Motion with Parametric & Vector Functions - Calculus 2

The acceleration of a particle is $\vec{a}(t) = \langle 0, -2 \rangle$ . At time $t=2$ , its velocity is $\vec{v}(2) = \langle 3, -4 \rangle$ . What is the tangential component of the particle's acceleration at $t=2$ ?
Answer: $8/5$ 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: $a_T = \frac{\vec{a} \cdot \vec{v}}{|\vec{v}|}$ , where you take the dot product of acceleration and velocity vectors, then divide by the speed. First, find the velocity vector at $t = 2$ . Since acceleration is constant at $\vec{a}(t) = \langle 0, -2 \rangle$ , integrate to get $\vec{v}(t) = \langle C_1, -2t + C_2 \rangle$ . Using the given condition $\vec{v}(2) = \langle 3, -4 \rangle$ , you get $C_1 = 3$ and $C_2 = 0$ , so $\vec{v}(2) = \langle 3, -4 \rangle$ . Now calculate: $\vec{a} \cdot \vec{v} = \langle 0, -2 \rangle \cdot \langle 3, -4 \rangle = 0(3) + (-2)(-4) = 8$ The speed is: $|\vec{v}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5$ Therefore: $a_T = \frac{8}{5}$ , which is answer B. Answer A ( $6/5$ ) likely comes from incorrectly calculating the dot product. Answer C ( $2$ ) is just the magnitude of the acceleration vector, ignoring the directional component. Answer D ( $8$ ) 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.
A particle moves in 3-space with position $\vec{r}(t) = \langle 2t, t^2, 8-t^3 \rangle$ . What is the particle's acceleration vector at the moment it crosses the xy-plane?
Answer: $\langle 0, 2, -12 \rangle$ The particle crosses the xy-plane when its z-coordinate is 0. Set $8 - t^3 = 0$ , which gives $t^3 = 8$ , so $t=2$ . To find the acceleration, we differentiate the position vector twice. The velocity is $\vec{v}(t) = \vec{r}'(t) = \langle 2, 2t, -3t^2 \rangle$ . The acceleration is $\vec{a}(t) = \vec{v}'(t) = \langle 0, 2, -6t \rangle$ . Evaluate the acceleration at $t=2$ : $\vec{a}(2) = \langle 0, 2, -6(2) \rangle = \langle 0, 2, -12 \rangle$ . Distractor B is the velocity vector at t=2. Distractor C is the position vector at t=2.
The velocity of a particle is given by $\vec{v}(t) = \langle 3t^2, 4t^3 \rangle$ . What is the rate of change of the particle's speed at $t=1$ ?
Answer: $66/5$ The rate of change of speed is given by the formula $a_T = \frac{\vec{v}(t) \cdot \vec{a}(t)}{||\vec{v}(t)||}$ . First, find the necessary vectors at $t=1$ . $\vec{v}(1) = \langle 3(1)^2, 4(1)^3 \rangle = \langle 3, 4 \rangle$ . The acceleration is $\vec{a}(t) = \langle 6t, 12t^2 \rangle$ , so $\vec{a}(1) = \langle 6, 12 \rangle$ . Now, calculate the components of the formula: $||\vec{v}(1)|| = \sqrt{3^2+4^2} = 5$ , and $\vec{v}(1) \cdot \vec{a}(1) = (3)(6) + (4)(12) = 18+48=66$ . The rate of change of speed is $66/5$ .
A particle moves with position vector $\mathbf{r}(t) = \langle \ln(t+1), \sqrt{t+4}, \arctan(t) \rangle$ for $t \geq 0$ . The particle's acceleration vector is perpendicular to its velocity vector at time $t = t_0$ . Which of the following must be true about $t_0$ ?
Answer: $t_0$ is a critical point of the speed function, but may be a local minimum, maximum, or neither When acceleration is perpendicular to velocity, $\mathbf{v}(t) \cdot \mathbf{a}(t) = 0$ . The speed is $s(t) = |\mathbf{v}(t)|$ , so $s'(t) = \frac{\mathbf{v}(t) \cdot \mathbf{a}(t)}{|\mathbf{v}(t)|}$ . When $\mathbf{v}(t) \cdot \mathbf{a}(t) = 0$ , we have $s'(t) = 0$ , making $t_0$ 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 $t_0$ must be a critical point, its nature requires additional investigation.
