9.6 Solving Motion Problems Using Parametric and Vector-Valued Functions

Planar motion uses a position vector $\mathbf r(t)=\langle x(t), y(t)\rangle$, where the velocity vector is $\mathbf v(t)=\mathbf r'(t)$ and the acceleration vector is $\mathbf a(t)=\mathbf v'(t)$. To get displacement, integrate the velocity vector component by component; to get total distance traveled, integrate the speed $|\mathbf v(t)|$.

Planar motion with vector-valued and parametric functions is BC only and sits in Unit 9. It builds on derivatives, definite integrals, and the arc length idea from earlier units, applied to objects moving along curves in the plane.

Differentiating and Integrating Planar Motion Vectors

• Position is $\mathbf r(t)=\langle x(t), y(t)\rangle$; differentiate component by component to get velocity $\mathbf v(t)=\mathbf r'(t)$ and acceleration $\mathbf a(t)=\mathbf v'(t)$.
• Speed is the magnitude of velocity: $|\mathbf v(t)|=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}$.
• Displacement over $[a,b]$ is the definite integral of the velocity vector, taken one component at a time.
• Total distance traveled over $[a,b]$ is $\int_a^b |\mathbf v(t)|\, dt$, the integral of speed.
• To go from velocity back to position, integrate and use an initial condition like $\mathbf r(t_0)=\mathbf r_0$ to solve for the constant.
• Decide differentiation versus integration first: derivatives move toward velocity and acceleration, integrals move toward displacement, distance, and position.

Recap of Motion Terms

Position, velocity, acceleration, displacement, and distance traveled work the same way in the plane as on a line, just with two components. If you want a fuller refresher on the line version, this guide covers it.

Position

Position is where the object is in the coordinate plane at time $t$. For planar motion it is given by a vector $\mathbf r(t)=\langle x(t), y(t)\rangle$.

Velocity

Velocity is the rate of change of position with respect to time, and it carries both speed and direction. It is the derivative of position: $\mathbf v(t)=\mathbf r'(t)=\left\langle \frac{dx}{dt}, \frac{dy}{dt}\right\rangle$.

Speed

Speed is the magnitude of the velocity vector, so it is a single positive number with no direction:

$$|\mathbf v(t)|=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}$$

Acceleration

Acceleration is the rate of change of velocity with respect to time. It is the derivative of velocity and the second derivative of position:

$$\mathbf a(t)=\mathbf v'(t)=\mathbf r''(t)$$

Displacement

Displacement is the net change in position from start to end, regardless of the path taken. As a vector it equals the final position minus the initial position. You can find it by integrating the velocity vector over the time interval.

Two different paths: distance versus displacement. Image Courtesy of Wikipedia.

Distance Traveled

Distance traveled is the total length of the path the object covers. Unlike displacement, it ignores direction and just adds up all the ground covered.

Here is a quick analogy. Say you leave home, go a mile, realize you forgot your homework, and return. Your displacement is zero because you end where you started, but your distance traveled is 1 mile + 1 mile = 2 miles.

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Calculus with Vector-Valued Functions

A vector-valued function is written $\mathbf r(t)=\langle f(t), g(t)\rangle$, where $f(t)$ and $g(t)$ give the position at time $t$. Differentiating each component gives velocity and acceleration.

Take the position $\mathbf r(t)=\langle t^2,\sin(t)\rangle$. Differentiating each component gives the velocity vector $\mathbf v(t)=\langle 2t,\cos(t)\rangle$, so the velocity in the x-direction is $2t$ and in the y-direction is $\cos(t)$.

Differentiating again gives the acceleration vector $\mathbf a(t)=\langle 2,-\sin(t)\rangle$. The x-acceleration is a constant $2$ units (often $m/s^2$) and the y-acceleration is $-\sin(t)$.

You can always reverse this with integration to recover velocity or position, but you will need initial values to pin down the constant and get a particular solution.

Integrating to Find Displacement

To find displacement, take the definite integral of the velocity vector, one component at a time. Using $\mathbf v(t)=\langle3t^2,2t\rangle$, find the displacement from $t=2$ to $t=5$:

$$\Delta x=\int_2^5 3t^2 \, dt = [t^3]_2^5$$ $$\Delta y=\int_2^5 2t \, dt = [t^2]_2^5$$

Evaluating:

$$\Delta x=5^3-2^3=125-8=117$$ $$\Delta y=5^2-2^2=25-4=21$$

So the displacement vector is $\langle 117,21 \rangle$, meaning the object ended 117 units in the positive x-direction and 21 units in the positive y-direction from where it started.

An example of using the area under the velocity-time graph to calculate displacement. Image courtesy of Resourceaholic.

Using the Speed Integral to Find Distance Traveled

Distance traveled is the total length of the path, which is exactly the arc length of the curve. For a vector-valued function the formula is

$$S=\int_a^b |\mathbf r'(t)|\, dt$$

where $\mathbf r'(t)$ is the velocity vector. In words, integrate the magnitude of the velocity, which is the speed. Note that this is the same idea as integrating speed over time. For a more detailed look at how arc length connects to distance traveled, see this guide.

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Calculus with Parametric Functions

Parametric functions work the same way; the only difference is notation. A parametric description uses the parameter $t$ to drive each coordinate:

$$x(t)=f(t), \quad y(t)=g(t)$$

Each coordinate is its own function of $t$, which is just the component form of $\mathbf r(t)$. Take $x(t)=4t^3$ and $y(t)=2t^4$. Differentiate each with respect to $t$ to get velocity:

$$\frac{dx}{dt}=12t^2, \quad \frac{dy}{dt}=8t^3$$

Differentiate again for acceleration:

$$\frac{d}{dt}(12t^2)=24t, \quad \frac{d}{dt}(8t^3)=24t^2$$

Each of these tells you how fast that coordinate is changing and in which direction.

