Calculus 2 Quiz: Motion With Parametric And Vector Functions
20 questions · exam conditions
0:00
Motion With Parametric And Vector FunctionsQuestion 1 of 20

The velocity of a particle is given by v⃗(t)=⟨3t2,4t3⟩\vec{v}(t) = \langle 3t^2, 4t^3 \rangle. What is the rate of change of the particle's speed at t=1t=1?

66/566/5
656\sqrt{5}
6666
55
← Back to quizzes

Calculus 2 Quiz

Calculus 2 Quiz: Motion With Parametric And Vector Functions

Practice Motion With Parametric And Vector Functions in Calculus 2 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 Motion With Parametric And Vector Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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

The velocity of a particle is given by v⃗(t)=⟨3t2,4t3⟩\vec{v}(t) = \langle 3t^2, 4t^3 \rangle. What is the rate of change of the particle's speed at t=1t=1?

  1. 66/566/5 (correct answer)
  2. 656\sqrt{5}
  3. 6666
  4. 55
Explanation: The rate of change of speed is given by the formula aT=v⃗(t)⋅a⃗(t)∣∣v⃗(t)∣∣a_T = \frac{\vec{v}(t) \cdot \vec{a}(t)}{||\vec{v}(t)||}. First, find the necessary vectors at t=1t=1. v⃗(1)=⟨3(1)2,4(1)3⟩=⟨3,4⟩\vec{v}(1) = \langle 3(1)^2, 4(1)^3 \rangle = \langle 3, 4 \rangle. The acceleration is a⃗(t)=⟨6t,12t2⟩\vec{a}(t) = \langle 6t, 12t^2 \rangle, so a⃗(1)=⟨6,12⟩\vec{a}(1) = \langle 6, 12 \rangle. Now, calculate the components of the formula: ∣∣v⃗(1)∣∣=32+42=5||\vec{v}(1)|| = \sqrt{3^2+4^2} = 5, and v⃗(1)⋅a⃗(1)=(3)(6)+(4)(12)=18+48=66\vec{v}(1) \cdot \vec{a}(1) = (3)(6) + (4)(12) = 18+48=66. The rate of change of speed is 66/566/5.

Question 2

A particle has acceleration a⃗(t)=⟨6t,12t2⟩\vec{a}(t) = \langle 6t, 12t^2 \rangle. Its initial velocity at t=0t=0 is v⃗(0)=⟨1,−1⟩\vec{v}(0) = \langle 1, -1 \rangle and its initial position is r⃗(0)=⟨2,3⟩\vec{r}(0) = \langle 2, 3 \rangle. What is the particle's position at t=1t=1?

  1. ⟨4,3⟩\langle 4, 3 \rangle (correct answer)
  2. ⟨2,0⟩\langle 2, 0 \rangle
  3. ⟨3,4⟩\langle 3, 4 \rangle
  4. ⟨1,1⟩\langle 1, 1 \rangle
Explanation: First, integrate a⃗(t)\vec{a}(t) to find v⃗(t)\vec{v}(t): v⃗(t)=⟨3t2+C1,4t3+C2⟩\vec{v}(t) = \langle 3t^2+C_1, 4t^3+C_2 \rangle. Using v⃗(0)=⟨1,−1⟩\vec{v}(0) = \langle 1, -1 \rangle, we find C1=1C_1=1 and C2=−1C_2=-1. So v⃗(t)=⟨3t2+1,4t3−1⟩\vec{v}(t) = \langle 3t^2+1, 4t^3-1 \rangle. Next, integrate v⃗(t)\vec{v}(t) to find r⃗(t)\vec{r}(t): r⃗(t)=⟨t3+t+D1,t4−t+D2⟩\vec{r}(t) = \langle t^3+t+D_1, t^4-t+D_2 \rangle. Using r⃗(0)=⟨2,3⟩\vec{r}(0) = \langle 2, 3 \rangle, we find D1=2D_1=2 and D2=3D_2=3. So r⃗(t)=⟨t3+t+2,t4−t+3⟩\vec{r}(t) = \langle t^3+t+2, t^4-t+3 \rangle. Finally, evaluate at t=1t=1: r⃗(1)=⟨1+1+2,1−1+3⟩=⟨4,3⟩\vec{r}(1) = \langle 1+1+2, 1-1+3 \rangle = \langle 4, 3 \rangle.

Question 3

The acceleration of a particle is a⃗(t)=⟨0,−2⟩\vec{a}(t) = \langle 0, -2 \rangle. At time t=2t=2, its velocity is v⃗(2)=⟨3,−4⟩\vec{v}(2) = \langle 3, -4 \rangle. What is the tangential component of the particle's acceleration at t=2t=2?

