IB Mathematics: Applications and Interpretation Quiz: Parametric Models
20 questions · exam conditions
0:00
Parametric ModelsQuestion 1 of 20

The path of a particle is given by x(t)=t+1x(t) = \sqrt{t+1} and y(t)=t−1y(t) = t-1 for t≥0t \ge 0. Which of the following is the Cartesian equation for the path with its correct domain?

y=x2−2y = x^2 - 2 for all real xx
y=x2−2y = x^2 - 2 for x≥0x \ge 0
y=x−2y = x - 2 for x≥1x \ge 1
y=x2−2y = x^2 - 2 for x≥1x \ge 1
← Back to quizzes

IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Parametric Models

Practice Parametric Models in IB Mathematics: Applications and Interpretation 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 Parametric Models, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Applications and Interpretation.

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 path of a particle is given by x(t)=t+1x(t) = \sqrt{t+1} and y(t)=t−1y(t) = t-1 for t≥0t \ge 0. Which of the following is the Cartesian equation for the path with its correct domain?

  1. y=x2−2y = x^2 - 2 for all real xx
  2. y=x2−2y = x^2 - 2 for x≥0x \ge 0
  3. y=x−2y = x - 2 for x≥1x \ge 1
  4. y=x2−2y = x^2 - 2 for x≥1x \ge 1 (correct answer)
Explanation: When you encounter parametric equations, you need to eliminate the parameter to find the Cartesian equation, then carefully determine the domain based on the parameter's constraints. To eliminate parameter tt from x(t)=t+1x(t) = \sqrt{t+1} and y(t)=t−1y(t) = t-1, start with the simpler equation. From y=t−1y = t-1, you get t=y+1t = y+1. Substituting this into the x-equation: x=(y+1)+1=y+2x = \sqrt{(y+1)+1} = \sqrt{y+2}. Squaring both sides gives x2=y+2x^2 = y+2, so y=x2−2y = x^2 - 2. Now for the crucial domain step. Since t≥0t \geq 0, you have x=t+1≥0+1=1x = \sqrt{t+1} \geq \sqrt{0+1} = 1. Therefore, x≥1x \geq 1. Option A is wrong because it claims the domain is all real xx, but we've shown xx must be at least 1. Option B incorrectly states x≥0x \geq 0 when the actual constraint is x≥1x \geq 1. This is a common error—students sometimes assume that since t≥0t \geq 0, then x≥0x \geq 0, but you must calculate the actual minimum value of xx. Option C has the wrong function entirely (y=x−2y = x - 2 instead of y=x2−2y = x^2 - 2), likely from incorrectly manipulating the parametric equations. Option D correctly gives y=x2−2y = x^2 - 2 for x≥1x \geq 1. Study tip: When converting parametric to Cartesian equations, always check the domain by finding the range of each variable based on the parameter's constraints. Don't assume the domain starts at zero just because the parameter does.

Question 2

The path of a particle is given by x(t)=t+1x(t) = \sqrt{t+1} and y(t)=t−1y(t) = t-1 for t≥0t \ge 0. Which of the following is the Cartesian equation for the path with its correct domain?

  1. y=x2−2y = x^2 - 2 for all real xx
  2. y=x2−2y = x^2 - 2 for x≥0x \ge 0
  3. y=x−2y = x - 2 for x≥1x \ge 1
  4. y=x2−2y = x^2 - 2 for x≥1x \ge 1 (correct answer)
Explanation: When you encounter parametric equations, you need to eliminate the parameter to find the Cartesian equation, then carefully determine the domain based on the parameter's constraints. To eliminate parameter tt from x(t)=t+1x(t) = \sqrt{t+1} and y(t)=t−1y(t) = t-1, start with the simpler equation. From y=t−1y = t-1, you get t=y+1t = y+1. Substituting this into the x-equation: x=(y+1)+1=y+2x = \sqrt{(y+1)+1} = \sqrt{y+2}. Squaring both sides gives x2=y+2x^2 = y+2, so y=x2−2y = x^2 - 2. Now for the crucial domain step. Since t≥0t \geq 0, you have x=t+1≥0+1=1x = \sqrt{t+1} \geq \sqrt{0+1} = 1. Therefore, x≥1x \geq 1. Option A is wrong because it claims the domain is all real xx, but we've shown xx must be at least 1. Option B incorrectly states x≥0x \geq 0 when the actual constraint is x≥1x \geq 1. This is a common error—students sometimes assume that since t≥0t \geq 0, then x≥0x \geq 0, but you must calculate the actual minimum value of xx. Option C has the wrong function entirely (y=x−2y = x - 2 instead of y=x2−2y = x^2 - 2), likely from incorrectly manipulating the parametric equations. Option D correctly gives y=x2−2y = x^2 - 2 for x≥1x \geq 1. Study tip: When converting parametric to Cartesian equations, always check the domain by finding the range of each variable based on the parameter's constraints. Don't assume the domain starts at zero just because the parameter does.

