Parametric Functions Modeling Planar Motion - AP PreCalculus Study Guide

The Core Idea: Parametric Functions Modeling Planar Motion

Parametric functions provide a powerful way to describe the motion of an object in a two-dimensional plane. Instead of defining the object's vertical position ($y$) as a function of its horizontal position ($x$), we define both $x$ and $y$ as separate functions of a third variable, typically time ($t$). This approach allows us to determine the precise coordinates of the object, $(x(t),y(t))$, at any given moment.

This topic extends the concept of position to include motion. By analyzing the rates of change of the position functions, $x^{\prime }(t)$ and $y^{\prime }(t)$, we can determine the object's velocity—a vector that tells us both how fast and in what direction the object is moving. From this velocity vector, we can then calculate the object's speed, which is a scalar quantity representing the overall magnitude of its motion at an instant in time.

Key Formulas

The study of planar motion relies on three fundamental concepts, each with a specific formula derived from the parametric position functions $x(t)$ and $y(t)$.

1. Position: The location of an object in the plane at a specific time $t$ is given by an ordered pair.

   $$\text{Position}=(x(t),y(t))$$

2. Velocity Vector: The velocity of the object is a vector quantity that describes the instantaneous rate of change of its horizontal and vertical positions. The components are $x^{\prime }(t)$ and $y^{\prime }(t)$.

   $$\text{Velocity}=⟨x^{\prime }(t),y^{\prime }(t)⟩$$

   • $x^{\prime }(t)$ represents the horizontal velocity.

   • $y^{\prime }(t)$ represents the vertical velocity.

3. Speed: The speed of the object is the magnitude of its velocity vector. It is a scalar quantity, meaning it has a size but no direction. It is calculated using a formula analogous to the Pythagorean theorem.

   $$\text{Speed}=\sqrt{(x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2}$$

Understanding Position, Velocity, and Speed

It is critical to distinguish between the concepts of position, velocity, and speed, as they describe different aspects of an object's motion.

• Position is simply a location. The pair of functions $(x(t),y(t))$ tells you where the object is at time $t$. The output is a coordinate pair.

• Velocity describes how the position is changing. It is a vector, $⟨x^{\prime }(t),y^{\prime }(t)⟩$, which means it has both magnitude and direction. The signs of the components are crucial for interpretation:

  • If $x^{\prime }(t)>0$, the object is moving to the right.

  • If $x^{\prime }(t)<0$, the object is moving to the left.

  • If $y^{\prime }(t)>0$, the object is moving up.

  • If $y^{\prime }(t)<0$, the object is moving down.

• Speed describes how fast the object is moving, irrespective of its direction. It is the length (or magnitude) of the velocity vector and is always a non-negative scalar value. An object can have a high speed while moving left, right, up, down, or in any combination of those directions.

Core Concepts & Rules

• The motion of an object in a plane can be modeled by a pair of parametric equations, $x=x(t)$ and $y=y(t)$, where $t$ represents time.

• The ordered pair $(x(t),y(t))$ specifies the object's exact coordinates in the plane at time $t$.

• The velocity of the object is a vector, $⟨x^{\prime }(t),y^{\prime }(t)⟩$. The components $x^{\prime }(t)$ and $y^{\prime }(t)$ represent the instantaneous rates of change of the horizontal and vertical positions, respectively.

• The sign of $x^{\prime }(t)$ indicates horizontal direction (positive for right, negative for left), while the sign of $y^{\prime }(t)$ indicates vertical direction (positive for up, negative for down).

• Speed is a scalar quantity representing the magnitude of the velocity vector.

• The formula for speed at time $t$ is $\surd ((x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2)$. Speed can never be negative.

Step-by-Step Example 1: Calculating Position, Velocity, and Speed

Problem: The position of a particle moving in a plane is given by the parametric equations $x(t)=3t^2-5$ and $y(t)=2t+1$. The rates of change of the position are $x^{\prime }(t)=6t$ and $y^{\prime }(t)=2$. Find the position, velocity vector, and speed of the particle at time $t=2$.

Step 1: Find the position at $t=2$

Substitute $t=2$ into the position functions $x(t)$ and $y(t)$.

• $x(2)=3(2{)}^2-5=3(4)-5=12-5=7$

• $y(2)=2(2)+1=4+1=5$

The position of the particle at $t=2$ is $(7,5)$.

Step 2: Find the velocity vector at $t=2$

Substitute $t=2$ into the rate of change functions $x^{\prime }(t)$ and $y^{\prime }(t)$.

• $x^{\prime }(2)=6(2)=12$

• $y^{\prime }(2)=2$

The velocity vector at $t=2$ is $⟨12,2⟩$. This means the particle is moving to the right at a rate of 12 units per unit of time and up at a rate of 2 units per unit of time.

Step 3: Find the speed at $t=2$

Use the speed formula with the velocity components found in Step 2.

• $\text{Speed}=\sqrt{(x^{\prime }(2){)}^2+(y^{\prime }(2){)}^2}$

• $\text{Speed}=\sqrt{(12{)}^2+(2{)}^2}$

• $\text{Speed}=\sqrt{144+4}$

• $\text{Speed}=\sqrt{148}$

The speed of the particle at $t=2$ is $\surd 148$ units per unit of time.

