This question evaluates computing the distance traveled via arc length for a parametric curve with negative t values. Distance is the integral of speed, the magnitude of velocity, not signed displacement, to account for the entire path regardless of direction. For $r(t) = \langle t^2, t^2 \rangle$, $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 2t$, so $\int_{-1}^{2} \sqrt{(2t)^2 + (2t)^2} , dt$ gives the distance. A tempting distractor is choice E, which uses positions $t^2$ instead of derivatives $2t$ in the integrand. The magnitude-first strategy is key: always derive to get velocity, compute its magnitude with the square root, then integrate over the interval for total distance.

The skill being tested is calculating the arc length or distance traveled by a particle along a parametric curve. The distance traveled is found by integrating the speed, which is the magnitude of the velocity vector, over the time interval. This differs from displacement, which is the straight-line distance from start to end and can be signed. For r(t) = $<e^t$, $e^{-t}$>, the velocity components are $e^t$ and $-e^{-t}$, so the integral is ∫ from -1 to 1 of $sqrt((e^t$)² + $(-e^{-t}$)²) dt, matching choice C. A tempting distractor is choice D, which reverses the limits, yielding a negative value unsuitable for distance. Always compute the magnitude inside the integral first to capture the instantaneous speed. This magnitude-first strategy ensures accurate distance for any parametric path.

The skill being tested is calculating the arc length or distance traveled by a particle along a parametric curve. The distance traveled is found by integrating the speed, which is the magnitude of the velocity vector, over the time interval. This differs from displacement, which is the straight-line distance from start to end and can be signed. For the two particles, particle 2's path along the parabola has greater length since ∫ sqrt(1 + 4t²) dt > ∫ 1 dt, matching choice B. A tempting distractor is choice C, which equates distances based on x-displacement alone, ignoring the y-component. Always compute the magnitude inside the integral first to capture the instantaneous speed. This magnitude-first strategy ensures accurate distance for any parametric path.

This particle moves along a circle of radius 2, with velocity components $ \frac{dx}{dt} = 2\cos(t) $ and $ \frac{dy}{dt} = -2\sin(t) $. The speed is $ \sqrt{(2\cos t)^2 + (-2\sin t)^2} = \sqrt{4\cos^2 t + 4\sin^2 t} = 2\sqrt{\cos^2 t + \sin^2 t} = 2 $. With constant speed 2, the distance from t=0 to t=π/2 is $ \int_0^{\pi/2} 2 , dt = 2(\pi/2) = \pi $. Choice B ($2\pi$) would be a full circle, while we only travel a quarter circle. The key is recognizing that for circular motion with radius r, the constant speed equals r times the angular velocity (here $ r=2 $, $ \omega=1 $).

The skill being tested is calculating the arc length or distance traveled by a particle along a parametric curve. The distance traveled is found by integrating the speed, which is the magnitude of the velocity vector, over the time interval. This differs from displacement, which is the straight-line distance from start to end and can be signed. For r(t) = <3 cos t, 4 sin t>, the velocity components are -3 sin t and 4 cos t, so the integral is ∫ from 0 to π/2 of sqrt(9 sin² t + 16 cos² t) dt, matching choice B. A tempting distractor is choice A, which swaps the coefficients, altering the ellipse axes. Always compute the magnitude inside the integral first to capture the instantaneous speed. This magnitude-first strategy ensures accurate distance for any parametric path.

This problem involves finding the distance traveled along a circular path given by parametric equations. The velocity components are dx/dt = -sin(t) and dy/dt = cos(t), giving speed = √[(-sin t)² + (cos t)²] = √[sin²t + cos²t] = 1. Since the speed is constant at 1, the distance traveled from t=0 to t=π equals ∫₀^π 1 dt = π. Choice B (2π) would be the full circumference, but we only travel half the circle. The key insight is recognizing that for uniform circular motion, speed equals the radius (1 in this case), making distance calculations straightforward.

The skill tested is formulating the arc length integral for distance traveled with non-polynomial parametrics. Distance traveled integrates the magnitude of velocity, distinguishing it from signed displacement which could net zero for closed paths. For x(t) = ln(t+1) and y(t) = √t, dx/dt = 1/(t+1) and dy/dt = 1/(2√t), yielding ∫ √[(1/(t+1))² + (1/(2√t))²] dt from 0 to 3. Choice C is tempting, using positions ln(t+1) and √t inside the square root instead of their derivatives. Adopt the magnitude-first approach: calculate derivatives for velocity, find the magnitude via square root of summed squares, and integrate to get arc length.

This problem asks for the total distance traveled, which requires finding the arc length of the parametric curve. For parametric equations x(t) and y(t), distance is calculated using the speed formula: ∫√[(dx/dt)² + (dy/dt)²]dt. Here, dx/dt = 2t and dy/dt = 3, so we need ∫₀² √[(2t)² + 3²]dt. Choice A incorrectly adds the derivatives instead of using the Pythagorean formula for speed. Remember: distance traveled always uses the magnitude of velocity, found by taking the square root of the sum of squared component derivatives.

Particle 1 has velocity v₁ = ⟨1,1⟩ with speed |v₁| = √[1² + 1²] = √2. Particle 2 has velocity v₂ = ⟨1,2⟩ with speed |v₂| = √[1² + 2²] = √5. Since √5 > √2 and both speeds are constant, particle 2 travels farther: distance₂ = √5 × 1 > √2 × 1 = distance₁. Choice A confuses velocity magnitude with individual components. Choice C incorrectly assumes equal speeds. For distance comparisons, always calculate speed as the magnitude of the velocity vector, not just looking at components.

The skill being tested is calculating the arc length or distance traveled by a particle along a parametric curve. The distance traveled is found by integrating the speed, which is the magnitude of the velocity vector, over the time interval. This differs from displacement, which is the straight-line distance from start to end and can be signed. For x=t³-3t, y=4t, the velocity components are 3t²-3 and 4, so the integral is ∫ from -1 to 2 of sqrt((3t²-3)² + 4²) dt, matching choice A. A tempting distractor is choice E, which computes the magnitude of the integrals separately, giving displacement instead. Always compute the magnitude inside the integral first to capture the instantaneous speed. This magnitude-first strategy ensures accurate distance for any parametric path.