Question 3

A ball's movement on a table is modelled by x(t)=10t−2t2x(t) = 10t - 2t^2 and y(t)=8ty(t) = 8t, for t≥0t \ge 0, with distances in cm. The table has a wall at x=12x=12. Find the speed of the ball at the moment it first hits this wall.

  1. 8.00 cm/s
  2. 8.25 cm/s (correct answer)
  3. 16.0 cm/s
  4. 20.0 cm/s
Explanation: First, find the time tt when the ball hits the wall by setting x(t)=12x(t) = 12: 10t−2t2=12  ⟹  2t2−10t+12=0  ⟹  t2−5t+6=0  ⟹  (t−2)(t−3)=010t - 2t^2 = 12 \implies 2t^2 - 10t + 12 = 0 \implies t^2 - 5t + 6 = 0 \implies (t-2)(t-3) = 0. The first time is t=2t=2. Next, find the speed at t=2t=2. The velocity components are dxdt=10−4t\frac{dx}{dt} = 10 - 4t and dydt=8\frac{dy}{dt} = 8. At t=2t=2, dxdt=10−4(2)=2\frac{dx}{dt} = 10 - 4(2) = 2 and dydt=8\frac{dy}{dt} = 8. The speed is 22+82=4+64=68≈8.25\sqrt{2^2 + 8^2} = \sqrt{4 + 64} = \sqrt{68} \approx 8.25 cm/s.

Question 4

A robot vacuum cleaner moves on a large floor. Its path is modelled by x(t)=3cos⁡(0.5t)x(t) = 3\cos(0.5t) and y(t)=5sin⁡(0.5t)y(t) = 5\sin(0.5t) for 0≤t≤100 \le t \le 10, where tt is in seconds and distances are in metres. Calculate the total distance the robot travels in the first 10 seconds.

  1. 5.25 m
  2. 15.9 m
  3. 21.0 m (correct answer)
  4. 25.9 m
Explanation: The total distance travelled is the integral of the speed. First, find the velocity components: dxdt=−1.5sin⁡(0.5t)\frac{dx}{dt} = -1.5\sin(0.5t) and dydt=2.5cos⁡(0.5t)\frac{dy}{dt} = 2.5\cos(0.5t). The speed is S(t)=(dxdt)2+(dydt)2=(−1.5sin⁡(0.5t))2+(2.5cos⁡(0.5t))2S(t) = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} = \sqrt{(-1.5\sin(0.5t))^2 + (2.5\cos(0.5t))^2}. The total distance is D=∫010S(t) dtD = \int_{0}^{10} S(t) \,dt. Using a calculator's numerical integration function, D=∫0102.25sin⁡2(0.5t)+6.25cos⁡2(0.5t) dt≈21.01D = \int_{0}^{10} \sqrt{2.25\sin^2(0.5t) + 6.25\cos^2(0.5t)} \,dt \approx 21.01 m. The closest answer is 21.0 m.

Question 5

Particle P has a path given by xP(t)=t+1,yP(t)=t2x_P(t) = t+1, y_P(t) = t^2. Particle Q has a path given by xQ(s)=2s−1,yQ(s)=4s−4x_Q(s) = 2s-1, y_Q(s) = 4s-4, with independent time parameters t,s≥0t,s \ge 0. It is known that the particles collide at the point (3, 4).

At which other point do their paths cross?