  1. 6/56/5
  2. 8/58/5 (correct answer)
  3. 22
  4. 88
Explanation: When you encounter problems about tangential acceleration, you're dealing with how acceleration affects the particle's speed (not direction). The tangential component measures acceleration along the direction of motion. To find the tangential acceleration, you need the formula: aT=a⃗⋅v⃗∣v⃗∣a_T = \frac{\vec{a} \cdot \vec{v}}{|\vec{v}|}, where you take the dot product of acceleration and velocity vectors, then divide by the speed. First, find the velocity vector at t=2t = 2. Since acceleration is constant at a⃗(t)=⟨0,−2⟩\vec{a}(t) = \langle 0, -2 \rangle, integrate to get v⃗(t)=⟨C1,−2t+C2⟩\vec{v}(t) = \langle C_1, -2t + C_2 \rangle. Using the given condition v⃗(2)=⟨3,−4⟩\vec{v}(2) = \langle 3, -4 \rangle, you get C1=3C_1 = 3 and C2=0C_2 = 0, so v⃗(2)=⟨3,−4⟩\vec{v}(2) = \langle 3, -4 \rangle. Now calculate: a⃗⋅v⃗=⟨0,−2⟩⋅⟨3,−4⟩=0(3)+(−2)(−4)=8\vec{a} \cdot \vec{v} = \langle 0, -2 \rangle \cdot \langle 3, -4 \rangle = 0(3) + (-2)(-4) = 8 The speed is: ∣v⃗∣=32+(−4)2=9+16=5|\vec{v}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5 Therefore: aT=85a_T = \frac{8}{5}, which is answer B. Answer A (6/56/5) likely comes from incorrectly calculating the dot product. Answer C (22) is just the magnitude of the acceleration vector, ignoring the directional component. Answer D (88) is the numerator before dividing by the speed. Remember: tangential acceleration always requires both the dot product with velocity and division by speed—missing either step leads to common wrong answers.

Question 4

The velocity of a particle moving in the xy-plane is v⃗(t)=⟨f(t),g(t)⟩\vec{v}(t) = \langle f(t), g(t) \rangle. The particle is at the point (x0,y0)(x_0, y_0) when t=at=a. Which of the following expressions gives the x-coordinate of the particle at time t=bt=b?

  1. ∫abf(t)dt\int_a^b f(t) dt
  2. x0+∫abf(t)dtx_0 + \int_a^b f(t) dt (correct answer)
  3. x0+f(b)−f(a)x_0 + f(b) - f(a)
  4. x0+(b−a)f(b)x_0 + (b-a)f(b)
Explanation: This question tests your understanding of the relationship between velocity and position in parametric motion. When a particle moves in the xy-plane with velocity vector v⃗(t)=⟨f(t),g(t)⟩\vec{v}(t) = \langle f(t), g(t) \rangle, you need to integrate the velocity components to find position. Since velocity is the derivative of position, we have dxdt=f(t)\frac{dx}{dt} = f(t). To find the x-coordinate at time t=bt = b, you integrate the x-component of velocity: x(b)=x(a)+∫abf(t)dtx(b) = x(a) + \int_a^b f(t) dt. Since the particle is at (x0,y0)(x_0, y_0) when t=at = a, we know x(a)=x0x(a) = x_0. Therefore, x(b)=x0+∫abf(t)dtx(b) = x_0 + \int_a^b f(t) dt. Choice B is correct because it properly accounts for both the initial position x0x_0 and the displacement ∫abf(t)dt\int_a^b f(t) dt over the time interval. Choice A gives only the displacement (change in x-coordinate) from time aa to bb, but ignores the initial position x0x_0. This would only be correct if the particle started at the origin. Choice C treats f(t)f(t) as if it were the position function rather than the velocity function. This applies the fundamental theorem of calculus incorrectly. Choice D assumes constant velocity f(b)f(b) over the entire time interval, which is generally false since f(t)f(t) typically varies with time. Study tip: Remember that position equals initial position plus displacement. Always integrate velocity to get displacement, then add the initial position to get the final position.

Question 5

Find the minimum speed of a particle whose position is given by r⃗(t)=⟨t2,t3−3t⟩\vec{r}(t) = \langle t^2, t^3-3t \rangle.

  1. 22
  2. 423\frac{4\sqrt{2}}{3} (correct answer)
  3. 33
  4. 73\frac{\sqrt{7}}{3}
Explanation: When you encounter a position vector and need to find minimum speed, remember that speed is the magnitude of velocity. You'll need to find the velocity vector, calculate its magnitude (speed), then minimize that function. Start by finding the velocity vector by differentiating the position: v⃗(t)=r⃗′(t)=⟨2t,3t2−3⟩\vec{v}(t) = \vec{r}'(t) = \langle 2t, 3t^2-3 \rangle Speed is the magnitude of velocity: ∣v⃗(t)∣=(2t)2+(3t2−3)2=4t2+9t4−18t2+9=9t4−14t2+9|\vec{v}(t)| = \sqrt{(2t)^2 + (3t^2-3)^2} = \sqrt{4t^2 + 9t^4 - 18t^2 + 9} = \sqrt{9t^4 - 14t^2 + 9} To minimize speed, you can minimize the expression under the square root. Let f(t)=9t4−14t2+9f(t) = 9t^4 - 14t^2 + 9. Taking the derivative: f′(t)=36t3−28t=4t(9t2−7)f'(t) = 36t^3 - 28t = 4t(9t^2 - 7) Setting f′(t)=0f'(t) = 0: either t=0t = 0 or t2=79t^2 = \frac{7}{9}, giving t=±73t = \pm\frac{\sqrt{7}}{3} Evaluating f(t)f(t) at these critical points:
  • At t=0t = 0: f(0)=9f(0) = 9, so speed = 33
  • At t=±73t = \pm\frac{\sqrt{7}}{3}: f=9⋅4981−14⋅79+9=329f = 9 \cdot \frac{49}{81} - 14 \cdot \frac{7}{9} + 9 = \frac{32}{9}, so speed = 423\frac{4\sqrt{2}}{3}
Since 423<3\frac{4\sqrt{2}}{3} < 3, the minimum speed is 423\frac{4\sqrt{2}}{3}, which is choice (B). Choice (A) gives 22, which is too small. Choice (C) gives 33, which is the speed at t=0t=0 but not the minimum. Choice (D) gives 73\frac{\sqrt{7}}{3}, which might result from confusing the critical point value with the minimum speed. Strategy tip: Always check critical points systematically and compare function values—don't assume the first critical point you find gives the minimum.