A particle moves along the parabola $y=x^3$ . At time $t>0$ , its x-coordinate is given by $x(t) = \sqrt{t}$ . What is the speed of the particle as a function of $t$ ?
Answer: $\frac{\sqrt{1+9t^2}}{2\sqrt{t}}$ When a particle moves along a curve in parametric form, you need to find both $\frac{dx}{dt}$ and $\frac{dy}{dt}$ to calculate speed. The speed formula is $\text{speed} = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}$ . Given $x(t) = \sqrt{t}$ , we have $\frac{dx}{dt} = \frac{1}{2\sqrt{t}}$ . Since the particle moves along $y = x^3$ , we get $y(t) = (\sqrt{t})^3 = t^{3/2}$ . Therefore, $\frac{dy}{dt} = \frac{3}{2}t^{1/2} = \frac{3\sqrt{t}}{2}$ . The speed is: $\sqrt{\left(\frac{1}{2\sqrt{t}}\right)^2 + \left(\frac{3\sqrt{t}}{2}\right)^2} = \sqrt{\frac{1}{4t} + \frac{9t}{4}} = \sqrt{\frac{1 + 9t^2}{4t}} = \frac{\sqrt{1 + 9t^2}}{2\sqrt{t}}$ This confirms answer B is correct. Answer A ( $\sqrt{1+9t^2}$ ) represents just the numerator without the denominator factor, suggesting someone forgot to account for the $\frac{1}{2\sqrt{t}}$ term properly. Answer C ( $\frac{\sqrt{t+9t^3}}{2t}$ ) comes from incorrectly factoring inside the square root, writing $\sqrt{\frac{1}{4t} + \frac{9t}{4}}$ as $\sqrt{\frac{t + 9t^3}{4t^2}}$ instead of $\sqrt{\frac{1 + 9t^2}{4t}}$ . Answer D ( $\sqrt{\frac{1}{4t} + \frac{9}{4}t^2}$ ) shows the expression before taking the final square root step, missing that $\frac{9t}{4} = \frac{9}{4}t$ , not $\frac{9}{4}t^2$ . Remember: for parametric motion problems, always find both component derivatives first, then apply the speed formula carefully, checking your algebra at each step.
The position of a particle moving in the plane is given by $\vec{r}(t) = \langle t^2, t^3 - 4t \rangle$ . For which interval of $t>0$ is the speed of the particle increasing?
Answer: $(\frac{\sqrt{10}}{3}, \infty)$ The speed is increasing when the dot product of the velocity and acceleration vectors is positive. $\vec{v}(t) = \langle 2t, 3t^2-4 \rangle$ and $\vec{a}(t) = \langle 2, 6t \rangle$ . Their dot product is $\vec{v}(t) \cdot \vec{a}(t) = (2t)(2) + (3t^2-4)(6t) = 4t + 18t^3 - 24t = 18t^3 - 20t$ . We need to find where $18t^3 - 20t > 0$ . Factoring gives $2t(9t^2 - 10) > 0$ . Since $t>0$ , we must have $9t^2 - 10 > 0$ , which means $t^2 > 10/9$ , or $t > \sqrt{10}/3$ .
A particle has acceleration $\vec{a}(t) = \langle 6t, 12t^2 \rangle$ . Its initial velocity at $t=0$ is $\vec{v}(0) = \langle 1, -1 \rangle$ and its initial position is $\vec{r}(0) = \langle 2, 3 \rangle$ . What is the particle's position at $t=1$ ?