  1. (-1, -4)
  2. The paths do not cross at any other point.
  3. (2, 1)
  4. (1, 0) (correct answer)
Explanation: When you encounter particle path problems, you're dealing with parametric equations where different particles may traverse the same geometric points at different times. The key distinction is between when paths intersect (same location, different times) versus when particles collide (same location, same time). To find where the paths cross, you need to solve for when xP(t)=xQ(s)x_P(t) = x_Q(s) and yP(t)=yQ(s)y_P(t) = y_Q(s) simultaneously. Setting up the equations: t+1=2s−1t + 1 = 2s - 1 and t2=4s−4t^2 = 4s - 4. From the first equation: t=2s−2t = 2s - 2. Substituting into the second: (2s−2)2=4s−4(2s - 2)^2 = 4s - 4. Expanding: 4s2−8s+4=4s−44s^2 - 8s + 4 = 4s - 4, which simplifies to 4s2−12s+8=04s^2 - 12s + 8 = 0, or s2−3s+2=0s^2 - 3s + 2 = 0. Factoring gives (s−1)(s−2)=0(s-1)(s-2) = 0, so s=1s = 1 or s=2s = 2. When s=1s = 1: t=0t = 0, giving point (1,0)(1, 0) When s=2s = 2: t=2t = 2, giving point (3,4)(3, 4) Since we know the collision occurs at (3,4)(3, 4), the other intersection point is (1,0)(1, 0), making D correct. Answer A gives (−1,−4)(-1, -4), which doesn't satisfy either parametric equation. Answer B incorrectly assumes only one intersection exists. Answer C provides (2,1)(2, 1), which you can verify doesn't lie on particle Q's path. Remember: always distinguish between path intersections (geometric crossings) and particle collisions (same place, same time). Set up your parametric equations carefully and solve systematically.

Question 6

The position of a ship is given by x(t)=5t−20x(t) = 5t - 20 and y(t)=3t−30y(t) = 3t - 30, where tt is hours after noon and distances are in kilometres. A lighthouse is at the origin (0,0). At what time tt is the ship closest to the lighthouse?

  1. 4.00 hours
  2. 5.59 hours (correct answer)
  3. 7.00 hours
  4. 10.0 hours
Explanation: The distance DD from the origin is given by D=x(t)2+y(t)2D = \sqrt{x(t)^2 + y(t)^2}. To minimize DD, we can minimize its square, D2=(5t−20)2+(3t−30)2D^2 = (5t-20)^2 + (3t-30)^2. Expanding this gives D2=(25t2−200t+400)+(9t2−180t+900)=34t2−380t+1300D^2 = (25t^2 - 200t + 400) + (9t^2 - 180t + 900) = 34t^2 - 380t + 1300. This is a quadratic in tt, and its minimum occurs at the vertex, t=−b2a=−−3802(34)=38068≈5.588t = -\frac{b}{2a} = -\frac{-380}{2(34)} = \frac{380}{68} \approx 5.588. The closest time is 5.59 hours.

Question 7

A financial model tracks two economic indicators, I1 and I2, over time tt (in years, t≥0t \ge 0). Their values are given by x(t)=100+5tx(t) = 100 + 5t and y(t)=50e0.1tln⁡(t+1)y(t) = 50e^{0.1t} \ln(t+1). Find the rate of change of indicator I2 with respect to indicator I1 at t=5t=5 years.

  1. 1.18
  2. 5.70 (correct answer)
  3. 14.8
  4. 28.5
Explanation: The rate of change of I2 with respect to I1 is dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}. First, dxdt=5\frac{dx}{dt} = 5. For dydt\frac{dy}{dt}, we use the product rule: dydt=(50e0.1t)′ln⁡(t+1)+50e0.1t(ln⁡(t+1))′=5e0.1tln⁡(t+1)+50e0.1tt+1\frac{dy}{dt} = (50e^{0.1t})'\ln(t+1) + 50e^{0.1t}(\ln(t+1))' = 5e^{0.1t}\ln(t+1) + \frac{50e^{0.1t}}{t+1}. At t=5t=5, dxdt=5\frac{dx}{dt}=5 and dydt=5e0.5ln⁡(6)+50e0.56≈14.773+13.739=28.512\frac{dy}{dt} = 5e^{0.5}\ln(6) + \frac{50e^{0.5}}{6} \approx 14.773 + 13.739 = 28.512. Therefore, dydx=28.5125≈5.702\frac{dy}{dx} = \frac{28.512}{5} \approx 5.702.