Question 6

The position of a particle is given by a vector function r⃗(t)\vec{r}(t) such that the particle is always moving. If the speed of the particle is constant, which of the following must be true about its velocity vector v⃗(t)\vec{v}(t) and acceleration vector a⃗(t)\vec{a}(t) for all tt?

  1. The magnitude of the acceleration vector is constant.
  2. The acceleration vector is always the zero vector.
  3. The velocity and acceleration vectors are always parallel.
  4. The velocity and acceleration vectors are always orthogonal. (correct answer)
Explanation: When analyzing motion with constant speed, you need to understand the relationship between velocity and acceleration vectors. The key insight is that constant speed means the magnitude of velocity is unchanging, which places a specific constraint on how acceleration can act. Since speed is constant, we have ∣v⃗(t)∣=c|\vec{v}(t)| = c for some constant c>0c > 0. Taking the derivative of both sides: ddt∣v⃗(t)∣2=ddt(v⃗⋅v⃗)=2v⃗⋅a⃗=0\frac{d}{dt}|\vec{v}(t)|^2 = \frac{d}{dt}(\vec{v} \cdot \vec{v}) = 2\vec{v} \cdot \vec{a} = 0. This means v⃗⋅a⃗=0\vec{v} \cdot \vec{a} = 0, so the velocity and acceleration vectors are always orthogonal (perpendicular). The correct answer is D. Let's examine why the other options fail. Option A claims acceleration magnitude is constant, but this isn't necessarily true—while acceleration must be perpendicular to velocity, its magnitude can vary. Think of a particle moving on an ellipse at constant speed: the acceleration changes magnitude as curvature changes. Option B suggests acceleration is always zero, but this would mean the particle moves in a straight line. Since the problem states the particle is "always moving" (implying non-trivial motion), acceleration typically exists to change direction while maintaining constant speed. Option C claims velocity and acceleration are parallel, but this would change the speed—parallel acceleration components either increase or decrease speed. Remember this key principle: constant speed motion requires acceleration to be purely perpendicular to velocity. Any parallel component would change the speed, violating the constraint. This appears frequently in circular motion and curve-following problems.

Question 7

A particle's velocity is given by the vector function v⃗(t)=⟨3cos⁡(t),−2sin⁡(2t)⟩\vec{v}(t) = \langle 3\cos(t), -2\sin(2t) \rangle. If the particle is at the point (1,2)(1, 2) at time t=0t=0, what is its position at t=π/2t=\pi/2?

  1. (4,0)(4, 0) (correct answer)
  2. (3,−2)(3, -2)
  3. (4,4)(4, 4)
  4. (1,1)(1, 1)
Explanation: To find the position vector r⃗(t)\vec{r}(t), we integrate the velocity vector v⃗(t)\vec{v}(t) and use the initial condition to find the constants of integration. x(t)=∫3cos⁡(t)dt=3sin⁡(t)+C1x(t) = \int 3\cos(t) dt = 3\sin(t) + C_1. Since x(0)=1x(0)=1, 3sin⁡(0)+C1=1⇒C1=13\sin(0) + C_1 = 1 \Rightarrow C_1 = 1. So, x(t)=3sin⁡(t)+1x(t) = 3\sin(t) + 1. y(t)=∫−2sin⁡(2t)dt=cos⁡(2t)+C2y(t) = \int -2\sin(2t) dt = \cos(2t) + C_2. Since y(0)=2y(0)=2, cos⁡(0)+C2=2⇒C2=1\cos(0) + C_2 = 2 \Rightarrow C_2 = 1. So, y(t)=cos⁡(2t)+1y(t) = \cos(2t) + 1. At t=π/2t=\pi/2, the position is r⃗(π/2)=⟨3sin⁡(π/2)+1,cos⁡(2(π/2))+1⟩=⟨3(1)+1,cos⁡(π)+1⟩=⟨4,−1+1⟩=⟨4,0⟩\vec{r}(\pi/2) = \langle 3\sin(\pi/2) + 1, \cos(2(\pi/2)) + 1 \rangle = \langle 3(1)+1, \cos(\pi)+1 \rangle = \langle 4, -1+1 \rangle = \langle 4, 0 \rangle.

Question 8

A projectile is fired from the origin with an initial velocity of v⃗(0)=⟨20,40⟩\vec{v}(0) = \langle 20, 40 \rangle m/s. Assuming the only force is gravity, so that acceleration is a⃗(t)=⟨0,−10⟩\vec{a}(t) = \langle 0, -10 \rangle m/s2^2, what is the speed of the projectile at its maximum height?