Answer: $\langle 4, 3 \rangle$ First, integrate $\vec{a}(t)$ to find $\vec{v}(t)$ : $\vec{v}(t) = \langle 3t^2+C_1, 4t^3+C_2 \rangle$ . Using $\vec{v}(0) = \langle 1, -1 \rangle$ , we find $C_1=1$ and $C_2=-1$ . So $\vec{v}(t) = \langle 3t^2+1, 4t^3-1 \rangle$ . Next, integrate $\vec{v}(t)$ to find $\vec{r}(t)$ : $\vec{r}(t) = \langle t^3+t+D_1, t^4-t+D_2 \rangle$ . Using $\vec{r}(0) = \langle 2, 3 \rangle$ , we find $D_1=2$ and $D_2=3$ . So $\vec{r}(t) = \langle t^3+t+2, t^4-t+3 \rangle$ . Finally, evaluate at $t=1$ : $\vec{r}(1) = \langle 1+1+2, 1-1+3 \rangle = \langle 4, 3 \rangle$ .
The position of a particle is given by a vector function $\vec{r}(t)$ 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 $\vec{v}(t)$ and acceleration vector $\vec{a}(t)$ for all $t$ ?
Answer: The velocity and acceleration vectors are always orthogonal. 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 $|\vec{v}(t)| = c$ for some constant $c > 0$ . Taking the derivative of both sides: $\frac{d}{dt}|\vec{v}(t)|^2 = \frac{d}{dt}(\vec{v} \cdot \vec{v}) = 2\vec{v} \cdot \vec{a} = 0$ . This means $\vec{v} \cdot \vec{a} = 0$ , 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.
A particle moves in the xy-plane with velocity $\vec{v}(t) = \langle 6t, 3t^2 - 3 \rangle$ for $t \ge 0$ . What is the total distance traveled by the particle from $t=0$ to $t=2$ ?
Answer: $14$ The total distance traveled is the integral of the speed. The speed is the magnitude of the velocity vector: $||\vec{v}(t)|| = \sqrt{(6t)^2 + (3t^2-3)^2} = \sqrt{36t^2 + 9t^4 - 18t^2 + 9} = \sqrt{9t^4 + 18t^2 + 9} = \sqrt{9(t^2+1)^2} = 3(t^2+1)$ . The total distance is $\int_0^2 3(t^2+1) dt = 3[\frac{t^3}{3} + t]_0^2 = 3((\frac{8}{3} + 2) - 0) = 3(\frac{14}{3}) = 14$ . Distractor B is the magnitude of the displacement vector. Distractor C results from forgetting the factor of 3 when simplifying the speed.
A particle moves along a curve with parametric equations $x(t) = t^3 - 6t$ and $y(t) = t^2 + 2t$ . The particle changes direction (velocity becomes zero) at certain values of $t$ . At how many distinct points does the particle change direction?
Answer: The particle never changes direction during its motion For the particle to change direction, the velocity vector must be zero, meaning both $x'(t) = 0$ and $y'(t) = 0$ simultaneously. We have $x'(t) = 3t^2 - 6$ and $y'(t) = 2t + 2$ . Setting $y'(t) = 0$ : $2t + 2 = 0$ , so $t = -1$ . Setting $x'(t) = 0$ : $3t^2 - 6 = 0$ , so $t^2 = 2$ , giving $t = \pm\sqrt{2}$ . For both components to be zero simultaneously, we need a common solution, but $t = -1$ doesn't satisfy $x'(t) = 0$ since $x'(-1) = 3(-1)^2 - 6 = 3 - 6 = -3 \neq 0$ . Similarly, $t = \pm\sqrt{2}$ don't satisfy $y'(t) = 0$ since $y'(\sqrt{2}) = 2\sqrt{2} + 2 \neq 0$ . 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.
A particle's position is described by $\vec{r}(t) = \langle t^2-1, 2t - \frac{1}{4}t^4 \rangle$ for $t \ge 0$ . What is the speed of the particle when it reaches its highest point?