Question 8

A robot vacuum cleaner moves on a large floor. Its path is modelled by x(t)=3cos⁡(0.5t)x(t) = 3\cos(0.5t) and y(t)=5sin⁡(0.5t)y(t) = 5\sin(0.5t) for 0≤t≤100 \le t \le 10, where tt is in seconds and distances are in metres. Calculate the total distance the robot travels in the first 10 seconds.

  1. 5.25 m
  2. 15.9 m
  3. 21.0 m (correct answer)
  4. 25.9 m
Explanation: The total distance travelled is the integral of the speed. First, find the velocity components: dxdt=−1.5sin⁡(0.5t)\frac{dx}{dt} = -1.5\sin(0.5t) and dydt=2.5cos⁡(0.5t)\frac{dy}{dt} = 2.5\cos(0.5t). The speed is S(t)=(dxdt)2+(dydt)2=(−1.5sin⁡(0.5t))2+(2.5cos⁡(0.5t))2S(t) = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} = \sqrt{(-1.5\sin(0.5t))^2 + (2.5\cos(0.5t))^2}. The total distance is D=∫010S(t) dtD = \int_{0}^{10} S(t) \,dt. Using a calculator's numerical integration function, D=∫0102.25sin⁡2(0.5t)+6.25cos⁡2(0.5t) dt≈21.01D = \int_{0}^{10} \sqrt{2.25\sin^2(0.5t) + 6.25\cos^2(0.5t)} \,dt \approx 21.01 m. The closest answer is 21.0 m.

Question 9

A firework is launched and its trajectory is given by x(t)=30tx(t) = 30t and y(t)=50t−4.9t2y(t) = 50t - 4.9t^2, for t≥0t \ge 0. Find the angle, in degrees, that the firework's path makes with the horizontal at t=4t=4 seconds.

  1. 19.8° (correct answer)
  2. 45.4°
  3. 70.2°
  4. 88.6°
Explanation: The angle of the path is the angle of the velocity vector. First find the velocity components: dxdt=30\frac{dx}{dt} = 30 and dydt=50−9.8t\frac{dy}{dt} = 50 - 9.8t. At t=4t=4, dxdt=30\frac{dx}{dt} = 30 and dydt=50−9.8(4)=10.8\frac{dy}{dt} = 50 - 9.8(4) = 10.8. The angle θ\theta with the horizontal is given by tan⁡(θ)=dy/dtdx/dt=10.830=0.36\tan(\theta) = \frac{dy/dt}{dx/dt} = \frac{10.8}{30} = 0.36. Therefore, θ=arctan⁡(0.36)≈19.8∘\theta = \arctan(0.36) \approx 19.8^{\circ}.

Question 10

The path of a thrown javelin is modelled by x(t)=25tx(t) = 25t and y(t)=2+15t−4.9t2y(t) = 2 + 15t - 4.9t^2, where t≥0t \ge 0 is time in seconds and distances are in meters. What is the maximum height reached by the javelin?

  1. 1.53 m
  2. 13.5 m (correct answer)
  3. 25.0 m
  4. 38.3 m
Explanation: The maximum height occurs when the vertical velocity, dydt\frac{dy}{dt}, is zero. dydt=15−9.8t\frac{dy}{dt} = 15 - 9.8t. Setting this to zero gives 15−9.8t=0  ⟹  t=159.8≈1.530615 - 9.8t = 0 \implies t = \frac{15}{9.8} \approx 1.5306 seconds. To find the maximum height, substitute this time back into the y(t)y(t) equation: y(1.5306)=2+15(1.5306)−4.9(1.5306)2≈2+22.959−11.479≈13.48y(1.5306) = 2 + 15(1.5306) - 4.9(1.5306)^2 \approx 2 + 22.959 - 11.479 \approx 13.48 m. The closest answer is 13.5 m.

Question 11

The path of a remote-controlled car is given by x(t)=t3−12t+5x(t) = t^3 - 12t + 5 and y(t)=t2−8ty(t) = t^2 - 8t, for t≥0t \ge 0. For what positive value of tt is the car momentarily moving only horizontally?