  1. 8080 m/s
  2. 20520\sqrt{5} m/s
  3. 00 m/s
  4. 2020 m/s (correct answer)
Explanation: When analyzing projectile motion, you need to understand how velocity components behave independently. The horizontal component remains constant (no horizontal forces), while the vertical component changes due to gravity. Starting with the initial velocity v⃗(0)=⟨20,40⟩\vec{v}(0) = \langle 20, 40 \rangle and constant acceleration a⃗(t)=⟨0,−10⟩\vec{a}(t) = \langle 0, -10 \rangle, you can find the velocity at any time using v⃗(t)=v⃗(0)+a⃗t=⟨20,40−10t⟩\vec{v}(t) = \vec{v}(0) + \vec{a}t = \langle 20, 40 - 10t \rangle. At maximum height, the vertical velocity component becomes zero (the projectile momentarily stops rising before falling). Setting the vertical component to zero: 40−10t=040 - 10t = 0, so t=4t = 4 seconds. At this moment, v⃗(4)=⟨20,0⟩\vec{v}(4) = \langle 20, 0 \rangle. The speed is the magnitude: ∣v⃗∣=202+02=20|\vec{v}| = \sqrt{20^2 + 0^2} = 20 m/s. Looking at the wrong answers: Choice A (80 m/s) likely comes from adding the initial velocity components (20 + 40 + 20 = 80), which has no physical meaning. Choice B (20520\sqrt{5} m/s) equals the initial speed 202+402=205\sqrt{20^2 + 40^2} = 20\sqrt{5}, but this ignores that the vertical component changes. Choice C (0 m/s) assumes the entire velocity becomes zero at maximum height, confusing the vertical component stopping with total motion stopping. The correct answer is D (20 m/s). Key insight: At maximum height in projectile motion, only the vertical velocity component is zero—the horizontal component continues unchanged. Always analyze velocity components separately, then find the magnitude for speed.

Question 9

The position of a particle is given by r⃗(t)=⟨2cos⁡(t),ksin⁡(t)⟩\vec{r}(t) = \langle 2\cos(t), k\sin(t) \rangle for some constant kk. If the particle's speed at t=π/4t=\pi/4 is 10\sqrt{10}, what is a possible value of kk?

  1. 222\sqrt{2}
  2. 10\sqrt{10}
  3. 44 (correct answer)
  4. 1616
Explanation: When you encounter a particle motion problem involving position vectors, you need to connect position to velocity to speed. The key insight is that speed is the magnitude of the velocity vector, and velocity is the derivative of position. First, find the velocity by differentiating the position vector: v⃗(t)=r⃗′(t)=⟨−2sin⁡(t),kcos⁡(t)⟩\vec{v}(t) = \vec{r}'(t) = \langle -2\sin(t), k\cos(t) \rangle Next, calculate the speed, which is the magnitude of velocity: ∣v⃗(t)∣=(−2sin⁡(t))2+(kcos⁡(t))2=4sin⁡2(t)+k2cos⁡2(t)|\vec{v}(t)| = \sqrt{(-2\sin(t))^2 + (k\cos(t))^2} = \sqrt{4\sin^2(t) + k^2\cos^2(t)} At t=π/4t = \pi/4, we have sin⁡(π/4)=cos⁡(π/4)=22\sin(\pi/4) = \cos(\pi/4) = \frac{\sqrt{2}}{2}. Substituting these values: ∣v⃗(π/4)∣=4⋅12+k2⋅12=2+k22|\vec{v}(\pi/4)| = \sqrt{4 \cdot \frac{1}{2} + k^2 \cdot \frac{1}{2}} = \sqrt{2 + \frac{k^2}{2}} Since the speed equals 10\sqrt{10}: 2+k22=10\sqrt{2 + \frac{k^2}{2}} = \sqrt{10} Squaring both sides: 2+k22=102 + \frac{k^2}{2} = 10, so k22=8\frac{k^2}{2} = 8, giving us k2=16k^2 = 16, therefore k=±4k = \pm 4. Choice C) 44 is correct. Choice A) 222\sqrt{2} would give k2=8k^2 = 8, making the speed 6\sqrt{6}. Choice B) 10\sqrt{10} would give k2=10k^2 = 10, making the speed 7\sqrt{7}. Choice D) 1616 would give k2=256k^2 = 256, making the speed 130\sqrt{130}. Remember: in particle motion problems, always work systematically from position → velocity → speed, and don't forget that speed is always non-negative while components of velocity can be negative.

Question 10

A particle moves with position vector r(t)=⟨ln⁡(t+1),t+4,arctan⁡(t)⟩\mathbf{r}(t) = \langle \ln(t+1), \sqrt{t+4}, \arctan(t) \rangle for t≥0t \geq 0. The particle's acceleration vector is perpendicular to its velocity vector at time t=t0t = t_0. Which of the following must be true about t0t_0?

