Derivatives and Motion
In This Chapter
- What is a position equation?
- The relationship between position, velocity, and acceleration
- Speed versus velocity
- Understanding projectile motion
Mathematics can actually be applied in the real world. This may shock and appall you, but it’s true. It’s probably shocking because most of the problems we’ve dealt with have been purely computational in nature, devoid of correlation to real life. (For example, estimating gas mileage is a useful mathematical real-life skill, whereas factoring difference of perfect cube polynomials is not as useful in a straightforward way.) Most people hate real-life application problems because they are (insert scary wolf howl here) word problems!
Factoring and equation solving may be rote, repetitive, and a little boring, but at least they’re predictable. How many nights have you gone to sleep haunted by problems like this: “If Train A is going from Pittsburgh to Los Angeles at a rate of 110 kilometers per hour, Train B is traveling 30 kilometers less than half the number of male passengers in Train A, and the heading of Train B is 3 degrees less than the difference of the prices of a club sandwich on each train, then at what time will the conductor of the first train remember that he forgot to set the DVR to record Jeopardy?”
Position equations are a nice transition into calculus word problems. Even though they are slightly bizarre, they follow clear patterns. Furthermore, they give you the chance to show off your new derivative skills.
A position equation is an equation that mathematically models something in real life. Specifically, it gives the position of an object at a specified time. Different books and teachers use different notation, but I will always indicate a position equation with the notation s(t), to stay consistent. By plugging values of t into the equation, you can determine where the object in question was at that moment in time. Just in case you’re starting to get stressed out, I’ll insert something cute and cuddly into the mix—my cat, Peanut.
Definition
A position equation is a mathematical model that outputs an object’s position at a given time t. Position is usually given with relation to some fixed landmark, like the ground or the origin, so that a negative position means something. For example, s(5) = –6 may mean that the object in question is 6 feet below the origin after 5 seconds have passed.
Peanut pretty much has the run of my basement, and her favorite pastime (apart from her strange habit of chewing on my eyeglasses) is batting a ball back and forth along one of the basement walls. For the sake of ease, let’s say the wall in question is 20 feet long; we’ll call the exact middle of the wall position 0, the left edge of the wall (our left, not her left) position –10, and the right edge of the wall position 10, as in Figure 12.1.
Figure 12.1
The domain of Peanut the cat. For sake of reference, I have labeled the middle and edges of the room.
Let’s examine the kitty’s position versus time in a simple example. We’ll keep returning to this example throughout the chapter as we compound our knowledge of derivatives and motion (and cat recreation).
Example 1: During the first four seconds of a particularly frisky playtime, Peanut’s position (in feet at time = t seconds) along the wall is given by the equation s(t) = t3 – 3t2 – 2t + 1. Evaluate and explain what is meant by s(0), s(2), and s(4).
Solution: Plug each number into s(t). A positive answer means she is toward the right of the room, whereas a negative answer means she is to the left of center. The larger the number, the farther she is to the right or left:
Kelley’s Cautions
The position given by s(t) in Example 1 is the horizontal position of the cat—had it meant vertical position, that negative answer would have been disturbing. Every position problem should infer what is meant by its output and will usually include units (such as feet and seconds) as well.
Therefore, when t = 0 (i.e., before you start measuring elapsed time), s(0) = 1 tells you the cat began 1 foot to the right of the center. Two seconds later (t = 2), she had used her lightning-fast kitty movements to travel 8 feet left, meaning she was then only 3 feet from the left wall, and 7 feet left of center. Two seconds after that (t = 4), she had moved 16 feet right, now only 1 foot away from the right-hand wall. That is one fast-moving cat.
Critical Point
The value s(0) is often called initial position, because it gives the position of the object before you start measuring time. Similarly, v(0) and a(0) are the initial velocity and initial acceleration.
There’s nothing really fancy about the position equation; given a time input, it tells you where the object was at that moment. Notice that the position equation in Example 1 is a nice, continuous, and differentiable polynomial. You can find the derivative awfully easily, but what does the derivative of the position equation represent?
You’ve got problems
Problem 1: A particle moves vertically (in inches) along the y-axis according to the position equation 
, where t represents seconds. At what time(s) is the particle 30 inches below the origin?