  1. 2.00
  2. 2.65
  3. 4.00 (correct answer)
  4. 8.00
Explanation: Moving only horizontally means the vertical component of velocity, dydt\frac{dy}{dt}, is zero, while the horizontal component, dxdt\frac{dx}{dt}, is non-zero. The vertical velocity is dydt=2t−8\frac{dy}{dt} = 2t - 8. Setting this to zero gives 2t−8=0  ⟹  t=42t - 8 = 0 \implies t=4. We must check that the horizontal velocity is not zero at this time. dxdt=3t2−12\frac{dx}{dt} = 3t^2 - 12. At t=4t=4, dxdt=3(42)−12=48−12=36≠0\frac{dx}{dt} = 3(4^2) - 12 = 48 - 12 = 36 \neq 0. Thus, the car is moving horizontally at t=4t=4.

Question 12

A particle's velocity components are given by dxdt=t−3\frac{dx}{dt} = t-3 and dydt=t\frac{dy}{dt} = \sqrt{t} for t≥0t \ge 0. At what time tt is its speed at a minimum?

  1. 0
  2. 1.66
  3. 2.50 (correct answer)
  4. 3.00
Explanation: The speed S(t)S(t) is given by S(t)=(dxdt)2+(dydt)2=(t−3)2+(t)2=t2−6t+9+t=t2−5t+9S(t) = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} = \sqrt{(t-3)^2 + (\sqrt{t})^2} = \sqrt{t^2 - 6t + 9 + t} = \sqrt{t^2 - 5t + 9}. To find the minimum speed, we can minimize the function inside the square root, f(t)=t2−5t+9f(t) = t^2 - 5t + 9. This is a parabola opening upwards. Its minimum occurs at the vertex, where t=−b2a=−−52(1)=2.5t = -\frac{b}{2a} = -\frac{-5}{2(1)} = 2.5. So the minimum speed occurs at t=2.5t=2.5.

Question 13

The position of a hovercraft is modelled by x(t)=100t+5x(t) = \frac{100}{t+5} and y(t)=20ln⁡(t+1)y(t) = 20\ln(t+1) for t≥0t \ge 0. Find the initial velocity vector of the hovercraft.

  1. (-4, 20) (correct answer)
  2. (4, 20)
  3. (20, 0)
  4. (20.4, 0)
Explanation: Initial velocity means the velocity at t=0t=0. First, find the velocity components by differentiation: x(t)=100(t+5)−1  ⟹  dxdt=−100(t+5)−2=−100(t+5)2x(t) = 100(t+5)^{-1} \implies \frac{dx}{dt} = -100(t+5)^{-2} = \frac{-100}{(t+5)^2}. y(t)=20ln⁡(t+1)  ⟹  dydt=20t+1y(t) = 20\ln(t+1) \implies \frac{dy}{dt} = \frac{20}{t+1}. Now evaluate at t=0t=0: dxdt(0)=−100(0+5)2=−4\frac{dx}{dt}(0) = \frac{-100}{(0+5)^2} = -4. dydt(0)=200+1=20\frac{dy}{dt}(0) = \frac{20}{0+1} = 20. The initial velocity vector is ⟨−4,20⟩\langle -4, 20 \rangle.

Question 14

A drone's flight path is modelled by the parametric equations x(t)=2t+sin⁡(t)x(t) = 2t + \sin(t) and y(t)=10−t2y(t) = 10 - t^2, where t≥0t \ge 0 is the time in seconds, and xx and yy are displacements in meters. What is the speed of the drone at t=3t=3 seconds, correct to three significant figures?

  1. 6.00 m/s
  2. 6.08 m/s (correct answer)
  3. 6.22 m/s
  4. 7.01 m/s
Explanation: The velocity components are found by differentiating the position equations with respect to tt: dxdt=2+cos⁡(t)\frac{dx}{dt} = 2 + \cos(t) and dydt=−2t\frac{dy}{dt} = -2t. At t=3t=3, dxdt=2+cos⁡(3)≈1.0100\frac{dx}{dt} = 2 + \cos(3) \approx 1.0100 and dydt=−2(3)=−6\frac{dy}{dt} = -2(3) = -6. The speed is the magnitude of the velocity vector: Speed =(dxdt)2+(dydt)2=(1.0100)2+(−6)2≈1.0201+36=37.0201≈6.084= \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} = \sqrt{(1.0100)^2 + (-6)^2} \approx \sqrt{1.0201 + 36} = \sqrt{37.0201} \approx 6.084 m/s. To three significant figures, this is 6.08 m/s.

