Difficulty: Intermediate | Prerequisites: Derivatives, implicit differentiation, polynomial curve fitting
This recitation models an aircraft's descent path as a cubic polynomial and uses calculus to derive a safe-landing constraint. It ties together polynomial boundary conditions, implicit differentiation with respect to time, and real-world unit conversion. If you are comfortable with taking derivatives and setting up systems of equations from boundary conditions, you are ready for this material. The problem originates from Stewart Calculus and appears frequently in applied-calculus courses.
You model an aircraft's descent as a cubic polynomial, pin it down with four boundary conditions (altitude and slope at both endpoints), then use implicit differentiation to turn a spatial curve into a time-dependent acceleration constraint. The result is a clean inequality, 6hv²/ℓ² ≤ k, that tells a pilot how far from the airport to begin descending.
Cubic polynomial (descent path)
A third-degree polynomial P(x) = ax³ + bx² + cx + d used to describe the aircraft's altitude y as a function of horizontal distance x. Think of it as the mathematical shape of the glide slope.
Boundary conditions
Constraints imposed at the endpoints of the descent (x = 0 for touchdown, x = ℓ for the start of descent). They fix P(0) = 0, P'(0) = 0, P(ℓ) = h, and P'(ℓ) = 0, giving four equations for four unknowns.
In simple terms, these say: the plane is on the ground and level at touchdown, and at cruising altitude and level when descent begins.
Implicit differentiation (with respect to time)
Differentiating both sides of y = P(x) with respect to t (time) rather than x, using the chain rule: dy/dt = P'(x) · dx/dt. This converts a spatial relationship into a time-based one so you can talk about vertical acceleration.
Vertical acceleration constraint
The requirement that |y''(t)| ≤ k, where k is a maximum tolerable vertical acceleration (well below gravitational acceleration g). This is a passenger-comfort and structural-safety condition.
Cruising altitude (h)
The constant altitude at which the aircraft flies before beginning its descent. In the worked example, h = 35,000 ft.
Horizontal distance to touchdown (ℓ)
The ground-level distance from the point where descent begins to the touchdown point (the origin). The main result of the problem is finding the minimum value of ℓ.
Constant horizontal speed (v)
The aircraft maintains dx/dt = v throughout the descent. This simplifies the chain rule calculations and means x and t are linearly related.
An aircraft descends from cruising altitude h to the runway (the origin). The descent path y = P(x) must satisfy:
(i) The altitude is h when descent starts at horizontal distance ℓ from the origin, and 0 at touchdown.
(ii) The pilot holds a constant horizontal speed v throughout.
(iii) The absolute vertical acceleration stays below a constant k (much less than g ≈ 9.8 m/s²).
We want P(x) = ax³ + bx² + cx + d. Condition (i) gives four boundary equations:
P(0) = 0: at touchdown, the plane is on the ground. This forces d = 0.
P'(0) = 0: the path is level at touchdown (no sudden nose-up or nose-down). This forces c = 0.
P(ℓ) = h: at the start of descent, the altitude equals h. This gives aℓ³ + bℓ² = h.
P'(ℓ) = 0: the path is level at the start of descent (smooth transition from cruising). This gives 3aℓ² + 2bℓ = 0.
From P'(ℓ) = 0, solve for b in terms of a:
3aℓ² + 2bℓ = 0, so b = −3aℓ/2.
Substitute into P(ℓ) = h:
aℓ³ + (−3aℓ/2)ℓ² = h
aℓ³ − 3aℓ³/2 = h
−aℓ³/2 = h
a = −2h/ℓ³
Then b = −3(−2h/ℓ³)(ℓ)/2 = 3h/ℓ².
The descent polynomial is:
P(x) = (−2h/ℓ³)x³ + (3h/ℓ²)x²
Or equivalently: P(x) = h(3x²/ℓ² − 2x³/ℓ³).
Condition (ii) says dx/dt = v (constant). We need the vertical acceleration y''(t).
Start from y = P(x) and differentiate with respect to time t:
y'(t) = P'(x) · (dx/dt) = P'(x) · v
Differentiate again:
y''(t) = P''(x) · (dx/dt)² + P'(x) · (d²x/dt²)
Since v is constant, d²x/dt² = 0, so y''(t) = P''(x) · v².
Now compute P''(x) from P(x) = (−2h/ℓ³)x³ + (3h/ℓ²)x²:
P'(x) = (−6h/ℓ³)x² + (6h/ℓ²)x
P''(x) = (−12h/ℓ³)x + 6h/ℓ²
So y''(t) = [(−12h/ℓ³)x + 6h/ℓ²] · v².
The maximum of |y''(t)| occurs at the endpoints. Evaluate:
At x = 0: y''(t) = (6h/ℓ²) · v² = 6hv²/ℓ²
At x = ℓ: y''(t) = [(−12h/ℓ³)ℓ + 6h/ℓ²] · v² = (−6h/ℓ²) · v² = −6hv²/ℓ²
Both endpoints give |y''(t)| = 6hv²/ℓ². Condition (iii) requires:
6hv²/ℓ² ≤ k
This is the central result.
Given: h = 35,000 ft, v = 300 mi/h, k = 860 mi/h².
Convert h to miles: 35,000/5,280 ≈ 6.629 mi.
From 6hv²/ℓ² ≤ k, rearrange for ℓ:
ℓ² ≥ 6hv²/k
ℓ ≥ √(6hv²/k)
Plug in:
ℓ ≥ √(6 × 6.629 × 300² / 860)
ℓ ≥ √(6 × 6.629 × 90,000 / 860)
ℓ ≥ √(3,579,660 / 860)
ℓ ≥ √(4,162.40)
ℓ ≥ 64.52 miles
The pilot should start the descent at least 64.52 miles from the airport.