  1. t0t_0 is the time when the particle's speed is minimized along its entire trajectory
  2. t0t_0 is the time when the particle's speed is maximized along its entire trajectory
  3. t0t_0 is a critical point of the speed function, but may be a local minimum, maximum, or neither (correct answer)
  4. t0t_0 has no special relationship to the particle's speed function at that time
Explanation: When acceleration is perpendicular to velocity, v(t)⋅a(t)=0\mathbf{v}(t) \cdot \mathbf{a}(t) = 0. The speed is s(t)=∣v(t)∣s(t) = |\mathbf{v}(t)|, so s′(t)=v(t)⋅a(t)∣v(t)∣s'(t) = \frac{\mathbf{v}(t) \cdot \mathbf{a}(t)}{|\mathbf{v}(t)|}. When v(t)⋅a(t)=0\mathbf{v}(t) \cdot \mathbf{a}(t) = 0, we have s′(t)=0s'(t) = 0, making t0t_0 a critical point of the speed function. However, this critical point could be a local minimum, local maximum, or neither (inflection point), depending on the second derivative test and the specific nature of the motion. We cannot determine without further analysis whether it's a global minimum or maximum. Choice A incorrectly assumes all such points are global minima. Choice B incorrectly assumes all such points are global maxima. Choice D ignores the fundamental relationship between perpendicular acceleration-velocity and critical points of speed. The correct answer recognizes that while t0t_0 must be a critical point, its nature requires additional investigation.

Question 11

A particle moves along a curve with parametric equations x(t)=t3−6tx(t) = t^3 - 6t and y(t)=t2+2ty(t) = t^2 + 2t. The particle changes direction (velocity becomes zero) at certain values of tt. At how many distinct points does the particle change direction?

  1. The particle never changes direction during its motion (correct answer)
  2. The particle changes direction at exactly one distinct point
  3. The particle changes direction at exactly two distinct points
  4. The particle changes direction at exactly three distinct points
Explanation: For the particle to change direction, the velocity vector must be zero, meaning both x′(t)=0x'(t) = 0 and y′(t)=0y'(t) = 0 simultaneously. We have x′(t)=3t2−6x'(t) = 3t^2 - 6 and y′(t)=2t+2y'(t) = 2t + 2. Setting y′(t)=0y'(t) = 0: 2t+2=02t + 2 = 0, so t=−1t = -1. Setting x′(t)=0x'(t) = 0: 3t2−6=03t^2 - 6 = 0, so t2=2t^2 = 2, giving t=±2t = \pm\sqrt{2}. For both components to be zero simultaneously, we need a common solution, but t=−1t = -1 doesn't satisfy x′(t)=0x'(t) = 0 since x′(−1)=3(−1)2−6=3−6=−3≠0x'(-1) = 3(-1)^2 - 6 = 3 - 6 = -3 \neq 0. Similarly, t=±2t = \pm\sqrt{2} don't satisfy y′(t)=0y'(t) = 0 since y′(2)=22+2≠0y'(\sqrt{2}) = 2\sqrt{2} + 2 \neq 0. Therefore, the velocity vector is never zero, and the particle never changes direction. Choice A is correct. Choice B assumes students find where one component is zero. Choice C assumes students think each zero of each component corresponds to a direction change. Choice D assumes students count all zeros of both components separately.

Question 12

A particle moves in the xy-plane with velocity v⃗(t)=⟨6t,3t2−3⟩\vec{v}(t) = \langle 6t, 3t^2 - 3 \rangle for t≥0t \ge 0. What is the total distance traveled by the particle from t=0t=0 to t=2t=2?

  1. 1414 (correct answer)
  2. 2372\sqrt{37}
  3. 14/314/3
  4. 1515
Explanation: The total distance traveled is the integral of the speed. The speed is the magnitude of the velocity vector: ∣∣v⃗(t)∣∣=(6t)2+(3t2−3)2=36t2+9t4−18t2+9=9t4+18t2+9=9(t2+1)2=3(t2+1)||\vec{v}(t)|| = \sqrt{(6t)^2 + (3t^2-3)^2} = \sqrt{36t^2 + 9t^4 - 18t^2 + 9} = \sqrt{9t^4 + 18t^2 + 9} = \sqrt{9(t^2+1)^2} = 3(t^2+1). The total distance is ∫023(t2+1)dt=3[t33+t]02=3((83+2)−0)=3(143)=14\int_0^2 3(t^2+1) dt = 3[\frac{t^3}{3} + t]_0^2 = 3((\frac{8}{3} + 2) - 0) = 3(\frac{14}{3}) = 14. Distractor B is the magnitude of the displacement vector. Distractor C results from forgetting the factor of 3 when simplifying the speed.

Question 13

A particle's position is given by r⃗(t)=⟨t3−12t,t2−2t⟩\vec{r}(t) = \langle t^3 - 12t, t^2 - 2t \rangle. What is the speed of the particle at the first time t>0t>0 when the tangent to its path is vertical?

