Parametric Models - IB Mathematics Applications and Interpretation

1. A ball's movement on a table is modelled by $x(t) = 10t - 2t^2$ and $y(t) = 8t$, for $t \ge 0$, with distances in cm. The table has a wall at $x=12$. Find the speed of the ball at the moment it first hits this wall.
   Answer: 8.25 cm/s First, find the time $t$ when the ball hits the wall by setting $x(t) = 12$: $10t - 2t^2 = 12 \implies 2t^2 - 10t + 12 = 0 \implies t^2 - 5t + 6 = 0 \implies (t-2)(t-3) = 0$. The first time is $t=2$. Next, find the speed at $t=2$. The velocity components are $\frac{dx}{dt} = 10 - 4t$ and $\frac{dy}{dt} = 8$. At $t=2$, $\frac{dx}{dt} = 10 - 4(2) = 2$ and $\frac{dy}{dt} = 8$. The speed is $\sqrt{2^2 + 8^2} = \sqrt{4 + 64} = \sqrt{68} \approx 8.25$ cm/s.
2. A drone's flight path is modelled by the parametric equations $x(t) = 2t + \sin(t)$ and $y(t) = 10 - t^2$, where $t \ge 0$ is the time in seconds, and $x$ and $y$ are displacements in meters. What is the speed of the drone at $t=3$ seconds, correct to three significant figures?
   Answer: 6.08 m/s The velocity components are found by differentiating the position equations with respect to $t$: $\frac{dx}{dt} = 2 + \cos(t)$ and $\frac{dy}{dt} = -2t$. At $t=3$, $\frac{dx}{dt} = 2 + \cos(3) \approx 1.0100$ and $\frac{dy}{dt} = -2(3) = -6$. The speed is the magnitude of the velocity vector: Speed $= \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} = \sqrt{(1.0100)^2 + (-6)^2} \approx \sqrt{1.0201 + 36} = \sqrt{37.0201} \approx 6.084$ m/s. To three significant figures, this is 6.08 m/s.
3. A financial model tracks two economic indicators, I1 and I2, over time $t$ (in years, $t \ge 0$). Their values are given by $x(t) = 100 + 5t$ and $y(t) = 50e^{0.1t} \ln(t+1)$. Find the rate of change of indicator I2 with respect to indicator I1 at $t=5$ years.
   Answer: 5.70 The rate of change of I2 with respect to I1 is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. First, $\frac{dx}{dt} = 5$. For $\frac{dy}{dt}$, we use the product rule: $\frac{dy}{dt} = (50e^{0.1t})'\ln(t+1) + 50e^{0.1t}(\ln(t+1))' = 5e^{0.1t}\ln(t+1) + \frac{50e^{0.1t}}{t+1}$. At $t=5$, $\frac{dx}{dt}=5$ and $\frac{dy}{dt} = 5e^{0.5}\ln(6) + \frac{50e^{0.5}}{6} \approx 14.773 + 13.739 = 28.512$. Therefore, $\frac{dy}{dx} = \frac{28.512}{5} \approx 5.702$.
4. A firework is launched and its trajectory is given by $x(t) = 30t$ and $y(t) = 50t - 4.9t^2$, for $t \ge 0$. Find the angle, in degrees, that the firework's path makes with the horizontal at $t=4$ seconds.
   Answer: 19.8° The angle of the path is the angle of the velocity vector. First find the velocity components: $\frac{dx}{dt} = 30$ and $\frac{dy}{dt} = 50 - 9.8t$. At $t=4$, $\frac{dx}{dt} = 30$ and $\frac{dy}{dt} = 50 - 9.8(4) = 10.8$. The angle $\theta$ with the horizontal is given by $\tan(\theta) = \frac{dy/dt}{dx/dt} = \frac{10.8}{30} = 0.36$. Therefore, $\theta = \arctan(0.36) \approx 19.8^{\circ}$.
