4)+0×6; 0×(-4)+(1/2)×6] = [-8+0; 0+3] = [-8; 3], showing the x-coordinate doubled and y-coordinate halved. Choice D is incorrect because it miscalculates the y-component as 12 instead of 3, possibly by multiplying by 2 instead of 1/2. To help students: Emphasize that diagonal matrices perform independent scaling on each axis, practice matrix multiplication step-by-step, and visualize transformations geometrically. Watch for: Confusion about which diagonal entry affects which coordinate, arithmetic errors with fractions, and misunderstanding scaling directions.

Question 14

A robot is commanded by r⃗(t)=⟨3t−2, −t+5⟩\vec{r}(t)=\langle 3t-2,\,-t+5\rangler(t)=⟨3t−2,−t+5⟩ meters; determine its coordinates at t=4t=4t=4.

  1. (10, 1)(10,\,1)(10,1) (correct answer)
  2. (14, 1)(14,\,1)(14,1)
  3. (10, 9)(10,\,9)(10,9)
  4. (14, 9)(14,\,9)(14,9)

Explanation: This question tests AP Precalculus skills in parametric functions, vectors, and matrices, focusing on evaluating linear parametric equations. Parametric functions use parameters to express coordinates, vectors represent direction and magnitude, and matrices perform transformations. In this scenario, the robot's position vector r(t) = ⟨3t-2, -t+5⟩ describes linear motion where x and y coordinates change linearly with time t. Choice A is correct because substituting t=4 gives x(4)=3(4)-2=12-2=10 and y(4)=-4+5=1, resulting in coordinates (10,1). Choice D is incorrect because it miscalculates both components, getting x=14 instead of 10 and y=9 instead of 1, likely from arithmetic errors or misreading the equations. To help students: Emphasize careful substitution of parameter values, practice evaluating each component separately, and verify results by checking if they satisfy the original equations. Watch for: Sign errors in negative terms, order of operations mistakes, and confusion between parameter t and coordinates.

Question 15

A linear map uses M=[20012]M=\begin{bmatrix}2&0\\0&\tfrac12\end{bmatrix}M=[20​021​​] on v⃗=[−46]\vec{v}=\begin{bmatrix}-4\\6\end{bmatrix}v=[−46​]. Find the transformed vector Mv⃗M\vec{v}Mv.

  1. [−83]\begin{bmatrix}-8\\3\end{bmatrix}[−83​] (correct answer)
  2. [−212]\begin{bmatrix}-2\\12\end{bmatrix}[−212​]
  3. [−812]\begin{bmatrix}-8\\12\end{bmatrix}[−812​]
  4. [−8−3]\begin{bmatrix}-8\\-3\end{bmatrix}[−8−3​]

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 16

An AC model is v(t)=V0cos⁡(100πt)v(t)=V_0\cos(100\pi t)v(t)=V0​cos(100πt) with v(0)=V0v(0)=V_0v(0)=V0​. How does increasing V0V_0V0​ affect the voltage function?

  1. It increases amplitude, scaling voltage values by the same factor. (correct answer)
  2. It increases frequency, shortening the period of oscillation.
  3. It shifts phase by π/2\pi/2π/2, swapping sine and cosine.
  4. It changes time units so ttt must be in degrees.

Explanation: This question tests AP Precalculus skills in parametric functions, vectors, and matrices, focusing on understanding amplitude in sinusoidal functions. Parametric functions use parameters to express coordinates, vectors represent direction and magnitude, and matrices perform transformations. In this scenario, an AC voltage model v(t)=V₀cos(100πt) has amplitude parameter V₀ that scales the entire function. Choice A is correct because increasing V₀ multiplies all voltage values by the same factor, making the oscillation larger without changing its frequency or phase. Choice B is incorrect because frequency is determined by the coefficient 100π in the argument, not by V₀, showing confusion between amplitude and frequency parameters. To help students: Use graphical representations to show how amplitude affects the vertical stretch of sinusoidal curves. Watch for: Confusion between different parameters' roles - amplitude (V₀) affects size, angular frequency (100π) affects period.