Velocity
Remember that the derivative describes the rate of change of a function. Therefore, s′(t) describes the velocity of the object in question at any given instant. It makes sense that velocity is equivalent to the rate of change of position, because velocity measures how quickly you move from one position to another. Speed also measures how quickly something moves, but speed and velocity are not the same thing.
Critical Point
If an object is moving downward at a rate of 15 feet per second, you could say that its velocity is –15 feet/second, whereas its speed is 15 feet/second. Speed is always the absolute value of velocity.
Velocity combines an object’s speed with its direction, whereas speed just gives you the rate at which the object is traveling. Practically speaking, this means that velocity can be negative, but speed cannot. What does a negative velocity mean? It depends on the problem. In a horizontal motion problem (like the Peanut the cat problem), it means velocity towards the left (because the left was defined as the negative direction). In a vertical motion problem, a negative velocity typically means that the object is dropping.
To find the velocity of an object at any instant, calculate the derivative and plug in the desired time for t. If, however, you want an object’s average velocity (i.e., average rate of change), remember that this value comes from the slope of the secant line. Remember how quickly Peanut was darting around in Example 1? Let’s get those exact speeds and velocities using the derivative.
Example 2: Peanut the cat’s position, in feet, for 0 ≤ t ≤ 4 seconds is given by s(t) = t3 –3t2 – 2t + 1. Find her velocity and speed at times t = 1 and t = 3.5 seconds, and give her average velocity over the t interval [1,3.5].
Solution: There are lots of parts to this problem, but none are hard. Start by calculating her velocity at the given times. Remember that the velocity is the first derivative of the position equation, so s′(t) = v(t) = 3t2 – 6t – 2:
Peanut is moving at a speed of 5 feet/second to the left (because the velocity is negative) at t = 1 second, and she is moving much faster, at a speed of 13.75 feet/second to the right, when t = 3.5 seconds. Now to find the average velocity on [1,3.5]—it’s equal to the slope of the line segment connecting the points on the position graph where t = 1 and t = 3.5. To find these points, plug those values into s(t):
Kelley’s Cautions
The slope of a position equation’s tangent line equals the instantaneous velocity at the point of tangency. The slope of a position equation’s secant line gives the average velocity over that interval. Notice that instantaneous and average rates of change are both based on linear slopes drawn on the position equation, not its derivative.
Calculate the secant slope using the points (1,–3) and (3.5, 0.125):
Therefore, even though she runs left and right at varying speeds over the time interval [1,3.5], she averages a rightward speed of 1.25 feet/second.
You’ve got problems
Problem 2: A particle moves vertically (in inches) along the y-axis according to the position equation 
, where t represents seconds. Rank the following from least to greatest: the speed when t = 3, the velocity when t = 7, and the average velocity on the interval [2,6].
Acceleration
As velocity is to position, so is acceleration to velocity. In other words, acceleration is the rate of change of velocity. Think about it—if you’re driving in a car that suddenly speeds up, the sense of being pushed back in your seat is due to the effects of acceleration. It is not the high rate of speed that makes roller coasters so scary. Aside from their height, it is the sudden acceleration and deceleration of the rides that causes the passengers to experience dizzying effects (and occasionally their previous meal).
Critical Point
The units for acceleration will be the same as the units for velocity, except the denominator will be squared. For example, if velocity is measured in feet per second (ft/sec), then acceleration is measured in feet per second per second, or ft/sec2.
To calculate the acceleration of an object, evaluate the second derivative of the position equation (or the first derivative of velocity). To calculate average acceleration, find the slope of the secant line on the velocity function (for the same reasons that average velocity is the secant slope on the position function). Let’s head back to the cat of mathematical mysteries one last time.
Critical Point
If the first derivative of position represents velocity and the second derivative represents acceleration, the third derivative represents “jerk,” the rate of change of acceleration. Think of jerk as that feeling you get as you switch gears in your car and the acceleration changes. I’ve never seen a problem concerning jerk, but I have known a few mathematicians who were pretty jerky.