Step-by-Step Example 2: Interpreting Motion from Given Rates

Problem: A remote-controlled car moves in a field. Its velocity vector at time $t$ is given by $⟨x^{\prime }(t),y^{\prime }(t)⟩$, where $x^{\prime }(t)=10-4t$ and $y^{\prime }(t)=3\cos (t)$. Describe the car's direction of motion and find its speed at time $t=5$.

Step 1: Determine the velocity vector at $t=5$

Calculate the values of $x^{\prime }(5)$ and $y^{\prime }(5)$.

• $x^{\prime }(5)=10-4(5)=10-20=-10$

• $y^{\prime }(5)=3\cos (5)$

Using a calculator in radian mode, $\cos (5)\approx 0.2837$.

• $y^{\prime }(5)\approx 3(0.2837)\approx 0.851$

The velocity vector at $t=5$ is approximately $⟨-10,0.851⟩$.

Step 2: Interpret the direction of motion

Analyze the signs of the velocity components.

• $x^{\prime }(5)=-10$, which is negative. This means the car is moving to the left.

• $y^{\prime }(5)\approx 0.851$, which is positive. This means the car is moving up.

At $t=5$, the car is moving to the left and up.

Step 3: Calculate the speed at $t=5$

Use the speed formula with the velocity components from Step 1.

• $\text{Speed}=\sqrt{(x^{\prime }(5){)}^2+(y^{\prime }(5){)}^2}$

• $\text{Speed}\approx \sqrt{(-10{)}^2+(0.851{)}^2}$

• $\text{Speed}\approx \sqrt{100+0.724}$

• $\text{Speed}\approx \sqrt{100.724}\approx 10.036$

The speed of the car at $t=5$ is approximately $10.036$ units per unit of time.

Using Your Calculator

A graphing calculator is essential for finding rates of change (derivatives) for complex functions and for performing calculations accurately.

Task: Given $x(t)=\sin (t^2)$ and $y(t)=\ln (t+1)$, find the speed of an object at $t=2$.

Step 1: Enter Parametric Mode

• Press the [MODE] key.

• Navigate down to the FUNCTION line and change it from FUNC to PAR (for Parametric).

Step 2: Calculate the Horizontal Velocity Component $x^{\prime }(2)$

• Go to the home screen.

• Use the numerical derivative function. On a TI-84, press [MATH] and select `nDeriv((or $8).

• Enter the expression, variable, and value: nDeriv(sin(T^2), T, 2)

• The calculator will return a value. Let's say it is $A\approx -2.625$.

  (Note: Use the variable $T$ as it corresponds to the parametric variable key $[X,T,\theta ,n]$)

Step 3: Calculate the Vertical Velocity Component $y^{\prime }(2)$

• Repeat the process for y(t)`. - Enter `nDeriv(ln(T+1), T, 2)`. - The calculator will return a value. Let's say it is $B \approx 0.333.

Step 4: Calculate the Speed

• On the home screen, use the values you found to calculate the magnitude of the velocity vector.

• Enter √((A)^2 + (B)^2).

• $\surd ((-2.625{)}^2+(0.333{)}^2)\approx \sqrt{6.891+0.111}=\sqrt{7.002}\approx 2.646$.

• The speed at $t=2$ is approximately $2.646$.

AP Exam Quick Hit

Common Question Types

• Calculating Speed at a Point: You will be given $x(t)$ and $y(t)$ and asked to find the speed at a specific time $t=c$. This is a direct application of the speed formula and often requires a calculator to find $x^{\prime }(c)$ and $y^{\prime }(c)$.

  • Example: "A particle's position is given by $x(t)=5\cos (t)$ and $y(t)=t^2$. What is the speed of the particle at $t=\pi$?"

• Describing Motion: You will be given information about $x^{\prime }(t)$ and $y^{\prime }(t)$ (either as functions, in a table, or at a specific point) and asked to describe the object's direction of motion.

  • Example: "The velocity of a particle is $⟨-4,1.5⟩$. At this instant, is the particle moving primarily horizontally or vertically? In which directions?"

Common Mistakes

• Confusing Velocity and Speed: Reporting the speed (a single number) when asked for the velocity (a vector), or vice versa. Remember: velocity is $⟨x^{\prime }(t),y^{\prime }(t)⟩$; speed is $\surd (...)$.

• Incorrectly Calculating Speed: Forgetting to square one or both of the velocity components inside the square root. For example, calculating $\surd (x^{\prime }(t)+y^{\prime }(t))$ instead of $\surd ((x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2)$.

• Sign Errors When Squaring: Forgetting that squaring a negative number results in a positive number. For example, if $x^{\prime }(t)=-4$, then $(x^{\prime }(t){)}^2=(-4{)}^2=16$, not $-16$. Speed can never be imaginary.

• Using Position Instead of Velocity: Plugging the position values $(x(t),y(t))$ into the speed formula instead of the velocity components $⟨x^{\prime }(t),y^{\prime }(t)⟩$.

• Calculator Mode Error: Using the calculator in Degree mode when the problem implies Radian mode (which is the standard for calculus-based questions involving trigonometric functions).