Descent polynomial
P(x) = \frac{-2h}{\ell^3}\,x^3 + \frac{3h}{\ell^2}\,x^2First derivative (slope of descent path)
P'(x) = \frac{-6h}{\ell^3}\,x^2 + \frac{6h}{\ell^2}\,xSecond derivative (curvature of descent path)
P''(x) = \frac{-12h}{\ell^3}\,x + \frac{6h}{\ell^2}Vertical acceleration in terms of time
y''(t) = P''(x) \cdot v^2 = \left(\frac{-12h}{\ell^3}\,x + \frac{6h}{\ell^2}\right) v^2The key inequality (acceleration constraint)
\frac{6hv^2}{\ell^2} \leq kMinimum descent distance
\ell \geq \sqrt{\frac{6hv^2}{k}}Students often confuse differentiating with respect to x and differentiating with respect to t. P'(x) gives the slope of the path in space. To get vertical acceleration you need y''(t), which requires the chain rule and the constant-speed condition dx/dt = v.
Some students forget that d²x/dt² = 0 when v is constant. If you leave that term in, the algebra gets messy and the clean inequality 6hv²/ℓ² ≤ k does not drop out.
A common error is setting P(0) = h and P(ℓ) = 0. The origin is the touchdown point, so P(0) = 0 and P(ℓ) = h. Getting this backwards flips the entire problem.
Students sometimes assume the maximum vertical acceleration occurs at the midpoint of the descent. It does not. P''(x) is linear in x, so its extreme values occur at the endpoints x = 0 and x = ℓ, where |P''(x)| is largest.
⚠️ The derivation of 6hv²/ℓ² ≤ k combines three core Calc I skills in a single problem: polynomial curve-fitting from boundary conditions, the chain rule, and implicit differentiation with respect to time. Expect exam questions that test all three.
⚠️ Unit conversion is a classic place to lose marks. In Problem 3, h must be in miles (not feet) because v and k are in miles and mi/h². Always check dimensional consistency before plugging in.
⚠️ The boundary conditions P'(0) = 0 and P'(ℓ) = 0 encode the physical requirement that the transition is smooth (no sudden change in slope). Examiners may ask you to explain what these conditions mean physically, not just algebraically.
⚠️ This type of applied problem, where you model a real scenario with a polynomial and then differentiate with constraints, is a Stewart Calculus staple. Variations include modelling roads over hills and roller-coaster tracks.
True or False: The polynomial P(x) is degree 4 because there are four boundary conditions. ___ (Answer: False. Four conditions pin down four coefficients of a degree-3 polynomial.)
Fill in the blank: P'(0) = 0 and P'(ℓ) = 0 ensure that the descent path is ___ at both endpoints. (Answer: level / horizontal / flat.)
True or False: Because horizontal speed v is constant, d²x/dt² = 0. ___ (Answer: True.)
Fill in the blank: The vertical acceleration y''(t) equals P''(x) times ___. (Answer: v².)
True or False: The maximum vertical acceleration along the descent path occurs at the midpoint x = ℓ/2. ___ (Answer: False. It occurs at the endpoints x = 0 and x = ℓ.)
Q: Write the four boundary conditions for the descent polynomial P(x) = ax³ + bx² + cx + d, and state what each one means physically.
A: P(0) = 0 (plane is on the ground at touchdown), P'(0) = 0 (path is level at touchdown), P(ℓ) = h (altitude equals cruising height at start of descent), P'(ℓ) = 0 (path is level at start of descent). These give d = 0, c = 0, aℓ³ + bℓ² = h, and 3aℓ² + 2bℓ = 0.
Q: Derive the coefficients a and b of the descent polynomial in terms of h and ℓ.
A: From 3aℓ² + 2bℓ = 0, we get b = −3aℓ/2. Substituting into aℓ³ + bℓ² = h gives aℓ³ − 3aℓ³/2 = h, so −aℓ³/2 = h, hence a = −2h/ℓ³. Then b = −3(−2h/ℓ³)(ℓ/2) = 3h/ℓ².
Q: Why does the vertical acceleration simplify to y''(t) = P''(x) · v² rather than y''(t) = P''(x) · v² + P'(x) · (d²x/dt²)?
A: Because the horizontal speed v is constant, d²x/dt² = 0. The second term vanishes, leaving only the first.
Q: An aircraft cruises at 35,000 ft with a speed of 300 mi/h. If the maximum tolerable vertical acceleration is 860 mi/h², find the minimum horizontal distance ℓ to begin descent. (1 mile = 5,280 ft.)
A: Convert h = 35,000/5,280 ≈ 6.629 mi. Then ℓ ≥ √(6 × 6.629 × 300²/860) = √(4,162.4) ≈ 64.52 miles.
Q: If the cruising altitude doubles but speed and k stay the same, what happens to the minimum descent distance ℓ?
A: Since ℓ ≥ √(6hv²/k), doubling h multiplies ℓ by √2 ≈ 1.41. The pilot needs about 41% more distance.
This problem connects directly to related rates problems in Calc I: you are converting a spatial derivative into a time derivative via the chain rule, which is the same technique used in shadow problems, sliding-ladder problems, and expanding-volume problems.
The boundary-condition setup (four unknowns, four equations from endpoint values and derivatives) reappears in spline interpolation, a major topic in numerical analysis and computer graphics where smooth curves are pieced together from polynomials.
The constraint 6hv²/ℓ² ≤ k is an optimisation-flavoured inequality. Rearranging for ℓ gives the minimum safe descent distance, which is the kind of "find the minimum value satisfying a constraint" question that bridges into Calc I optimisation problems.
Related Terms / Search Tags
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