Motion Problems, Parametric, Vector-Valued Functions - AP Calculus BC

1. A particle moves with $x(t)=\ln(t+1)$ and $y(t)=t$; what is its speed at $t=0$?
   Answer: $\sqrt{2}$ To find speed, we need the velocity vector components. Differentiating gives vₓ(t) = d/dt[ln(t+1)] = 1/(t+1) and vᵧ(t) = 1. At t = 0, we get vₓ(0) = 1/(0+1) = 1 and vᵧ(0) = 1, so v(0) = ⟨1, 1⟩. The speed is |v(0)| = √(1² + 1²) = √2. Students might report just one component value of 1 (choice A) or the velocity vector itself (choice C). Always remember to find the magnitude of the velocity vector to get speed.
2. A particle follows $r(t)=\langle t-2,\,(t-2)^2\rangle$; what is its speed at $t=2$?
   Answer: $1$ This problem involves analyzing the motion of a particle using parametric equations. To find the speed, differentiate: dx/dt = 1 and dy/dt = 2(t-2). At t=2, velocity is <1,0>, and speed is √(1+0)=1. This indicates motion only in the x-direction at that point. A tempting distractor might be √2, perhaps from evaluating at t=1 or t=3 where the y-component is nonzero. Always remember to integrate velocity to find position or differentiate position to find velocity and speed when analyzing parametric motion.
3. A particle moves as $x=t^2+1$ and $y=4-t$; what is its velocity at $t=3$?
   Answer: $\langle 6,-1\rangle$ This problem involves analyzing the motion of a particle using parametric equations. To find the velocity, differentiate the given x and y: dx/dt = 2t and dy/dt = -1. At t=3, this yields <6, -1>. The velocity vector shows both components of the motion at that instant. A tempting distractor might be √37, which is the speed instead of the velocity vector. Always remember to integrate velocity to find position or differentiate position to find velocity and speed when analyzing parametric motion.
4. A particle moves with $x(t)=2t-1$ and $y(t)=t^3$; what is its velocity at $t=0$?
   Answer: $\langle 2,0\rangle$ This problem requires finding the velocity vector from parametric equations. The velocity components are found by differentiating: vₓ(t) = dx/dt = 2 and vᵧ(t) = dy/dt = 3t². At t = 0, we get vₓ(0) = 2 and vᵧ(0) = 3(0)² = 0, so v(0) = ⟨2, 0⟩. The particle is moving horizontally to the right at this instant. A student might mistakenly evaluate the position at t = 0 to get ⟨-1, 0⟩ (choice A). Remember that velocity requires differentiation of position, not just evaluation.
5. A car's position is $r(t)=\langle t^3,\,2t^2\rangle$; what is its velocity at $t=1$?
   Answer: $\langle 3,4\rangle$ This problem involves analyzing the motion of a car using parametric equations. To find the velocity, differentiate: $ dx/dt = $3t^2$ $ and $ dy/dt = 4t $. At t=1, this is $\langle 3,4 \rangle$. The components show the rates in x and y directions. A tempting distractor might be 5, which is the speed $\sqrt{9+16} = 5$, confusing vector with scalar. Always remember to integrate velocity to find position or differentiate position to find velocity and speed when analyzing parametric motion.
6. A skater moves with $r(t)=\langle 3t-1,\;t^2+2\rangle$. What is the speed at $t=0$?
   Answer: $\left|\u007fr'(0)\right|$ This problem requires finding the speed of a skater at t=0 from parametric equations. Given r(t) = ⟨3t-1, t²+2⟩, we first find the velocity vector: r'(t) = ⟨3, 2t⟩. At t=0, the velocity is r'(0) = ⟨3, 0⟩. Speed is the magnitude of velocity: |r'(0)| = √(3² + 0²) = √9 = 3. Choice B (r'(0) = ⟨3, 0⟩) gives the velocity vector itself, not its magnitude—remember that speed is a scalar quantity. For parametric motion, always compute speed as |r'(t)|, the magnitude of the velocity vector.
7. For $r(t)=\langle t,\,t^2\rangle$, what is the displacement vector from $t=1$ to $t=3$?
   Answer: $\langle 2,\,8\rangle$ This problem involves analyzing the motion of a particle using parametric equations. To find the displacement vector from t = 1 to t = 3, evaluate the position at both times: r(3) = <3, 9> and r(1) = <1, 1>. Subtract the initial position from the final position to get <3 - 1, 9 - 1> = <2, 8>. This vector represents the net change in position over the interval. A tempting distractor is E, √68, which is the magnitude of <2, 8> or the distance traveled, but the question asks for the displacement vector, not its length. A transferable motion-analysis strategy is to compute displacement as the difference in position vectors for any time interval in parametric motion.