Question 17

An AC circuit uses v(t)=120sin⁡(120πt)v(t)=120\sin(120\pi t)v(t)=120sin(120πt) and i(t)=6sin⁡(120πt−π/3)i(t)=6\sin(120\pi t-\pi/3)i(t)=6sin(120πt−π/3). Find v ⁣(1240)v\!\left(\tfrac{1}{240}\right)v(2401​).

  1. 120120120 (correct answer)
  2. 000
  3. −120-120−120
  4. 606060

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 18

Circular motion is r⃗(t)=⟨Rcos⁡(3t), Rsin⁡(3t)⟩\vec{r}(t)=\langle R\cos(3t),\,R\sin(3t)\rangler(t)=⟨Rcos(3t),Rsin(3t)⟩ with r⃗(0)=⟨R,0⟩\vec{r}(0)=\langle R,0\rangler(0)=⟨R,0⟩. How does increasing RRR affect the path?

  1. It increases the circle’s radius, scaling all positions outward. (correct answer)
  2. It increases angular speed, making the point complete more rotations.
  3. It shifts the circle’s center to (R,0)(R,0)(R,0).
  4. It changes the period to 2πR\tfrac{2\pi}{R}R2π​.

Explanation: This question tests AP Precalculus skills in parametric functions, vectors, and matrices, focusing on understanding parameters in circular motion equations. Parametric functions use parameters to express coordinates, vectors represent direction and magnitude, and matrices perform transformations. In this scenario, circular motion is described by r⃗(t)=⟨Rcos(3t),Rsin(3t)⟩ where R is the radius parameter. Choice A is correct because increasing R scales both x and y coordinates proportionally, making the circle larger while maintaining its shape and center at the origin. Choice B is incorrect because the angular speed is determined by the coefficient 3 in the argument, not by R, showing confusion between radius and frequency parameters. To help students: Visualize how each parameter affects the motion - R controls size, the coefficient of t controls speed. Watch for: Confusion between different parameters' roles and misunderstanding how scaling affects parametric curves.

Question 19

A car’s position is r}(t)=\langle 2t-1,\,t^2+3\rangle with ttt in seconds; determine coordinates at t=3t=3t=3.

  1. ⟨5, 12⟩\langle 5,\,12\rangle⟨5,12⟩ (correct answer)
  2. ⟨7, 12⟩\langle 7,\,12\rangle⟨7,12⟩
  3. ⟨5, 10⟩\langle 5,\,10\rangle⟨5,10⟩
  4. ⟨−5, 12⟩\langle -5,\,12\rangle⟨−5,12⟩

Explanation: This question tests AP Precalculus skills in parametric functions, vectors, and matrices, focusing on evaluating position functions with polynomial parametric equations. Parametric functions use parameters to express coordinates, vectors represent direction and magnitude, and matrices perform transformations. In this scenario, a car's position follows r(t) = ⟨2t-1, t²+3⟩, with linear horizontal motion and quadratic vertical motion. Choice A is correct because at t=3, x(3) = 2(3)-1 = 6-1 = 5 and y(3) = 3²+3 = 9+3 = 12, giving coordinates ⟨5, 12⟩. Choice B is incorrect because it miscalculates the x-coordinate as 7, possibly computing 2(3)+1 = 7 instead of 2(3)-1 = 5, a sign error. To help students: Emphasize careful substitution in multi-term expressions, practice evaluating polynomial functions, and verify calculations by checking each component separately. Watch for: Sign errors with subtraction, order of operations mistakes, and confusion between linear and quadratic terms.

Question 20

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}.

  1. ⟨3,3⟩\langle 3,3\rangle⟨3,3⟩ (correct answer)
  2. ⟨5,−7⟩\langle 5,-7\rangle⟨5,−7⟩
  3. ⟨−3,7⟩\langle -3,7\rangle⟨−3,7⟩
  4. ⟨3,−3⟩\langle 3,-3\rangle⟨3,−3⟩

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.