Integrating Parametric Functions to Find Displacement

Just like with vector-valued functions, integrate the $x$ and $y$ rates separately. Take $x(t)=t-3t^2$ and $y(t)=t^2+4t^3$ and find the displacement from $t=1$ to $t=3$:

$$\Delta x = \int_1^3 (t-3t^2)\, dt, \quad \Delta y = \int_1^3 (t^2+4t^3)\, dt$$ $$\Delta x = \left[\frac{t^2}{2}-t^3\right]_1^3, \quad \Delta y = \left[\frac{t^3}{3}+t^4\right]_1^3$$ $$\Delta x = -22, \quad \Delta y = \frac{266}{3}$$

So the final x-position is 22 units left of the start, and the final y-position is $\frac{266}{3}$ units above the start.

Using Parametric Arc Length to Find Distance Traveled

The parametric arc length formula looks a little different from the vector form, but it is really the same thing. For more on where it comes from, see this guide on parametric arc length. The formula is:

$$S=\int_a^b \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt, \quad a\le t \le b$$

Read one operation at a time: you are finding the magnitude of the velocity (the speed) and integrating it over the interval. That is exactly $\int_a^b |\mathbf r'(t)|\, dt$ in component form.

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Choosing Between Derivatives and Integrals in Motion Problems

Problem Solving: Start every motion problem by asking which direction you are moving: toward derivatives or toward integrals.

• Given position, differentiate to get velocity, then differentiate again for acceleration.
• Given velocity, integrate for displacement or position, or take the magnitude for speed.
• Need total distance? Integrate the speed $\int_a^b |\mathbf v(t)|\, dt$.

Free Response:

• When you need position from velocity, write the antiderivative and then use the initial condition, such as $\mathbf r(t_0)=\mathbf r_0$, to solve for the constant in each component.
• For displacement, set up $\int_a^b \mathbf v(t)\, dt$ and handle the $x$ and $y$ components separately.
• Distance traveled problems usually let you use a calculator for the integral, so the main work is writing $\int_a^b \sqrt{(x'(t))^2+(y'(t))^2}\, dt$ correctly.
• Label answers with units and direction so your meaning is clear and easy to check.

Common Trap: Speed is a scalar, so do not integrate the velocity components separately and then try to call that distance. Distance traveled requires integrating the magnitude $|\mathbf v(t)|$, which combines both components under a square root before integrating.

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Worked Example

A particle moves along a curve defined by $\mathbf r(t)=\langle t, t^2 \rangle$ (in meters). Determine the total distance traveled from $t=1$ second to $t=3$ seconds.

Solution

This is a distance traveled problem, so use $S=\int_a^b |\mathbf r'(t)|\, dt$. First differentiate the position vector to get velocity:

$$\mathbf v(t) = \langle 1, 2t \rangle$$

Next find the magnitude using $\sqrt{x^2+y^2}$ (this is the Pythagorean Theorem applied to the components):

$$|\mathbf v(t)| = \sqrt{1^2+(2t)^2} = \sqrt{1+4t^2}$$

Notice this is the same as the parametric arc length setup, since $\sqrt{(x'(t))^2+(y'(t))^2}$ is just the speed. Now integrate the speed from $t=1$ to $t=3$:

$$S=\int_1^3 \sqrt{1+4t^2}\, dt$$

Evaluating with a calculator gives approximately $8.268$ meters.

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Common Misconceptions

• Misconception: Displacement is distance traveled.
  Actually: Displacement is not distance traveled. Displacement is the net change in position (integrate the velocity vector); distance traveled is the total path length (integrate the speed $|\mathbf v(t)|$). The magnitude of displacement equals distance traveled only when the object moves along a straight line without reversing direction.
• Misconception: Speed is a vector.
  Actually: Speed is not a vector. Speed is the magnitude of velocity, a single nonnegative number. Velocity has direction and components; speed does not.
• Misconception: You cannot integrate components and take the magnitude later for distance.
  Actually: Take the magnitude first, then integrate. Order matters because $\sqrt{\int (\,)^2}$ is not the same as $\int \sqrt{(\,)^2}$.
• Misconception: Integrating velocity gives the actual position without an initial condition.
  Actually: Recovering position needs an initial condition. Integrating velocity gives position only up to a constant vector. Without an initial value like $\mathbf r(t_0)=\mathbf r_0$, you only have the general solution, not the actual position.
• Misconception: The slope along a parametric path is $\frac{dy}{dt}$, and acceleration is the second derivative of $y$ with respect to $x$.
  Actually: For parametric motion the slope along the path is $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$, not $\frac{dy}{dt}$ alone, and acceleration is not just the second derivative of $y$ with respect to $x$.

Related AP Calculus Guides

• 9.1 Defining and Differentiating Parametric Equations
• 9.4 Defining and Differentiating Vector-Valued Functions
• Unit 9 Review: Parametric Equations, Polar Coordinates, and Vector-Valued Functions
• 9.2 Second Derivatives of Parametric Equations
• 9.3 Finding Arc Lengths of Curves Given by Parametric Equations
• 9.7 Defining Polar Coordinates and Differentiating in Polar Form