Topic 7.4: Reasoning Using Slope Fields

Follow the dashes

With the whole field on the water, let’s set a lantern down at zero, one and watch what it does. At the start the water tips it downward with slope negative one, but every new spot has a new slope, so the path keeps bending, flattens out, and then starts to climb. The path doesn’t only run forward, either. If we ask where this lantern came from, the field answers that too, and it came sweeping down from way up here.

A curve that follows the field at every point like this is a solution of the differential equation, and sketching one is something the AP exam asks you to do all the time. Let’s see if you can do what the water does, before the water does it. Draw the path you think a lantern set down at negative two, one will take, all the way to the right edge, and then let it go.

Let’s take that back to the board. The AP exam hands you exactly this, a slope field and a point, and asks you to sketch the solution through it. Your curve has to go through the point, follow the nearby dashes without copying any single one, stay smooth with no corners, and keep going both ways until it reaches the edge of the field.

One mistake the AP graders have actually reported is a single straight line drawn along the first dash, and that can’t be right, because one step away the slope is already different. Every starting point gives its own curve, so this one differential equation has a whole family of solutions, infinitely many of them.

To pick out just one, we need a starting point. If we call the solution y equals f of x, then f of zero equals one says that at x equals zero the output is one. A starting point like that is called an initial condition, and it gives us the particular solution through zero, one. A solution to a differential equation is a function, and without an initial condition the honest answer is a whole family of functions.

When only one letter matters

Some rules are simpler, like d y d x equals two x. The slope depends only on x, so every dash in the same column has the same tilt. That makes the solutions copies of one curve slid up and down, and you already know that curve, x squared plus C. What happens when the rule cares only about y? To see it, let’s go somewhere where height is everything.

This is sunrise over the valleys of Cappadocia, in Turkey, where on a good morning the sky fills with dozens of hot-air balloons. Let’s imagine a steady breeze that carries every balloon east at the same seventy-five meters each minute, so how far east a balloon has drifted works like a clock. That turns the sky into a graph, with time across and height up.

With the burner held steady, our model says d h d t, the climb rate, is six hundred minus the height, divided by three, in meters per minute. That’s two hundred meters a minute right off the ground, and slower and slower as the air thins. Watch these balloons lift off one minute apart, each crew seventy-five meters farther downwind than the last, which is exactly one minute farther along our clock.

They all climb the same way, so each curve is the first one slid one minute to the right, and that’s because the rule only uses the height, so every row of dashes matches. Drag this balloon up or down to the one height where it won’t climb or sink at all, and then let it go.

A balloon that starts too high, at nine hundred meters, sinks a hundred meters a minute at first, then slower and slower. It gets closer and closer to six hundred but never drops below it, and if you drew it crossing that line on the exam, you’d lose the point.

Back on the board, every row has matching dashes, the line h equals six hundred is an equilibrium solution, and solutions above it fall toward it while solutions below it rise toward it. That gives us three questions for matching a field to its equation, so ask which letters matter, find where the slopes are zero, and then check the sign at one point. Use those three questions on this field, and pick the equation that made it.

The lane

Now we can finally solve the river’s mystery, and it starts with this diagonal, where every single dash lies flat. It’s tempting to think a lantern could ride along that diagonal, but the line y equals x has slope one, while the rule says the slope there is zero, so it can’t be a path. Lanterns do cross it, though, and they cross it flat, which is exactly where each one reaches its lowest point.

Let’s prove that the way the AP exam wants it, for the lantern with f of one equals one. The slope there, d y d x at one, one, is one minus one, zero. To tell a low point from a high point, let’s differentiate the rule itself. The derivative of x is one, and the derivative of y is d y d x.

That makes the second derivative one minus d y d x. Here that’s one minus zero, which is one, a positive number. The full-credit sentence says, because d y d x equals zero and the second derivative is positive at one, one, f has a relative minimum at x equals one. Now you try one. At the point zero, one, where the slope is negative one, what is the second derivative?

Now let’s find the lane. Remember the diagonal one step below y equals x, where every dash has slope one? A straight line along that diagonal rises one for every one across, so its own slope matches every dash it passes through, and a lantern set on it would ride it forever. Let’s write that line as y equals x plus b. What does b have to be?

Above the lane, y is bigger than x minus one, so x minus y is less than one. Those paths climb more slowly than the lane, or even fall, so the gap keeps shrinking. Below the lane the slopes are bigger than one, so those paths climb faster and close the gap from underneath. Either way, every path gets pulled closer and closer to the lane.

They never actually touch it, though. If two lanterns ever reached the same spot, the water would carry them together from then on, and they’d have arrived together too, so they’d really be the same path.

That’s why every lantern picks the same arch. The lane y equals x minus one runs straight under it, and the river’s slopes squeeze every path closer and closer to it, until you can’t tell them apart. Look, there’s the lantern you predicted, from negative two, one, sliding under the arch with all the others. Real water does something like this too. Wherever two currents push together, foam and leaves collect in long lines on the surface.

On the exam

Let’s try one the way the AP exam writes it, with a new rule, d y d x equals x times the quantity two minus y, and nine points to sketch. Turn each of these nine dashes to the slope the equation gives at its point.

Now let y equals f of x be the solution with f of one equals three, so we start at x equals one, at a height of three. There the slope is one times, two minus three, which is negative one, so the curve heads down toward two. It gets closer and closer to two but never crosses, because y equals two is a solution, just like the balloon’s six hundred meters.

To the left, the curve climbs to a peak at x equals zero, where the slope is zero. Past zero, x is negative, so the dashes tip the other way and the curve falls back toward two.

Out to the right, the limit of f of x, as x goes to infinity, is two. The curve can’t cross that line, and to the right of zero, at any height above two, the dashes get steeper as you move right, so it can’t level off above two. You can read all of that from the field without solving anything.

The exam also loves the tangent line, and the equation hands us its slope directly, negative one. The line starts at height three and drops one for each step to the right, so it’s y equals three minus, x minus one. Use the tangent line to approximate f of one point two.

Your lantern

One last river, and this time I’m flipping the rule to y minus x plus two, so watch what happens to the lane. The lane y equals x minus one is still a path, because the rule gives x minus one, minus x, plus two, which is one. But y minus x plus two grows as y grows, so above the lane the slopes are now bigger than one.

Paths above it climb away from the lane, and paths below it slip farther and farther beneath it. Watch these three lanterns, set down just barely off the lane, as every one of them misses the arch.

Now it’s your lantern, so set it down on the dock line at x equals zero, in the one spot that keeps it riding the lane all the way under the lit arch. Let’s put your spot back on both rivers, with a neighbor just above it. On the first river, yours rides the lane and its neighbor closes in on it. On the flipped river, yours still rides it, and its neighbor peels away.

It’s the same straight line, a solution of both rules, but one river pulls every path toward it and the other pushes every path away. Think about everything you did today. You took a rule for the slope and sketched its field by hand, you followed the dashes to draw a solution, and you knew where a lantern would go before you let it go.

And the mystery of the five arches came down to one straight line hiding in the slopes, which you found with a pencil. Soon we’ll learn to solve some of these equations exactly, but you’ll keep coming back to fields like this one to check that your answers make sense.