Question 15

Particle A moves along a path defined by xA(t)=t+1x_A(t) = t+1 and yA(t)=t2y_A(t) = t^2. Particle B moves along a path defined by xB(t)=2t−1x_B(t) = 2t-1 and yB(t)=4t−4y_B(t) = 4t-4. Time t≥0t \ge 0. At what point do the particles collide?

  1. (1, 0)
  2. (2, 1)
  3. (3, 4) (correct answer)
  4. (5, 16)
Explanation: A collision occurs if the particles are at the same position at the same time tt. We set their coordinates equal: xA(t)=xB(t)  ⟹  t+1=2t−1  ⟹  t=2x_A(t) = x_B(t) \implies t+1 = 2t-1 \implies t=2. We must verify that their y-coordinates are also equal at this time: yA(2)=22=4y_A(2) = 2^2 = 4 and yB(2)=4(2)−4=4y_B(2) = 4(2)-4 = 4. Since both coordinates match at t=2t=2, a collision occurs. The location of the collision is found by substituting t=2t=2 into either set of equations: x(2)=2+1=3x(2) = 2+1 = 3 and y(2)=22=4y(2) = 2^2 = 4. The collision point is (3, 4).

Question 16

A firework is launched and its trajectory is given by x(t)=30tx(t) = 30t and y(t)=50t−4.9t2y(t) = 50t - 4.9t^2, for t≥0t \ge 0. Find the angle, in degrees, that the firework's path makes with the horizontal at t=4t=4 seconds.

  1. 19.8° (correct answer)
  2. 45.4°
  3. 70.2°
  4. 88.6°
Explanation: The angle of the path is the angle of the velocity vector. First find the velocity components: dxdt=30\frac{dx}{dt} = 30 and dydt=50−9.8t\frac{dy}{dt} = 50 - 9.8t. At t=4t=4, dxdt=30\frac{dx}{dt} = 30 and dydt=50−9.8(4)=10.8\frac{dy}{dt} = 50 - 9.8(4) = 10.8. The angle θ\theta with the horizontal is given by tan⁡(θ)=dy/dtdx/dt=10.830=0.36\tan(\theta) = \frac{dy/dt}{dx/dt} = \frac{10.8}{30} = 0.36. Therefore, θ=arctan⁡(0.36)≈19.8∘\theta = \arctan(0.36) \approx 19.8^{\circ}.

Question 17

The path of a thrown javelin is modelled by x(t)=25tx(t) = 25t and y(t)=2+15t−4.9t2y(t) = 2 + 15t - 4.9t^2, where t≥0t \ge 0 is time in seconds and distances are in meters. What is the maximum height reached by the javelin?

  1. 1.53 m
  2. 13.5 m (correct answer)
  3. 25.0 m
  4. 38.3 m
Explanation: The maximum height occurs when the vertical velocity, dydt\frac{dy}{dt}, is zero. dydt=15−9.8t\frac{dy}{dt} = 15 - 9.8t. Setting this to zero gives 15−9.8t=0  ⟹  t=159.8≈1.530615 - 9.8t = 0 \implies t = \frac{15}{9.8} \approx 1.5306 seconds. To find the maximum height, substitute this time back into the y(t)y(t) equation: y(1.5306)=2+15(1.5306)−4.9(1.5306)2≈2+22.959−11.479≈13.48y(1.5306) = 2 + 15(1.5306) - 4.9(1.5306)^2 \approx 2 + 22.959 - 11.479 \approx 13.48 m. The closest answer is 13.5 m.

Question 18

The trajectory of a comet is modelled by x(t)=(t−5)3+100x(t) = (t-5)^3 + 100 and y(t)=e0.2ty(t) = e^{0.2t} relative to a star at the origin, for time tt in years. At t=5t=5, the comet's horizontal velocity is zero. What can be concluded about its overall motion at this instant?