Answer: $2\sqrt[3]{2}$ When analyzing particle motion, finding the "highest point" means identifying where the vertical component of position reaches its maximum. This occurs when the vertical velocity component equals zero. Given $\vec{r}(t) = \langle t^2-1, 2t - \frac{1}{4}t^4 \rangle$ , you first find the velocity by taking the derivative: $\vec{v}(t) = \langle 2t, 2 - t^3 \rangle$ . The vertical component is $v_y(t) = 2 - t^3$ . At the highest point, $v_y(t) = 0$ , so $2 - t^3 = 0$ , which gives $t^3 = 2$ , or $t = \sqrt[3]{2}$ . Now you need the speed (magnitude of velocity) at this time. At $t = \sqrt[3]{2}$ : - $v_x = 2\sqrt[3]{2}$ - $v_y = 0$ The speed is $|\vec{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{(2\sqrt[3]{2})^2 + 0^2} = 2\sqrt[3]{2}$ . Looking at the wrong answers: Choice A gives $\frac{3}{2}\sqrt[3]{2}$ , which might result from incorrectly calculating the velocity components. Choice B gives just $\sqrt[3]{2}$ , which is the time value, not the speed—a common mix-up. Choice D gives 2, which you'd get if you forgot to include the $\sqrt[3]{2}$ factor when computing $v_x$ . The correct answer is C: $2\sqrt[3]{2}$ . Study tip: Always distinguish between when you need to find the time at an extreme point versus the value of some quantity at that time. Set the appropriate derivative to zero first, solve for time, then substitute back to find what the question actually asks for.
The position of a particle is given by the vector function $\vec{r}(t) = \langle e^t, e^{-2t} \rangle$ . At what time $t$ are the velocity and acceleration vectors orthogonal?
Answer: $t = \frac{\ln(2)}{2}$ Velocity is $\vec{v}(t) = \langle e^t, -2e^{-2t} \rangle$ and acceleration is $\vec{a}(t) = \langle e^t, 4e^{-2t} \rangle$ . The vectors are orthogonal when their dot product is zero. $\vec{v}(t) \cdot \vec{a}(t) = (e^t)(e^t) + (-2e^{-2t})(4e^{-2t}) = e^{2t} - 8e^{-4t} = 0$ . This simplifies to $e^{2t} = 8e^{-4t}$ . Multiplying both sides by $e^{4t}$ gives $e^{6t} = 8$ . Taking the natural logarithm gives $6t = \ln(8) = \ln(2^3) = 3\ln(2)$ . Therefore, $t = \frac{3\ln(2)}{6} = \frac{\ln(2)}{2}$ .
A particle's motion is described by the vector function $\vec{r}(t)$ . If the particle is momentarily at rest at time $t=c$ , which of the following must be true?
Answer: The velocity vector $\vec{v}(c)$ is the zero vector. This question tests your understanding of what "momentarily at rest" means in the context of vector-valued functions and particle motion. When analyzing particle motion, you need to distinguish between position, velocity, and acceleration vectors. A particle is "momentarily at rest" means it has zero velocity at that instant. Since velocity is the first derivative of the position vector, $\vec{v}(t) = \vec{r}'(t)$ , this condition requires $\vec{v}(c) = \vec{0}$ . This is precisely what choice B states. Let's examine why the other options are incorrect: Choice A is wrong because acceleration and velocity are independent at any given moment. A particle can have zero velocity while still experiencing acceleration - think of a ball thrown upward at the peak of its trajectory. It momentarily stops (zero velocity) but gravity still acts on it (non-zero acceleration). Choice C is incorrect because being at rest has nothing to do with location. A particle can be momentarily at rest at any point in space, not just the origin. The position vector $\vec{r}(c)$ and velocity vector $\vec{v}(c)$ are completely independent quantities. Choice D is misleading because "horizontal" implies a specific direction in 2D or 3D space. While it's true that when $\vec{v}(c) = \vec{0}$ , there's no well-defined tangent direction at that point, describing this as "horizontal" is imprecise and potentially wrong depending on the coordinate system. Remember: "At rest" always means zero velocity, regardless of position or acceleration. Don't confuse the three fundamental quantities of motion: position, velocity, and acceleration.
A particle's position is given by $\vec{r}(t) = \langle t^3 - 12t, t^2 - 2t \rangle$ . What is the speed of the particle at the first time $t>0$ when the tangent to its path is vertical?
