The Drone Controller and a Broken Inverse
The drone club at a high school programs a quadcopter to fly straight up and then back down along a guide wire. Its height above the launch pad, in metres, is modelled by h(t) = -2(t - 5)² + 50, where t is the time in seconds and the flight runs from t = 0 to t = 10. Priya wants to build a controller that accepts a target height and returns the time at which the drone reaches it, so she enters the inverse relation into a graphing app over the whole interval. The app returns an error for some heights and a single time for others, and Priya cannot explain the pattern. Before the club is allowed to test the controller, the faculty adviser wants a written justification of the domain restriction and a correct inverse rule.
1. Is the inverse of h on the interval 0 <= t <= 10 a function? Justify your answer both algebraically and by referring to the graph of h.
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No. The graph of h is a downward parabola with vertex (5, 50), passing through (0, 0) and (10, 0). A horizontal line drawn at any height between 0 and 50 cuts the parabola twice, so the horizontal line test fails and the inverse is not a function.
Algebraically, try the target height 32 m:
-2(t - 5)² + 50 = 32 (t - 5)² = 9 t = 2 or t = 8
Both times lie in 0 <= t <= 10, so the inverse relation sends 32 to two different outputs. The pattern Priya sees is exactly this: heights above 50 m give no solution (the app reports an error), heights of 0 or between 0 and 50 give two solutions, and only the maximum height 50 m gives a single solution.
2. Priya keeps only the descending part of the flight, 5 <= t <= 10. Give the rule for h⁻¹(x), state its domain and range, and explain why this restriction repairs the problem.
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On 5 <= t <= 10 the model is strictly decreasing, from h(5) = 50 down to h(10) = 0, so each height in the range is reached exactly once: the function is one-to-one and its inverse is a function.
Solve for t: x = -2(t - 5)² + 50 (t - 5)² = (50 - x)/2 t - 5 = ±√((50 - x)/2)
Because t >= 5, take the positive root: h⁻¹(x) = 5 + √((50 - x)/2)
Domain of h⁻¹ = range of h = {x | 0 <= x <= 50}. Range of h⁻¹ = restricted domain of h = {y | 5 <= y <= 10}.
Checks: h⁻¹(50) = 5, h⁻¹(32) = 5 + 3 = 8 (the descending time, not 2), h⁻¹(0) = 5 + 5 = 10. The restriction discards one of every pair of times, which is precisely what makes the relation single-valued.
3. The adviser states: 'The inverse fails to be a function only for heights strictly between 0 and 50 metres.' Evaluate this statement.
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The statement is almost right but false at one endpoint, and the flaw matters for Priya's controller.
At x = 50 (the vertex height) there is exactly one time, t = 5. At every height x with 0 < x < 50 there are two times, symmetrically placed about t = 5. At x = 0 the drone is on the pad twice, at t = 0 and at t = 10, so the inverse also fails there.
The correct statement is: the inverse is not a function for 0 <= x < 50, and is a function only at x = 50 on the unrestricted flight. A controller built on the unrestricted relation would therefore return two answers for a landing command (height 0) as well as for mid-flight heights, which is a safety issue, not a rounding issue.