  1. It is stationary.
  2. It is at its closest point to the star.
  3. It has reached its maximum height.
  4. It is moving vertically. (correct answer)
Explanation: When analyzing parametric motion, you need to examine both horizontal and vertical velocity components separately to understand the complete motion picture. The velocity components are found by taking derivatives of the position functions. For this comet's trajectory, the horizontal velocity is dxdt=3(t−5)2\frac{dx}{dt} = 3(t-5)^2 and the vertical velocity is dydt=0.2e0.2t\frac{dy}{dt} = 0.2e^{0.2t}. At t=5t = 5, the horizontal velocity becomes 3(5−5)2=03(5-5)^2 = 0, confirming the given information. However, the vertical velocity at t=5t = 5 is 0.2e0.2(5)=0.2e1≈0.540.2e^{0.2(5)} = 0.2e^1 \approx 0.54, which is definitely not zero. Since the comet has zero horizontal velocity but non-zero vertical velocity at this instant, it's moving purely in the vertical direction. This is answer D. Let's examine why the other options are incorrect: A) claims the comet is stationary, but this ignores that vertical motion continues even when horizontal motion stops. B) suggests this is the closest point to the star, but you'd need to calculate the actual distance to the origin and show it's minimized here. C) states maximum height is reached, but since y(t)=e0.2ty(t) = e^{0.2t} is always increasing (its derivative is always positive), the comet never reaches a maximum height. Remember: in parametric motion problems, always check both velocity components separately. Zero velocity in one direction doesn't mean the object stops moving entirely—it might just be changing direction or moving along the other axis.

Question 19

The trajectory of a comet is modelled by x(t)=(t−5)3+100x(t) = (t-5)^3 + 100 and y(t)=e0.2ty(t) = e^{0.2t} relative to a star at the origin, for time tt in years. At t=5t=5, the comet's horizontal velocity is zero. What can be concluded about its overall motion at this instant?

  1. It is stationary.
  2. It is at its closest point to the star.
  3. It has reached its maximum height.
  4. It is moving vertically. (correct answer)
Explanation: When analyzing parametric motion, you need to examine both horizontal and vertical velocity components separately to understand the complete motion picture. The velocity components are found by taking derivatives of the position functions. For this comet's trajectory, the horizontal velocity is dxdt=3(t−5)2\frac{dx}{dt} = 3(t-5)^2 and the vertical velocity is dydt=0.2e0.2t\frac{dy}{dt} = 0.2e^{0.2t}. At t=5t = 5, the horizontal velocity becomes 3(5−5)2=03(5-5)^2 = 0, confirming the given information. However, the vertical velocity at t=5t = 5 is 0.2e0.2(5)=0.2e1≈0.540.2e^{0.2(5)} = 0.2e^1 \approx 0.54, which is definitely not zero. Since the comet has zero horizontal velocity but non-zero vertical velocity at this instant, it's moving purely in the vertical direction. This is answer D. Let's examine why the other options are incorrect: A) claims the comet is stationary, but this ignores that vertical motion continues even when horizontal motion stops. B) suggests this is the closest point to the star, but you'd need to calculate the actual distance to the origin and show it's minimized here. C) states maximum height is reached, but since y(t)=e0.2ty(t) = e^{0.2t} is always increasing (its derivative is always positive), the comet never reaches a maximum height. Remember: in parametric motion problems, always check both velocity components separately. Zero velocity in one direction doesn't mean the object stops moving entirely—it might just be changing direction or moving along the other axis.

Question 20

A drone's flight path is modelled by the parametric equations x(t)=2t+sin⁡(t)x(t) = 2t + \sin(t) and y(t)=10−t2y(t) = 10 - t^2, where t≥0t \ge 0 is the time in seconds, and xx and yy are displacements in meters. What is the speed of the drone at t=3t=3 seconds, correct to three significant figures?

  1. 6.00 m/s
  2. 6.08 m/s (correct answer)
  3. 6.22 m/s
  4. 7.01 m/s
Explanation: The velocity components are found by differentiating the position equations with respect to tt: dxdt=2+cos⁡(t)\frac{dx}{dt} = 2 + \cos(t) and dydt=−2t\frac{dy}{dt} = -2t. At t=3t=3, dxdt=2+cos⁡(3)≈1.0100\frac{dx}{dt} = 2 + \cos(3) \approx 1.0100 and dydt=−2(3)=−6\frac{dy}{dt} = -2(3) = -6. The speed is the magnitude of the velocity vector: Speed =(dxdt)2+(dydt)2=(1.0100)2+(−6)2≈1.0201+36=37.0201≈6.084= \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} = \sqrt{(1.0100)^2 + (-6)^2} \approx \sqrt{1.0201 + 36} = \sqrt{37.0201} \approx 6.084 m/s. To three significant figures, this is 6.08 m/s.