Answer: $2$ A vertical tangent occurs when $dx/dt = 0$ and $dy/dt \neq 0$ . The velocity vector is $\vec{v}(t) = \langle 3t^2 - 12, 2t - 2 \rangle$ . Set $dx/dt = 3t^2 - 12 = 0$ , which gives $t^2=4$ . Since $t>0$ , we have $t=2$ . At $t=2$ , $dy/dt = 2(2)-2 = 2 \neq 0$ , so there is a vertical tangent. The speed at $t=2$ is $||\vec{v}(2)|| = \sqrt{(3(2^2)-12)^2 + (2(2)-2)^2} = \sqrt{0^2 + 2^2} = 2$ . Distractor B is the speed at the time of a horizontal tangent.
What is the total distance traveled by a particle from $t=0$ to $t=\pi$ if its position is given by $\vec{r}(t) = \langle 3t, 4\cos(t), 4\sin(t) \rangle$ ?
Answer: $5\pi$ When you see a parametric vector function and need to find total distance traveled, you're looking for arc length. This requires finding the magnitude of the velocity vector and integrating it over the given interval. First, find the velocity vector by taking the derivative: $\vec{v}(t) = \vec{r}'(t) = \langle 3, -4\sin(t), 4\cos(t) \rangle$ Next, calculate the speed (magnitude of velocity): $|\vec{v}(t)| = \sqrt{3^2 + (-4\sin(t))^2 + (4\cos(t))^2} = \sqrt{9 + 16\sin^2(t) + 16\cos^2(t)}$ Using the Pythagorean identity $\sin^2(t) + \cos^2(t) = 1$ : $|\vec{v}(t)| = \sqrt{9 + 16(1)} = \sqrt{25} = 5$ The total distance is: $\int_0^{\pi} 5 \, dt = 5\pi$ Answer A ( $5$ ) represents just the constant speed, not the total distance over the time interval. Answer B ( $\sqrt{9\pi^2+64}$ ) likely comes from incorrectly trying to find the magnitude of the displacement vector or making an error in the arc length calculation. Answer D ( $3\pi$ ) might result from only considering the x-component's contribution to the distance calculation. The key insight is recognizing that the speed is constant at 5 units per unit time. When speed is constant, total distance simply equals speed times elapsed time. Remember: for parametric curves, always find the velocity vector first, then its magnitude, then integrate to get arc length.
A projectile is fired from the origin with an initial velocity of $\vec{v}(0) = \langle 20, 40 \rangle$ m/s. Assuming the only force is gravity, so that acceleration is $\vec{a}(t) = \langle 0, -10 \rangle$ m/s $^2$ , what is the speed of the projectile at its maximum height?
Answer: $20$ m/s 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 $\vec{v}(0) = \langle 20, 40 \rangle$ and constant acceleration $\vec{a}(t) = \langle 0, -10 \rangle$ , you can find the velocity at any time using $\vec{v}(t) = \vec{v}(0) + \vec{a}t = \langle 20, 40 - 10t \rangle$ . At maximum height, the vertical velocity component becomes zero (the projectile momentarily stops rising before falling). Setting the vertical component to zero: $40 - 10t = 0$ , so $t = 4$ seconds. At this moment, $\vec{v}(4) = \langle 20, 0 \rangle$ . The speed is the magnitude: $|\vec{v}| = \sqrt{20^2 + 0^2} = 20$ 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 ( $20\sqrt{5}$ m/s) equals the initial speed $\sqrt{20^2 + 40^2} = 20\sqrt{5}$ , 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.
A particle's motion is described by $x(t) = t^2+1$ and $y(t) = \ln(2t+1)$ for $t \ge 0$ . What is the y-coordinate of the particle at the instant its path has a slope of $1/6$ ?
Answer: $\ln(4)$ The slope of the path is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ . We have $dx/dt = 2t$ and $dy/dt = \frac{2}{2t+1}$ . The slope is $\frac{2/(2t+1)}{2t} = \frac{1}{t(2t+1)}$ . We set the slope equal to $1/6$ : $\frac{1}{t(2t+1)} = \frac{1}{6}$ , which implies $t(2t+1) = 6$ , or $2t^2+t-6=0$ . Factoring gives $(2t-3)(t+2)=0$ . Since $t \ge 0$ , we must have $t=3/2$ . The y-coordinate at this time is $y(3/2) = \ln(2(3/2)+1) = \ln(4)$ .