5. At which other point do their paths cross?
   Answer: (1, 0) 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 $x_P(t) = x_Q(s)$ and $y_P(t) = y_Q(s)$ simultaneously. Setting up the equations: $t + 1 = 2s - 1$ and $t^2 = 4s - 4$. From the first equation: $t = 2s - 2$. Substituting into the second: $(2s - 2)^2 = 4s - 4$. Expanding: $4s^2 - 8s + 4 = 4s - 4$, which simplifies to $4s^2 - 12s + 8 = 0$, or $s^2 - 3s + 2 = 0$. Factoring gives $(s-1)(s-2) = 0$, so $s = 1$ or $s = 2$. When $s = 1$: $t = 0$, giving point $(1, 0)$ When $s = 2$: $t = 2$, giving point $(3, 4)$ Since we know the collision occurs at $(3, 4)$, the other intersection point is $(1, 0)$, making D correct. Answer A gives $(-1, -4)$, which doesn't satisfy either parametric equation. Answer B incorrectly assumes only one intersection exists. Answer C provides $(2, 1)$, 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.
6. The position of a ship is given by $x(t) = 5t - 20$ and $y(t) = 3t - 30$, where $t$ is hours after noon and distances are in kilometres. A lighthouse is at the origin (0,0). At what time $t$ is the ship closest to the lighthouse?
   Answer: 5.59 hours The distance $D$ from the origin is given by $D = \sqrt{x(t)^2 + y(t)^2}$. To minimize $D$, we can minimize its square, $D^2 = (5t-20)^2 + (3t-30)^2$. Expanding this gives $D^2 = (25t^2 - 200t + 400) + (9t^2 - 180t + 900) = 34t^2 - 380t + 1300$. This is a quadratic in $t$, and its minimum occurs at the vertex, $t = -\frac{b}{2a} = -\frac{-380}{2(34)} = \frac{380}{68} \approx 5.588$. The closest time is 5.59 hours.
7. The path of a remote-controlled car is given by $x(t) = t^3 - 12t + 5$ and $y(t) = t^2 - 8t$, for $t \ge 0$. For what positive value of $t$ is the car momentarily moving only horizontally?
   Answer: 4.00 Moving only horizontally means the vertical component of velocity, $\frac{dy}{dt}$, is zero, while the horizontal component, $\frac{dx}{dt}$, is non-zero. The vertical velocity is $\frac{dy}{dt} = 2t - 8$. Setting this to zero gives $2t - 8 = 0 \implies t=4$. We must check that the horizontal velocity is not zero at this time. $\frac{dx}{dt} = 3t^2 - 12$. At $t=4$, $\frac{dx}{dt} = 3(4^2) - 12 = 48 - 12 = 36 \neq 0$. Thus, the car is moving horizontally at $t=4$.
8. Particle A moves along a path defined by $x_A(t) = t+1$ and $y_A(t) = t^2$. Particle B moves along a path defined by $x_B(t) = 2t-1$ and $y_B(t) = 4t-4$. Time $t \ge 0$. At what point do the particles collide?
   Answer: (3, 4) A collision occurs if the particles are at the same position at the same time $t$. We set their coordinates equal: $x_A(t) = x_B(t) \implies t+1 = 2t-1 \implies t=2$. We must verify that their y-coordinates are also equal at this time: $y_A(2) = 2^2 = 4$ and $y_B(2) = 4(2)-4 = 4$. Since both coordinates match at $t=2$, a collision occurs. The location of the collision is found by substituting $t=2$ into either set of equations: $x(2) = 2+1 = 3$ and $y(2) = 2^2 = 4$. The collision point is (3, 4).
9. A particle's motion is given by $x(t) = \sin(2t)$ and $y(t) = \cos(t)$. Which statement correctly describes the acceleration of the particle at $t=\pi/4$?
   Answer: The particle is accelerating to the left and downwards. Acceleration is found by taking the second derivative of the position functions. Velocity: $\frac{dx}{dt} = 2\cos(2t)$, $\frac{dy}{dt} = -\sin(t)$. Acceleration: $\frac{d^2x}{dt^2} = -4\sin(2t)$, $\frac{d^2y}{dt^2} = -\cos(t)$. At $t=\pi/4$: $\frac{d^2x}{dt^2} = -4\sin(2(\pi/4)) = -4\sin(\pi/2) = -4$. This is negative, so acceleration is to the left. $\frac{d^2y}{dt^2} = -\cos(\pi/4) = -\frac{\sqrt{2}}{2}$. This is negative, so acceleration is downwards. Therefore, the particle is accelerating to the left and downwards.