  1. 22 (correct answer)
  2. 99
  3. 1616
  4. 00
Explanation: A vertical tangent occurs when dx/dt=0dx/dt = 0 and dy/dt≠0dy/dt \neq 0. The velocity vector is v⃗(t)=⟨3t2−12,2t−2⟩\vec{v}(t) = \langle 3t^2 - 12, 2t - 2 \rangle. Set dx/dt=3t2−12=0dx/dt = 3t^2 - 12 = 0, which gives t2=4t^2=4. Since t>0t>0, we have t=2t=2. At t=2t=2, dy/dt=2(2)−2=2≠0dy/dt = 2(2)-2 = 2 \neq 0, so there is a vertical tangent. The speed at t=2t=2 is ∣∣v⃗(2)∣∣=(3(22)−12)2+(2(2)−2)2=02+22=2||\vec{v}(2)|| = \sqrt{(3(2^2)-12)^2 + (2(2)-2)^2} = \sqrt{0^2 + 2^2} = 2. Distractor B is the speed at the time of a horizontal tangent.

Question 14

A particle's motion is described by x(t)=t2+1x(t) = t^2+1 and y(t)=ln⁡(2t+1)y(t) = \ln(2t+1) for t≥0t \ge 0. What is the y-coordinate of the particle at the instant its path has a slope of 1/61/6?

  1. ln⁡(4)\ln(4) (correct answer)
  2. 13/413/4
  3. 3/23/2
  4. ln⁡(5)\ln(5)
Explanation: The slope of the path is dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}. We have dx/dt=2tdx/dt = 2t and dy/dt=22t+1dy/dt = \frac{2}{2t+1}. The slope is 2/(2t+1)2t=1t(2t+1)\frac{2/(2t+1)}{2t} = \frac{1}{t(2t+1)}. We set the slope equal to 1/61/6: 1t(2t+1)=16\frac{1}{t(2t+1)} = \frac{1}{6}, which implies t(2t+1)=6t(2t+1) = 6, or 2t2+t−6=02t^2+t-6=0. Factoring gives (2t−3)(t+2)=0(2t-3)(t+2)=0. Since t≥0t \ge 0, we must have t=3/2t=3/2. The y-coordinate at this time is y(3/2)=ln⁡(2(3/2)+1)=ln⁡(4)y(3/2) = \ln(2(3/2)+1) = \ln(4).

Question 15

The position of a particle at time t≥0t \ge 0 is r⃗(t)=⟨4t+1,3t−2⟩\vec{r}(t) = \langle 4t+1, 3t-2 \rangle. At what time tt is the particle closest to the origin?

  1. 5/45/4
  2. 2/32/3
  3. 00
  4. 2/252/25 (correct answer)
Explanation: When you encounter a problem about finding when a particle is closest to a point, you're looking for a minimum distance problem that requires optimization techniques from calculus. To find when the particle is closest to the origin, you need to minimize the distance function. The distance from the particle to the origin is d(t)=(4t+1)2+(3t−2)2d(t) = \sqrt{(4t+1)^2 + (3t-2)^2}. Since the square root function is always increasing, minimizing d(t)d(t) is equivalent to minimizing d2(t)=(4t+1)2+(3t−2)2d^2(t) = (4t+1)^2 + (3t-2)^2. Expanding this: d2(t)=16t2+8t+1+9t2−12t+4=25t2−4t+5d^2(t) = 16t^2 + 8t + 1 + 9t^2 - 12t + 4 = 25t^2 - 4t + 5 To find the minimum, take the derivative and set it equal to zero: ddt[d2(t)]=50t−4=0\frac{d}{dt}[d^2(t)] = 50t - 4 = 0 Solving: t=450=225t = \frac{4}{50} = \frac{2}{25}, which is answer D. Now for the wrong answers: Answer A (54\frac{5}{4}) might come from incorrectly setting up the distance formula or making algebraic errors. Answer B (23\frac{2}{3}) could result from mixing up coefficients during differentiation. Answer C (00) is a common trap - students might think the particle starts closest to the origin, but the initial position ⟨1,−2⟩\langle 1, -2 \rangle isn't actually the closest point along the path. Strategy tip: Always remember that "closest to a point" means minimizing distance, which requires calculus optimization. Don't assume the minimum occurs at the boundary points (like t=0t = 0) without checking the critical points first.

Question 16

The position of a particle is given by r⃗(t)=⟨sin⁡(πt),t2⟩\vec{r}(t) = \langle \sin(\pi t), t^2 \rangle. What is the average velocity of the particle over the time interval [0,2][0, 2]?