8. A particle has $r(t)=\langle 1+2t,\,5-t\rangle$; what is its speed for any $t$?
   Answer: $\sqrt{5}$ This problem asks for the speed of a particle with linear parametric equations at any time t. The velocity vector is found by differentiating: $v(t) = r'(t) = \langle \frac{d}{dt}(1+2t), \frac{d}{dt}(5-t) \rangle = \langle 2, -1 \rangle$. Since the velocity is constant, the speed is also constant: $|v(t)| = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}$. Choice A gives the constant velocity vector, not the speed, while choice E attempts a general formula but uses position instead of velocity. For linear parametric equations, velocity is constant, making speed calculation straightforward.
9. A particle has position $r(t)=\langle 2\cos t,\,2\sin t\rangle$; which gives its velocity vector at $t=\pi/3$?
   Answer: $\langle -2\sin(\pi/3),\,2\cos(\pi/3)\rangle$ This problem involves analyzing the motion of a particle using parametric equations. To find the velocity vector, differentiate each component of r(t) = <2 cos t, 2 sin t> with respect to t, resulting in v(t) = <-2 sin t, 2 cos t>. Evaluate at t = π/3, where sin(π/3) = √3/2 and cos(π/3) = 1/2, giving v(π/3) = <-2(√3/2), 2(1/2)> = <-√3, 1>, which matches <-2 sin(π/3), 2 cos(π/3)>. Note that the expression is left in terms of sine and cosine as presented in the choices. A tempting distractor is B, <2 cos(π/3), 2 sin(π/3)>, which is actually the position vector at t = π/3, confusing position with velocity. A transferable motion-analysis strategy is to remember that velocity is the derivative of position, so always apply differentiation to parametric components before evaluation.
10. A particle's position is $r(t)=\langle e^{t},\,e^{-t}\rangle$; what is its speed at $t=0$?
    Answer: $\sqrt{2}$ This problem involves analyzing the motion of a particle using parametric equations. To find the speed at t = 0, differentiate r(t) = $\langle e^t, e^{-t} \rangle$ to obtain v(t) = $\langle e^t, -e^{-t} \rangle$. Evaluate at t = 0: v(0) = $\langle 1, -1 \rangle$. The speed is the magnitude $\sqrt{1^2 + (-1)^2} = \sqrt{2}$. A tempting distractor is B, 0, which might come from mistakenly thinking the components cancel out without squaring for magnitude. A transferable motion-analysis strategy is to use the magnitude formula after finding velocity to determine speed, especially for exponential parametrizations.
11. A particle has $r(t)=\langle t^2,\,2t-5\rangle$; what is its position at $t=2$?
    Answer: $\langle 4,-1\rangle$ This question simply asks for the position at a specific time. We evaluate the position function at t = 2: r(2) = ⟨2², 2(2) - 5⟩ = ⟨4, 4 - 5⟩ = ⟨4, -1⟩. This is a direct substitution problem. Students might make arithmetic errors, such as getting ⟨4, 1⟩ by incorrectly computing 4 - 5 = 1 (choice B). When finding position at a given time, simply substitute the t-value into both component functions.
12. A particle's velocity is $v(t)=\langle 3t^2,\,-4\rangle$; what is its speed at $t=1$?
    Answer: $5$ This question asks for speed when velocity is already given. At t = 1, the velocity is v(1) = ⟨3(1)², -4⟩ = ⟨3, -4⟩. Speed is the magnitude of this velocity vector: |v(1)| = √(3² + (-4)²) = √(9 + 16) = √25 = 5. Students might add the components incorrectly to get 7 (choice A) or forget to square before adding. When finding speed from velocity, always use the formula |v| = √(vₓ² + vᵧ²) for the magnitude of the vector.
13. A robot follows $r(t)=\langle \cos t,\,\sin t\rangle$; what is its velocity vector at $t=\pi/2$?
    Answer: $\langle -1,0\rangle$ This problem involves analyzing the motion of a robot using parametric equations. To find the velocity vector, differentiate the position components: $dx/dt = -\sin t$ and $dy/dt = \cos t$. At $t=\pi/2$, this gives $\langle -1, 0 \rangle$. The velocity vector indicates the instantaneous direction and rate of change at that point. A tempting distractor might be $\langle 0,1 \rangle$, which is the position at $t=\pi/2$, confusing position with velocity. Always remember to integrate velocity to find position or differentiate position to find velocity and speed when analyzing parametric motion.
14. For $r(t)=\langle 3t,\,\sin t\rangle$, what is the particle's speed at $t=0$?
    Answer: $\sqrt{10}$ This problem requires finding the speed of a particle with mixed polynomial and trigonometric parametric equations. First, we find the velocity vector by differentiating: $v(t) = r'(t) = \langle 3, \cos t \rangle$. At $t=0$, this becomes $v(0) = \langle 3, \cos(0) \rangle = \langle 3, 1 \rangle$. The speed is the magnitude of this velocity vector: $|v(0)| = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$. Choice C gives only the x-component of velocity, ignoring the y-component contribution to speed. For parametric motion, speed requires considering both components of velocity in the magnitude calculation.
