Case Study 1 — The Radar Controller's Closing-Speed Problem

Field: Aviation / air-traffic control and tracking systems Calculus used: Related rates via implicit differentiation of distance and angle relations (§8.7–§8.13)

A problem you must solve in seconds

It is 3:14 a.m. in a darkened approach-control room outside a mid-sized airport. On the radar display, two blips creep toward the same navigation fix from different directions. One is a regional jet descending on the approach; the other is a cargo flight that has just been cleared to cross. Both are at the same altitude — 7,000 feet — and the controller's trained eye registers something the automation has already flagged in amber: the two aircraft are converging.

The question the controller must answer is not "where are they now?" — the screen shows that. The urgent question is "how fast is the gap between them shrinking?" If the answer is a few hundred knots, there is time to issue a turn. If it is close to a thousand knots, the situation is already critical. That number — the rate at which a distance changes — is a related-rates problem, and the collision-avoidance logic inside every modern radar system computes it many times per second. This case study works that computation by hand, slowly, the way you would on paper, so that you can see the chain rule doing the life-or-safety arithmetic.

We will keep the geometry deliberately clean so the calculus stands out. The full operational system layers on curved earth, wind, and three dimensions; the core idea is exactly what we develop here.

Setting up the geometry

Place the navigation fix at the origin. Both aircraft are flying straight, level, toward the fix, along tracks that happen to be perpendicular to each other — a common crossing geometry, and the one that produces the cleanest equations.

Let

  • $x(t)$ = the regional jet's distance from the fix, decreasing as it approaches;
  • $y(t)$ = the cargo flight's distance from the fix, also decreasing;
  • $D(t)$ = the straight-line distance between the two aircraft.

Because the two tracks are perpendicular, the fix, jet, and cargo plane form a right angle at the origin, and the Pythagorean theorem holds at every instant:

$$D^2 = x^2 + y^2.$$

This is the relating equation — step 4 of the seven-step method from §8.7. Notice it is a geometric truth that does not depend on time explicitly; it is true now, a second from now, and a minute from now. That permanence is exactly what lets us differentiate it with respect to $t$.

The Key Insight. We square the distance on purpose. Writing $D = \sqrt{x^2 + y^2}$ and differentiating would drag a square root through the whole computation. Differentiating $D^2 = x^2 + y^2$ instead keeps every term polynomial, and we recover $D$ itself only once, at the end, by plugging in the instant's numbers. The same trick appeared in §8.10's approaching-cars example — it is a habit worth keeping.

The numbers for this instant

From the data tags on the radar screen, the controller reads:

  • The regional jet is $x = 8$ nautical miles from the fix, closing at $\dfrac{dx}{dt} = -320$ knots (the minus sign because the distance is decreasing).
  • The cargo flight is $y = 6$ nautical miles from the fix, closing at $\dfrac{dy}{dt} = -300$ knots.

(A knot is one nautical mile per hour; we will keep all rates in knots and all distances in nautical miles, so the units stay consistent.)

We want $\dfrac{dD}{dt}$ — the rate at which the gap between the aircraft is changing — at this instant.

Differentiating the relation

Apply $\dfrac{d}{dt}$ to both sides of $D^2 = x^2 + y^2$. Every variable is a function of time, so every term picks up a rate via the chain rule:

$$2D\,\frac{dD}{dt} = 2x\,\frac{dx}{dt} + 2y\,\frac{dy}{dt}.$$

Divide through by $2$:

$$D\,\frac{dD}{dt} = x\,\frac{dx}{dt} + y\,\frac{dy}{dt}.$$

This single equation is the heart of the matter. It says the rate of change of the gap, scaled by the gap itself, is the sum of each aircraft's distance times its own closing rate. Now — and only now, at step 6 — do we substitute the instant.

