AP Calculus BC Topic 9.1: Defining and Differentiating Parametric Equations

Topic 9.1 key terms

parametric equations: A pair of functions $x(t)$ and $y(t)$ of one parameter $t$ (often time). Each value of $t$ gives one point $(x(t), y(t))$ of the path.
slope of a parametric curve: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ , provided $dx/dt \neq 0$ : the slope of the line tangent to the curve at the point given by $t$ .
tangent line at a time t: The line through $(x(t_0), y(t_0))$ with slope $\left.\frac{dy}{dx}\right|_{t=t_0}$ ; point and slope both come from the same $t_0$ .
horizontal tangent: Where $dy/dt = 0$ and $dx/dt \neq 0$ : the path is momentarily neither rising nor falling (the object need not be stopped).
vertical tangent: Where $dx/dt = 0$ and $dy/dt \neq 0$ : $dy/dx$ is undefined and the tangent line is vertical. If both are 0, the quotient alone can't decide.
second derivative of a parametric curve: $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}$ : how fast the slope changes per unit of $x$ . It is not $y''(t)/x''(t)$ .
concavity of a parametric curve: Concave up where $d^2y/dx^2 > 0$ , concave down where $d^2y/dx^2 < 0$ , read along increasing $x$ (so the sign of $dx/dt$ matters).
ricochet (skip) rule: An engineers' rule of thumb: a sphere skips off water if it meets it at less than about $18°/\sqrt{\text{specific gravity}}$ , about 6.7° (a slope of about 0.12) for cast iron.

Transcript

1,262 words · 4 parts

Skipping iron

Picture the last light of the day out on the Atlantic, back in the age of sail, with a warship’s guns run out toward the enemy ship across the water. Their first shot goes up high and comes down steep, punching straight into the sea short of her. Next comes something that sounds a little crazy. The barrel comes down until it’s almost flat, aiming low, right across the top of the water.

The ball hits the water and bounces right off it, then bounces again, skipping along like a stone, until it smashes into her side. For a couple of hundred years, gunners skipped iron cannonballs off the sea on purpose, and they even had a name for it, ricochet fire.

Engineers have a rule of thumb for an iron ball like this one, which says it skips when it meets the water at less than about seven degrees, and that anything steeper digs in and sinks. By the end of this lesson, you’ll be able to take the two equations of a cannonball in flight and work out the exact angle it’ll hit the water at. You’ll know whether it skips before it ever gets there.

A flying ball has two numbers changing at once, how far along it is and how high it is, and one clock drives both. To get a slope out of that, I know just the place.

Two coordinates, one clock

If you took precalc with us, you’ve been here before. It’s the Leaning Tower of Pisa at sunset, with a toy monkey hanging forty-nine meters up and a launcher on the lawn with its laser aimed straight at it.

Let’s look at it from the side. The ball’s position comes from two functions of the same clock, x of t for how far along the lawn it is and y of t for how high it is, both in meters. We call these parametric equations, because both coordinates depend on the same parameter, t, which is our clock.

Watch the ball’s two shadows while it flies. The one on the ground slides along at a steady twenty-one meters every second, while the one on the y-axis shoots up fast, slows down, stops for an instant at the top, and falls. Let’s freeze it one second in. Right now the ball is sliding sideways at twenty-one meters per second and climbing at fourteen point seven, so which way is it actually heading?

In each tiny bit of time it rises fourteen point seven for every twenty-one it moves across. That tilts its path at a slope of fourteen point seven over twenty-one, which is zero point seven. That’s the slope of the line that just touches the path at this point, the tangent line, and it points the way the ball is heading at this one instant.

This is the moment of the catch, two seconds after launch, and the ball is climbing at four point nine meters per second while it slides at twenty-one. What’s the slope of its path at that instant, to two decimal places?

Now try the very first instant. At launch the ball climbs at twenty-four point five and slides at twenty-one, a slope of seven sixths, about one point one seven, and look where that tangent line goes. It lies right on top of the launcher’s laser. The laser climbs forty-nine meters over forty-two across, the same seven sixths, so the launcher really is aiming along the tangent line at the moment of launch.

A tangent line tells you which way the ball is heading right then, and at launch it’s heading straight for the monkey. From then on, gravity keeps turning it away. Let’s take this to the board, because I’d like to show you why dividing those two rates really does give you the slope.

The slope of a path

Let’s pick two points on the path, the ball at one second and at two seconds. Between them it moves twenty-one meters across and nine point eight meters up. The line through those two points has a slope of nine point eight over twenty-one, about zero point four seven, and that’s the ball’s average direction over that second.

You can divide the top and the bottom by the same change in time without changing the fraction. Then the top is the average rate of y, and the bottom is the average rate of x. The calculus part is shrinking the time gap. As the second point slides back toward the first, the slope goes from zero point four seven to zero point five eight to zero point six eight. The line settles onto the tangent at exactly zero point seven.

In the limit the two average rates turn into two derivatives, so the slope of the tangent line, dy dx, equals dy dt divided by dx dt, as long as dx dt isn’t zero. It’s really the chain rule in disguise, because meters up per second equals meters up per meter across, times meters across per second.

Let’s write the tangent line at one second the way the AP exam asks for it. A line needs a point and a slope, and both of them come from the same value of t. For the point, put t equals one into both equations. x is negative twenty-one, and y is nineteen point six. For the slope, differentiate each one with respect to t, with the same derivative rules you always use.

dx dt is twenty-one, and dy dt is twenty-four point five minus nine point eight t, which is fourteen point seven at one second. Fourteen point seven over twenty-one gives a slope of zero point seven. Point-slope form does the rest, so the tangent line is y minus nineteen point six equals zero point seven times the quantity x plus twenty-one.

Flat, and straight down

Back at the tower, let’s let the tangent line ride along with the ball. It starts out steep, flattens as the ball climbs, and tips downward as it falls. Now you get to drive. Grab the ball and slide it along its path until its tangent line is perfectly flat.

Now look at the monkey, falling straight down. Its x never changes, so dx dt is zero while dy dt isn’t, and that’s exactly when the tangent points straight down. On the board, then, a horizontal tangent needs dy dt to be zero while dx dt isn’t, and a vertical tangent is the other way around, with dx dt zero while dy dt isn’t. If both are zero at once, the fraction is zero over zero, and this test can’t decide.

Let’s try all of this on a curve with more going on, x equals cosine t and y equals sine of two t. It draws a figure eight, the kind of shape you can make on an oscilloscope.

You differentiate each coordinate the way you always would, so dx dt is negative sine t, and dy dt is two cosine two t, where the chain rule hands you that extra two. dy dt is zero whenever cosine of two t is zero, which happens at four different times. Sine t isn’t zero at any of them, so the curve has four horizontal tangents, at the tops and bottoms of its two loops.

Where do you think the figure eight has vertical tangents? Look at the middle, too, where the curve crosses itself. It passes through the origin twice, with a slope of two the first time and negative two the second, which is why you always find dy dx at a value of t.

Topic 9.1: Defining and Differentiating Parametric Equations