A particle's velocity is $\vec{v}(t) = \langle \frac{1}{t}, \sqrt{t} \rangle$ for $t > 0$ . Which statement accurately describes the particle's motion at $t=4$ ?
Answer: It is moving up and to the right, and its speed is increasing. 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 $t = 4$ , the velocity is $\vec{v}(4) = \langle \frac{1}{4}, \sqrt{4} \rangle = \langle 0.25, 2 \rangle$ . 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: $\vec{a}(t) = \vec{v}'(t) = \langle -\frac{1}{t^2}, \frac{1}{2\sqrt{t}} \rangle$ . At $t = 4$ : $\vec{a}(4) = \langle -\frac{1}{16}, \frac{1}{4} \rangle$ . 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: $\vec{v}(4) \cdot \vec{a}(4) = (0.25)(-\frac{1}{16}) + (2)(\frac{1}{4}) = -\frac{1}{64} + \frac{1}{2} = \frac{31}{64} > 0$ . 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.
A particle moves according to $\vec{r}(t) = \langle \frac{1}{3}t^3+t, t^2 \rangle$ . For what time $t > 0$ are the particle's velocity and acceleration vectors parallel but not zero?
Answer: $t=1$ 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, $\vec{v}(t) = k\vec{a}(t)$ for some scalar $k \neq 0$ . Start by finding the velocity and acceleration vectors through differentiation. Given $\vec{r}(t) = \langle \frac{1}{3}t^3+t, t^2 \rangle$ , the velocity is $\vec{v}(t) = \vec{r}'(t) = \langle t^2+1, 2t \rangle$ and the acceleration is $\vec{a}(t) = \vec{v}'(t) = \langle 2t, 2 \rangle$ . For these vectors to be parallel, their components must be proportional: $\frac{t^2+1}{2t} = \frac{2t}{2} = t$ . This gives us the equation $t^2+1 = 2t^2$ , which simplifies to $t^2 = 1$ . Since we need $t > 0$ , we get $t = 1$ . Let's verify: At $t = 1$ , $\vec{v}(1) = \langle 2, 2 \rangle$ and $\vec{a}(1) = \langle 2, 2 \rangle$ . These are indeed parallel (and equal), confirming answer D. Answer A is wrong because we found a valid solution. Answer B ( $t = \sqrt{3}$ ) fails the proportionality test: $\vec{v}(\sqrt{3}) = \langle 4, 2\sqrt{3} \rangle$ and $\vec{a}(\sqrt{3}) = \langle 2\sqrt{3}, 2 \rangle$ aren't proportional. Answer C ( $t = 2$ ) similarly fails: $\vec{v}(2) = \langle 5, 4 \rangle$ and $\vec{a}(2) = \langle 4, 2 \rangle$ have non-equal ratios $\frac{5}{4} \neq \frac{4}{2}$ . Remember: parallel vectors have proportional components, so set up the proportion equation and solve systematically.
A particle travels along a path defined by $\vec{r}(t) = \langle t - \sin(t), 1 - \cos(t) \rangle$ for $0 \le t \le \pi$ . What is the total distance the particle travels?
Answer: $4$ First, find the velocity vector: $\vec{v}(t) = \langle 1 - \cos(t), \sin(t) \rangle$ . Next, find the speed: $||\vec{v}(t)|| = \sqrt{(1-\cos(t))^2 + \sin^2(t)} = \sqrt{1 - 2\cos(t) + \cos^2(t) + \sin^2(t)} = \sqrt{2 - 2\cos(t)}$ . Using the half-angle identity $2\sin^2(t/2) = 1-\cos(t)$ , the speed becomes $\sqrt{4\sin^2(t/2)} = 2|\sin(t/2)|$ . On the interval $0 \le t \le \pi$ , $\sin(t/2) \ge 0$ , so the speed is $2\sin(t/2)$ . The distance is $\int_0^\pi 2\sin(t/2) dt = [-4\cos(t/2)]_0^\pi = -4\cos(\pi/2) - (-4\cos(0)) = 0 - (-4) = 4$ . Distractor B is the magnitude of the displacement.

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