15. A particle moves with $r(t)=\langle t^3,\,t^2\rangle$; which is its speed at $t=2$?
    Answer: $\sqrt{160}$ This problem involves analyzing the motion of a particle using parametric equations. To find the speed at $t = 2$, differentiate $r(t) = \langle t^3, t^2 \rangle$ to get $v(t) = \langle 3t^2, 2t \rangle$. Evaluate at $t = 2$: $v(2) = \langle 3(4), 4 \rangle = \langle 12, 4 \rangle$. The speed is the magnitude $\sqrt{12^2 + 4^2} = \sqrt{144 + 16} = \sqrt{160}$. A tempting distractor is B, 160, which comes from adding the squares without taking the square root, mistaking it for speed. A transferable motion-analysis strategy is to derive velocity from position derivatives and then find its magnitude to determine speed at any point.
16. A car's position is $r(t)=\langle t^3,\;4t\rangle$ (meters). What is its velocity vector at $t=2$?
    Answer: $\langle 12,\;4\rangle$ This problem requires computing the velocity vector of a car at a specific time using parametric equations. Given r(t) = ⟨t³, 4t⟩, we find velocity by differentiating each component: r'(t) = ⟨3t², 4⟩. At t=2, we substitute to get r'(2) = ⟨3(2)², 4⟩ = ⟨3(4), 4⟩ = ⟨12, 4⟩. The velocity vector tells us both the speed and direction of motion at that instant. Choice C (⟨6, 4⟩) might result from incorrectly computing the derivative as ⟨3t, 4⟩ instead of ⟨3t², 4⟩, a common error when differentiating powers. Always differentiate parametric components carefully: d/dt($t^n$) = $nt^{n-1}$.
17. A particle moves with $r(t)=\langle 1+2t,\,4-t\rangle$; which is its velocity vector at $t=5$?
    Answer: $\langle 2,-1\rangle$ For a linear parametric path, the velocity is constant. Differentiating r(t) = $\langle 1 + 2t, 4 - t \rangle$ gives v(t) = $\langle 2, -1 \rangle$ for all t. This constant velocity means the particle moves with uniform speed in a fixed direction. Students might substitute t = 5 into the position function to get $\langle 11, -1 \rangle$ (choice C) instead of differentiating. For motion problems, always distinguish between position r(t) and velocity v(t) = r'(t).
18. A swimmer moves with $r(t)=\langle 1+t^2,\;3-t\rangle$. What is the magnitude of velocity at $t=2$?
    Answer: $\sqrt{17}$ This problem requires finding the speed (magnitude of velocity) for a swimmer's parametric motion. Given r(t) = ⟨1+t², 3-t⟩, we differentiate to find velocity: r'(t) = ⟨2t, -1⟩. At t=2, the velocity is r'(2) = ⟨4, -1⟩. The magnitude is |r'(2)| = √(4² + (-1)²) = √(16 + 1) = √17. Choice E (√13) might result from using r'(t) = ⟨2t, -t⟩ instead of ⟨2t, -1⟩, incorrectly differentiating the linear term. When finding speed from parametric equations, carefully differentiate each component before computing the magnitude.
19. For $r(t)=\langle 2t-5,\,t^2\rangle$, what is the particle's displacement vector from $t=1$ to $t=3$?
    Answer: $\langle 4,8\rangle$ This problem requires finding the displacement vector for parametric motion over a time interval. Displacement is the change in position vector: $\Delta r = r(3) - r(1)$. First, we evaluate the position at each time: $r(1) = \langle 2(1)-5, 1^2 \rangle = \langle -3, 1 \rangle$ and $r(3) = \langle 2(3)-5, 3^2 \rangle = \langle 1, 9 \rangle$. The displacement vector is $\Delta r = \langle 1, 9 \rangle - \langle -3, 1 \rangle = \langle 4, 8 \rangle$. Choice E incorrectly gives the magnitude of displacement rather than the vector itself. Remember: displacement is a vector quantity showing net change in position, not distance traveled.
20. A robot's position is $r(t)=\langle t,\;t^2\rangle$ (meters). Which gives its velocity at $t=4$?
    Answer: $\u007fr'(4)$ This problem asks for the velocity vector of a robot following a parabolic parametric path. Given r(t) = ⟨t, t²⟩, velocity is found by differentiating each component: r'(t) = ⟨1, 2t⟩. This represents the instantaneous rate of change of position. At any time t, including t=4, the velocity vector is r'(t), not its magnitude |r'(t)|. Choice D (|r'(4)|) would give the speed √(1 + 16) = √17, losing the directional information that makes velocity a vector. For parametric motion, velocity r'(t) is a vector quantity that includes both magnitude and direction information.

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