10. A robot vacuum cleaner moves on a large floor. Its path is modelled by $x(t) = 3\cos(0.5t)$ and $y(t) = 5\sin(0.5t)$ for $0 \le t \le 10$, where $t$ is in seconds and distances are in metres. Calculate the total distance the robot travels in the first 10 seconds.
    Answer: 21.0 m The total distance travelled is the integral of the speed. First, find the velocity components: $\frac{dx}{dt} = -1.5\sin(0.5t)$ and $\frac{dy}{dt} = 2.5\cos(0.5t)$. The speed is $S(t) = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} = \sqrt{(-1.5\sin(0.5t))^2 + (2.5\cos(0.5t))^2}$. The total distance is $D = \int_{0}^{10} S(t) \,dt$. Using a calculator's numerical integration function, $D = \int_{0}^{10} \sqrt{2.25\sin^2(0.5t) + 6.25\cos^2(0.5t)} \,dt \approx 21.01$ m. The closest answer is 21.0 m.
11. The path of a thrown javelin is modelled by $x(t) = 25t$ and $y(t) = 2 + 15t - 4.9t^2$, where $t \ge 0$ is time in seconds and distances are in meters. What is the maximum height reached by the javelin?
    Answer: 13.5 m The maximum height occurs when the vertical velocity, $\frac{dy}{dt}$, is zero. $\frac{dy}{dt} = 15 - 9.8t$. Setting this to zero gives $15 - 9.8t = 0 \implies t = \frac{15}{9.8} \approx 1.5306$ seconds. To find the maximum height, substitute this time back into the $y(t)$ equation: $y(1.5306) = 2 + 15(1.5306) - 4.9(1.5306)^2 \approx 2 + 22.959 - 11.479 \approx 13.48$ m. The closest answer is 13.5 m.
12. The path of a particle is given by $x(t) = \sqrt{t+1}$ and $y(t) = t-1$ for $t \ge 0$. Which of the following is the Cartesian equation for the path with its correct domain?
    Answer: $y = x^2 - 2$ for $x \ge 1$ 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 $t$ from $x(t) = \sqrt{t+1}$ and $y(t) = t-1$, start with the simpler equation. From $y = t-1$, you get $t = y+1$. Substituting this into the x-equation: $x = \sqrt{(y+1)+1} = \sqrt{y+2}$. Squaring both sides gives $x^2 = y+2$, so $y = x^2 - 2$. Now for the crucial domain step. Since $t \geq 0$, you have $x = \sqrt{t+1} \geq \sqrt{0+1} = 1$. Therefore, $x \geq 1$. Option A is wrong because it claims the domain is all real $x$, but we've shown $x$ must be at least 1. Option B incorrectly states $x \geq 0$ when the actual constraint is $x \geq 1$. This is a common error—students sometimes assume that since $t \geq 0$, then $x \geq 0$, but you must calculate the actual minimum value of $x$. Option C has the wrong function entirely ($y = x - 2$ instead of $y = x^2 - 2$), likely from incorrectly manipulating the parametric equations. Option D correctly gives $y = x^2 - 2$ for $x \geq 1$. 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.
13. The path of an object is given parametrically by $x(t) = 3\sec(t)$ and $y(t) = 2\tan(t)$ for $0 \le t < \pi/2$. Which of the following is the Cartesian equation of the path?
    Answer: $\frac{x^2}{9} - \frac{y^2}{4} = 1$ To find the Cartesian equation, we eliminate the parameter $t$. We use the trigonometric identity $\sec^2(t) - \tan^2(t) = 1$. From the parametric equations, we have $\sec(t) = \frac{x}{3}$ and $\tan(t) = \frac{y}{2}$. Substituting these into the identity gives $(\frac{x}{3})^2 - (\frac{y}{2})^2 = 1$, which simplifies to $\frac{x^2}{9} - \frac{y^2}{4} = 1$.
14. A particle's path is given by $x(t) = t^2 - 7t + 10$ and $y(t) = 4t - t^2$, for $t \ge 0$. Find the speed of the particle the second time it crosses the x-axis.