Substituting the instant

First find $D$ at this moment from the relating equation:

$$D = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ nautical miles}.$$

(A $6$–$8$–$10$ triangle — chosen so the arithmetic stays transparent.) Substitute everything:

$$10\,\frac{dD}{dt} = (8)(-320) + (6)(-300) = -2560 - 1800 = -4360.$$

Therefore

$$\frac{dD}{dt} = \frac{-4360}{10} = -436 \text{ knots}.$$

Reading the answer honestly

The gap between the two aircraft is shrinking at 436 knots — more than seven nautical miles every minute. The negative sign is not a nuisance; it is the answer's meaning: a negative $\dfrac{dD}{dt}$ says the distance is decreasing, the planes are converging. Had the number come out positive, they would be separating and the controller could relax.

Step 7 of the method asks us to check units and sign. Units: nautical-miles-per-hour throughout, which is knots — correct for a speed. Sign: negative, matching the physical fact that both distances are shrinking. The magnitude, $436$ knots, is the closure rate the collision-avoidance system reports, and it is large enough that the controller will issue an immediate vector — perhaps turning the cargo flight thirty degrees right to open the geometry.

Common Pitfall. A tempting shortcut is to "just add the speeds": $320 + 300 = 620$ knots. That is wrong, and dangerously so. The aircraft are not flying directly at each other; they are converging on a shared point along perpendicular tracks. The true closure rate, $436$ knots, is what the geometry delivers. Adding speeds overestimates the danger here, but in other geometries it can underestimate it — which is why you differentiate the actual relation rather than guess. This is the §8.7 discipline paying its dividend.

A second rate: the angle the gap subtends

The controller often cares about a related quantity: the bearing of one aircraft as seen from the other, and how fast it swings. A swinging bearing means the geometry is changing favorably; a constant bearing with closing range is the classic signature of a collision course — "constant bearing, decreasing range," a phrase drilled into every mariner and pilot.

Let $\theta$ be the angle at the regional jet's position in the right triangle, so that $\tan\theta = \dfrac{y}{x}$ (cargo distance over jet distance, both measured from the fix). Differentiate implicitly with respect to $t$:

$$\sec^2\theta\,\frac{d\theta}{dt} = \frac{d}{dt}\!\left(\frac{y}{x}\right) = \frac{x\,\dfrac{dy}{dt} - y\,\dfrac{dx}{dt}}{x^2}.$$

At our instant, $\tan\theta = 6/8 = 0.75$, so $\sec^2\theta = 1 + \tan^2\theta = 1 + 0.5625 = 1.5625$. The numerator is

$$x\frac{dy}{dt} - y\frac{dx}{dt} = (8)(-300) - (6)(-320) = -2400 + 1920 = -480,$$

and $x^2 = 64$, so

$$1.5625\,\frac{d\theta}{dt} = \frac{-480}{64} = -7.5 \quad\Longrightarrow\quad \frac{d\theta}{dt} = \frac{-7.5}{1.5625} = -4.8 \text{ rad/h}.$$

The bearing is changing — about $-4.8$ radians per hour, roughly $-4.6^\circ$ per minute — so this is not a perfectly constant-bearing collision course; the geometry is slowly rotating. But the range is closing fast, so the controller acts anyway. The two related rates together — closure speed and bearing rate — are precisely the inputs a Traffic Collision Avoidance System (TCAS) blends to decide whether to bark "CLIMB, CLIMB" into both cockpits.

Real-World Application — TCAS and the tau criterion. Airborne collision-avoidance systems do not actually track distance directly; they track time to closest approach, estimated as $\tau \approx -D / \dfrac{dD}{dt}$ — the current gap divided by the closure rate. In our scenario, $\tau \approx -10 / (-436) \approx 0.023$ hours $\approx 83$ seconds. TCAS issues a "traffic advisory" around $\tau = 40$ seconds and a "resolution advisory" around $\tau = 25$ seconds, so our controller still has margin — but not much. The entire logic rests on the related-rate $\dfrac{dD}{dt}$ we computed by hand.