Example 3: Peanut the cat’s position, in feet, at any time 0 ≤ t ≤ 4 seconds is given by s(t) = t3 – 3t2 – 2t + 1. When, on the interval [0,10], is she decelerating?
Solution: Because the sign of the second derivative determines acceleration, you want to know when sʺ(x) is negative. So make an sʺ(x) wiggle graph by setting it equal to 0, finding critical numbers, and picking test points (as you did in Chapter 11). The wiggle graph for the second derivative is given in Figure 12.2.
Figure 12.2
Because sʺ(x) is negative on (0,1), the cat is decelerating on that interval.
The acceleration equation sʺ(t) is negative on the interval (0,1), so the cat decelerates only between t = 0 and t = 1.
You’ve got problems
Problem 3: A particle moves vertically (in inches) along the y-axis according to the position equation 
, where t represents seconds. At what time t is the acceleration of the particle equal to –1 in/sec2?
Vertical Projectile Motion
One of the easiest types of motion to model in elementary calculus is projectile motion, the motion of an object acted upon solely by gravity. Have you ever noticed that any thrown object follows a vertical path to the ground? It is very easy to write the position equation describing that path with only a tiny bit of information. Mind you, these equations can’t give you the exact position, because ignoring wind resistance and drag makes the problem much easier.
Scientists often pooh-pooh these little pseudoscientific math applications, saying that ignoring such factors as wind resistance and drag renders these examples worthless. Math people usually contend that, although not perfect, these examples show how useful even a simple mathematical concept can be.
Critical Point
Unquestionably, one of the grossest examples of projectile motion is in the movie The Exorcist. We will not be doing any examples involving pea soup.
The position equation of a vertical projectile looks like this:
You plug the object’s initial velocity into v0, the initial height into h0, and the appropriate gravitational constant into g (which stands for acceleration due to gravity)—if you are working in feet, use g = –32, whereas g = –9.8 if the problem contains meters. Once you create your position equation by plugging into the formula, it will work just like the other position equations from this chapter, outputting the vertical height of the object in relation to the ground at any time t (for example, a position of 12 translates to a position of 12 feet above ground).
Example 4: Here’s a throwback to 1970s television for you. A radio station called WKRP in Cincinnati is running a radio promotion. For Thanksgiving, they are dropping live turkeys from the station’s traffic helicopter into the city below, but little do they know that turkeys are not so good at the whole flying thing. Assuming that the turkeys were tossed with a miniscule initial velocity of 2 ft/sec straight up from a safe hovering height of 1,000 feet above ground, how long does it take a turkey to hit the road below, and at what speed will the turkey be traveling at that time?
Solution: The problem contains feet, so use g = 32 ft/sec2; you’re given v0 = 2 and h0 = 1,000, so plug these into the formula to get the position equation of s(t) = –16t2 + 2t + 1,000. You want to know when they hit the ground, which means they have a position of 0, so solve the equation –16t2 + 2t + 1,000 = 0. You can use a calculator or the quadratic formula to come up with the answer of t = 7.9684412 seconds. (The other answer of –7.84 doesn’t make sense—a negative answer suggests going back in time, and that’s never a good idea, especially with poultry.) If you plug that value into s′(x), you’ll find that the turkeys were falling at a velocity of –252.990 ft/sec. Oh, the humanity.
Critical Point
Notice that g always equals either –32 or –9.8, depending on the units of the problem. This is because the pull of gravity never changes.
You’ve got problems
Problem 4: A cannonball is fired straight up from a fortification 75 meters above ground with an initial velocity of 100 meters/second. Given this information, answer the following questions:
(a) When will the cannonball reach its maximum height? Round your answer to the nearest thousandth of a second.
(b) What is the maximum height of the cannonball? Round your answer to the nearest meter.
(c) Assuming the cannon is firing into flat terrain, how long does it take the cannonball to first hit the ground?
The Least You Need to Know
- The position equation tells you where an object is at any time t.
- The derivative of the position equation is the velocity equation, and the derivative of velocity is acceleration.
- If you plug 0 into position, velocity, or acceleration, you’ll get the initial value for that function.
- The formula for the position of a vertical projectile is
.