  1. ⟨0,4⟩\langle 0, 4 \rangle
  2. ⟨−π,2⟩\langle -\pi, 2 \rangle
  3. ⟨π,2⟩\langle \pi, 2 \rangle
  4. ⟨0,2⟩\langle 0, 2 \rangle (correct answer)
Explanation: When you encounter a vector-valued function representing position, remember that average velocity is found using the displacement over the time interval, not by averaging the velocity function itself. Average velocity is defined as v⃗avg=r⃗(t2)−r⃗(t1)t2−t1\vec{v}_{avg} = \frac{\vec{r}(t_2) - \vec{r}(t_1)}{t_2 - t_1}, where you're finding the total displacement divided by the elapsed time. First, evaluate the position function at the endpoints of the interval [0, 2]:
  • At t=0t = 0: r⃗(0)=⟨sin⁡(0),02⟩=⟨0,0⟩\vec{r}(0) = \langle \sin(0), 0^2 \rangle = \langle 0, 0 \rangle
  • At t=2t = 2: r⃗(2)=⟨sin⁡(2π),22⟩=⟨0,4⟩\vec{r}(2) = \langle \sin(2\pi), 2^2 \rangle = \langle 0, 4 \rangle
Now calculate the average velocity: v⃗avg=⟨0,4⟩−⟨0,0⟩2−0=⟨0,4⟩2=⟨0,2⟩\vec{v}_{avg} = \frac{\langle 0, 4 \rangle - \langle 0, 0 \rangle}{2 - 0} = \frac{\langle 0, 4 \rangle}{2} = \langle 0, 2 \rangle Choice A) ⟨0,4⟩\langle 0, 4 \rangle represents the displacement vector itself, but forgets to divide by the time interval. Choice B) ⟨−π,2⟩\langle -\pi, 2 \rangle incorrectly uses the derivative of the position function at some point rather than the displacement formula. Choice C) ⟨π,2⟩\langle \pi, 2 \rangle makes a similar error with the derivative but gets the wrong sign for the x-component. The key insight is that average velocity depends only on starting and ending positions, not on the path taken between them. Even though the particle oscillates in the x-direction during this interval, it starts and ends at the same x-coordinate, giving zero average velocity in that direction.

Question 17

A particle's velocity is v⃗(t)=⟨1t,t⟩\vec{v}(t) = \langle \frac{1}{t}, \sqrt{t} \rangle for t>0t > 0. Which statement accurately describes the particle's motion at t=4t=4?

  1. It is moving down and to the right, and its speed is increasing.
  2. It is moving up and to the right, and its speed is decreasing.
  3. It is moving up and to the right, and its speed is increasing. (correct answer)
  4. It is moving up and to the left, and its speed is decreasing.
Explanation: When analyzing particle motion from a velocity vector, you need to determine two things: direction (from the velocity components' signs) and whether speed is increasing or decreasing (by comparing velocity and acceleration vectors). At t=4t = 4, the velocity is v⃗(4)=⟨14,4⟩=⟨0.25,2⟩\vec{v}(4) = \langle \frac{1}{4}, \sqrt{4} \rangle = \langle 0.25, 2 \rangle. Since both components are positive, the particle moves right (positive x-direction) and up (positive y-direction). To determine if speed is increasing or decreasing, you need the acceleration vector. Taking the derivative: a⃗(t)=v⃗′(t)=⟨−1t2,12t⟩\vec{a}(t) = \vec{v}'(t) = \langle -\frac{1}{t^2}, \frac{1}{2\sqrt{t}} \rangle. At t=4t = 4: a⃗(4)=⟨−116,14⟩\vec{a}(4) = \langle -\frac{1}{16}, \frac{1}{4} \rangle. Speed increases when velocity and acceleration point in the "same general direction" (positive dot product) and decreases when they point in "opposite general directions" (negative dot product). Computing: v⃗(4)⋅a⃗(4)=(0.25)(−116)+(2)(14)=−164+12=3164>0\vec{v}(4) \cdot \vec{a}(4) = (0.25)(-\frac{1}{16}) + (2)(\frac{1}{4}) = -\frac{1}{64} + \frac{1}{2} = \frac{31}{64} > 0. Since the dot product is positive, speed is increasing. Answer A is wrong because the particle moves up, not down. Answer B correctly identifies the direction but incorrectly states speed is decreasing. Answer D is wrong because the particle moves right, not left, and speed isn't decreasing. Remember: velocity components tell you direction, but you need the dot product of velocity and acceleration to determine if speed is changing. Positive dot product means increasing speed, negative means decreasing speed.

Question 18

A particle moves along the parabola y=x3y=x^3. At time t>0t>0, its x-coordinate is given by x(t)=tx(t) = \sqrt{t}. What is the speed of the particle as a function of tt?

  1. 1+9t2\sqrt{1+9t^2}
  2. 1+9t22t\frac{\sqrt{1+9t^2}}{2\sqrt{t}} (correct answer)
  3. t+9t32t\frac{\sqrt{t+9t^3}}{2t}
  4. 14t+94t2\sqrt{\frac{1}{4t} + \frac{9}{4}t^2}
Explanation: When a particle moves along a curve in parametric form, you need to find both dxdt\frac{dx}{dt} and dydt\frac{dy}{dt} to calculate speed. The speed formula is speed=(dxdt)2+(dydt)2\text{speed} = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}. Given x(t)=tx(t) = \sqrt{t}, we have dxdt=12t\frac{dx}{dt} = \frac{1}{2\sqrt{t}}. Since the particle moves along y=x3y = x^3, we get y(t)=(t)3=t3/2y(t) = (\sqrt{t})^3 = t^{3/2}. Therefore, dydt=32t1/2=3t2\frac{dy}{dt} = \frac{3}{2}t^{1/2} = \frac{3\sqrt{t}}{2}. The speed is: (12t)2+(3t2)2=14t+9t4=1+9t24t=1+9t22t\sqrt{\left(\frac{1}{2\sqrt{t}}\right)^2 + \left(\frac{3\sqrt{t}}{2}\right)^2} = \sqrt{\frac{1}{4t} + \frac{9t}{4}} = \sqrt{\frac{1 + 9t^2}{4t}} = \frac{\sqrt{1 + 9t^2}}{2\sqrt{t}} This confirms answer B is correct. Answer A (1+9t2\sqrt{1+9t^2}) represents just the numerator without the denominator factor, suggesting someone forgot to account for the 12t\frac{1}{2\sqrt{t}} term properly. Answer C (t+9t32t\frac{\sqrt{t+9t^3}}{2t}) comes from incorrectly factoring inside the square root, writing 14t+9t4\sqrt{\frac{1}{4t} + \frac{9t}{4}} as t+9t34t2\sqrt{\frac{t + 9t^3}{4t^2}} instead of 1+9t24t\sqrt{\frac{1 + 9t^2}{4t}}. Answer D (14t+94t2\sqrt{\frac{1}{4t} + \frac{9}{4}t^2}) shows the expression before taking the final square root step, missing that 9t4=94t\frac{9t}{4} = \frac{9}{4}t, not 94t2\frac{9}{4}t^2. Remember: for parametric motion problems, always find both component derivatives first, then apply the speed formula carefully, checking your algebra at each step.