    Answer: 4.12 The particle crosses the x-axis when $y(t) = 0$. So, $4t - t^2 = 0 \implies t(4-t) = 0$, which gives $t=0$ and $t=4$. The second time is $t=4$. Now we find the speed at $t=4$. The velocity components are $\frac{dx}{dt} = 2t - 7$ and $\frac{dy}{dt} = 4 - 2t$. At $t=4$, $\frac{dx}{dt} = 2(4) - 7 = 1$ and $\frac{dy}{dt} = 4 - 2(4) = -4$. The speed is $\sqrt{1^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17} \approx 4.12$.
15. A particle's velocity components are given by $\frac{dx}{dt} = t-3$ and $\frac{dy}{dt} = \sqrt{t}$ for $t \ge 0$. At what time $t$ is its speed at a minimum?
    Answer: 2.50 The speed $S(t)$ is given by $S(t) = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} = \sqrt{(t-3)^2 + (\sqrt{t})^2} = \sqrt{t^2 - 6t + 9 + t} = \sqrt{t^2 - 5t + 9}$. To find the minimum speed, we can minimize the function inside the square root, $f(t) = t^2 - 5t + 9$. This is a parabola opening upwards. Its minimum occurs at the vertex, where $t = -\frac{b}{2a} = -\frac{-5}{2(1)} = 2.5$. So the minimum speed occurs at $t=2.5$.
16. The motion of a particle is described by $x(t) = 5\cos(\pi t)$ and $y(t) = t^2-4t$, for $t \ge 0$. Which of the following describes the particle's motion at $t=1.5$?
    Answer: Moving right and downwards We need to determine the signs of the velocity components at $t=1.5$. $\frac{dx}{dt} = -5\pi \sin(\pi t)$ and $\frac{dy}{dt} = 2t - 4$. At $t=1.5$: $\frac{dx}{dt} = -5\pi \sin(1.5\pi) = -5\pi(-1) = 5\pi > 0$, which means the particle is moving to the right. $\frac{dy}{dt} = 2(1.5) - 4 = 3 - 4 = -1 < 0$, which means the particle is moving downwards. Thus, the particle is moving right and downwards.
17. The trajectory of a comet is modelled by $x(t) = (t-5)^3 + 100$ and $y(t) = e^{0.2t}$ relative to a star at the origin, for time $t$ in years. At $t=5$, the comet's horizontal velocity is zero. What can be concluded about its overall motion at this instant?
    Answer: It is moving vertically. 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 $\frac{dx}{dt} = 3(t-5)^2$ and the vertical velocity is $\frac{dy}{dt} = 0.2e^{0.2t}$. At $t = 5$, the horizontal velocity becomes $3(5-5)^2 = 0$, confirming the given information. However, the vertical velocity at $t = 5$ is $0.2e^{0.2(5)} = 0.2e^1 \approx 0.54$, 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 $y(t) = e^{0.2t}$ 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.
18. The path of a particle is given by $x(t) = t^3 - 6t$ and $y(t) = e^{0.5t}$ for $t \ge 0$. Find the slope of the tangent to the path at $t=2$.
    Answer: 0.227 The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. First, we find the derivatives: $\frac{dx}{dt} = 3t^2 - 6$ and $\frac{dy}{dt} = 0.5e^{0.5t}$. At $t=2$, $\frac{dx}{dt} = 3(2^2) - 6 = 12 - 6 = 6$ and $\frac{dy}{dt} = 0.5e^{0.5(2)} = 0.5e \approx 1.359$. The slope is $\frac{1.359}{6} \approx 0.2265$. To three significant figures, the slope is 0.227.
19. The position of a hovercraft is modelled by $x(t) = \frac{100}{t+5}$ and $y(t) = 20\ln(t+1)$ for $t \ge 0$. Find the initial velocity vector of the hovercraft.
    Answer: (-4, 20) Initial velocity means the velocity at $t=0$. First, find the velocity components by differentiation: $x(t) = 100(t+5)^{-1} \implies \frac{dx}{dt} = -100(t+5)^{-2} = \frac{-100}{(t+5)^2}$. $y(t) = 20\ln(t+1) \implies \frac{dy}{dt} = \frac{20}{t+1}$. Now evaluate at $t=0$: $\frac{dx}{dt}(0) = \frac{-100}{(0+5)^2} = -4$. $\frac{dy}{dt}(0) = \frac{20}{0+1} = 20$. The initial velocity vector is $\langle -4, 20 \rangle$.

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