Why the hand computation matters

A controller does not solve this with a pencil at 3 a.m.; the computer does. But the engineer who wrote that computer's logic had to set up exactly the relating equation $D^2 = x^2 + y^2$, differentiate it correctly, and reason about signs and units — or the automation would report nonsense. Every safety-critical tracking system, from air-traffic radar to maritime ARPA to autonomous-vehicle sensors, contains this related-rates skeleton. The numbers are computed by machine; the correctness of the formula is a human responsibility, and it is the §8.7 method, done once, carefully, by hand.

Discussion Questions

  1. Why is the closure rate ($436$ kt) less than the sum of the two speeds ($620$ kt)? What geometry would make $\dfrac{dD}{dt}$ equal to the sum, and what geometry would make it nearly zero even while both aircraft are moving fast?

  2. The problem set $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ both negative. Reconstruct the formula $D\dfrac{dD}{dt} = x\dfrac{dx}{dt} + y\dfrac{dy}{dt}$ and explain why, once the aircraft pass the fix, the same formula automatically reports separation (positive $\dfrac{dD}{dt}$) without any change to the method.

  3. We squared the distance to avoid a square root. Redo the differentiation starting from $D = \sqrt{x^2+y^2}$ and confirm you reach the same $\dfrac{dD}{dt}$. Which version would you rather code, and why?

  4. The "constant bearing, decreasing range" rule says a steady $\theta$ with shrinking $D$ signals a collision course. Using the angle relation, what would $\dfrac{d\theta}{dt} = 0$ tell you about the ratio of the two closing rates?

  5. Real tracks are not perpendicular. If the two tracks crossed at $60^\circ$ instead of $90^\circ$, which relating equation would replace $D^2 = x^2 + y^2$, and why can the Pythagorean theorem no longer be used directly?

Your Turn

  1. Recompute $\dfrac{dD}{dt}$ if the cargo flight is instead departing the fix at $\dfrac{dy}{dt} = +300$ kt while the jet still closes at $-320$ kt (same positions). Interpret the sign.

  2. Find the instant's $\tau = -D/\dfrac{dD}{dt}$ and decide whether TCAS would have issued a resolution advisory yet.

  3. Suppose both aircraft hold their speeds. Write $x(t) = 8 - 320t$ and $y(t) = 6 - 300t$ (hours) and find the exact time of closest approach by minimizing $D^2(t)$ — a preview of Chapter 10's optimization.

Annotated Further Reading

  • Nolan, M. S. (2010). Fundamentals of Air Traffic Control (5th ed.), Cengage. The standard ATC textbook; Chapter 8's separation-standards material is the operational backdrop to the closure-rate computation done here.

  • Kuchar, J. K., and Drumm, A. C. (2007). "The Traffic Alert and Collision Avoidance System," Lincoln Laboratory Journal 16(2). A readable technical account of TCAS, including the $\tau$ (time-to-closest-approach) logic that turns our $\dfrac{dD}{dt}$ into an alert.

  • Stewart, J. (2020). Calculus: Early Transcendentals (9th ed.), Cengage, §3.9. The related-rates section whose perpendicular-motion examples this case study generalizes; work its airplane and radar problems for more practice.

  • Blackman, S., and Popoli, R. (1999). Design and Analysis of Modern Tracking Systems, Artech House. For readers who want the full multidimensional, noisy, Kalman-filtered version of the clean related-rates problem we solved — the same geometry, made operational.

Connections

  • §8.7–§8.10 — the seven-step method and the perpendicular-motion (approaching-cars) archetype this case study extends.
  • §8.13 — the changing-angle technique reused here for the bearing rate.
  • Chapter 10 — the closest-approach question in "Your Turn" #3 is an optimization, minimizing $D^2(t)$.
  • Chapter 28 — vector-valued functions give the fully three-dimensional, frame-independent version of aircraft tracking.