Question 19

The position of a particle moving in the plane is given by r⃗(t)=⟨t2,t3−4t⟩\vec{r}(t) = \langle t^2, t^3 - 4t \rangle. For which interval of t>0t>0 is the speed of the particle increasing?

  1. (103,∞)(\frac{\sqrt{10}}{3}, \infty) (correct answer)
  2. (0,103)(0, \frac{\sqrt{10}}{3})
  3. (233,∞)(\frac{2\sqrt{3}}{3}, \infty)
  4. (0,∞)(0, \infty)
Explanation: The speed is increasing when the dot product of the velocity and acceleration vectors is positive. v⃗(t)=⟨2t,3t2−4⟩\vec{v}(t) = \langle 2t, 3t^2-4 \rangle and a⃗(t)=⟨2,6t⟩\vec{a}(t) = \langle 2, 6t \rangle. Their dot product is v⃗(t)⋅a⃗(t)=(2t)(2)+(3t2−4)(6t)=4t+18t3−24t=18t3−20t\vec{v}(t) \cdot \vec{a}(t) = (2t)(2) + (3t^2-4)(6t) = 4t + 18t^3 - 24t = 18t^3 - 20t. We need to find where 18t3−20t>018t^3 - 20t > 0. Factoring gives 2t(9t2−10)>02t(9t^2 - 10) > 0. Since t>0t>0, we must have 9t2−10>09t^2 - 10 > 0, which means t2>10/9t^2 > 10/9, or t>10/3t > \sqrt{10}/3.

Question 20

A particle moves according to r⃗(t)=⟨13t3+t,t2⟩\vec{r}(t) = \langle \frac{1}{3}t^3+t, t^2 \rangle. For what time t>0t > 0 are the particle's velocity and acceleration vectors parallel but not zero?

  1. Never
  2. t=3t=\sqrt{3}
  3. t=2t=2
  4. t=1t=1 (correct answer)
Explanation: When you encounter vector motion problems asking about parallel vectors, you need to find when one vector is a scalar multiple of another. For velocity and acceleration to be parallel, v⃗(t)=ka⃗(t)\vec{v}(t) = k\vec{a}(t) for some scalar k≠0k \neq 0. Start by finding the velocity and acceleration vectors through differentiation. Given r⃗(t)=⟨13t3+t,t2⟩\vec{r}(t) = \langle \frac{1}{3}t^3+t, t^2 \rangle, the velocity is v⃗(t)=r⃗′(t)=⟨t2+1,2t⟩\vec{v}(t) = \vec{r}'(t) = \langle t^2+1, 2t \rangle and the acceleration is a⃗(t)=v⃗′(t)=⟨2t,2⟩\vec{a}(t) = \vec{v}'(t) = \langle 2t, 2 \rangle. For these vectors to be parallel, their components must be proportional: t2+12t=2t2=t\frac{t^2+1}{2t} = \frac{2t}{2} = t. This gives us the equation t2+1=2t2t^2+1 = 2t^2, which simplifies to t2=1t^2 = 1. Since we need t>0t > 0, we get t=1t = 1. Let's verify: At t=1t = 1, v⃗(1)=⟨2,2⟩\vec{v}(1) = \langle 2, 2 \rangle and a⃗(1)=⟨2,2⟩\vec{a}(1) = \langle 2, 2 \rangle. These are indeed parallel (and equal), confirming answer D. Answer A is wrong because we found a valid solution. Answer B (t=3t = \sqrt{3}) fails the proportionality test: v⃗(3)=⟨4,23⟩\vec{v}(\sqrt{3}) = \langle 4, 2\sqrt{3} \rangle and a⃗(3)=⟨23,2⟩\vec{a}(\sqrt{3}) = \langle 2\sqrt{3}, 2 \rangle aren't proportional. Answer C (t=2t = 2) similarly fails: v⃗(2)=⟨5,4⟩\vec{v}(2) = \langle 5, 4 \rangle and a⃗(2)=⟨4,2⟩\vec{a}(2) = \langle 4, 2 \rangle have non-equal ratios 54≠42\frac{5}{4} \neq \frac{4}{2}. Remember: parallel vectors have proportional components, so set up the proportion equation and solve systematically.