Smooth Skatepark Ramp Profile
A skatepark designer is modelling the cross-section of a new ramp. The profile is described by f(x) = 1.2x for 0 ≤ x ≤ 2, and f(x) = a(x − 8)^2 + k for 2 < x ≤ 8, where x is horizontal distance in metres from the start and f(x) is height in metres. The designer wants the ramp to be smooth at the join x = 2, meaning f must be continuous and differentiable there. The far end at x = 8 must be exactly 6 m high and horizontal, so that skaters leave the ramp travelling horizontally. A safety officer asks whether the profile is one-to-one over its full domain, and whether a skater at height 4.5 m on the curved section could have come from two different horizontal positions. A second phase may extend the ramp beyond x = 8 with a horizontal platform at 6 m height, but only if the height-to-distance inverse remains well defined on the curved section.
1. Find the constants a and k, and verify that the profile is smooth at x = 2. State the gradient at x = 8.
Read the worked answer
Continuity at x = 2: the linear part gives f(2) = 1.2 × 2 = 2.4. The quadratic part gives f(2) = a(2 − 8)^2 + k = 36a + k, so 36a + k = 2.4.
Differentiability at x = 2: the left derivative is 1.2. The right derivative is 2a(x − 8), which at x = 2 is 2a(−6) = −12a. Hence −12a = 1.2, so a = −0.1.
Substitute into 36a + k = 2.4: 36(−0.1) + k = 2.4, so −3.6 + k = 2.4 and k = 6.
Thus f(x) = −0.1(x − 8)^2 + 6 on the curved section. At x = 8, f′(8) = 2(−0.1)(8 − 8) = 0, so the ramp is horizontal at the far end as required.
2. Determine whether f is one-to-one on [0, 8]. A skater is at height 4.5 m on the curved section; find the corresponding horizontal distance and explain why only one value is valid. Hence give the inverse function for the curved section and its domain.
Read the worked answer
On [0, 2], f(x) = 1.2x increases from 0 to 2.4. On (2, 8], f′(x) = −0.2(x − 8) = 0.2(8 − x), which is positive for x < 8 and zero at x = 8. So the curved section increases from 2.4 to 6. Therefore f is strictly increasing on [0, 8], so it is one-to-one.
At height 4.5: 4.5 = −0.1(x − 8)^2 + 6, so (x − 8)^2 = 15. Thus x = 8 ± √15. Only x = 8 − √15 ≈ 4.13 lies in [2, 8]; the other value is greater than 8 and is outside the curved section. So there is exactly one valid horizontal distance.
Solving y = −0.1(x − 8)^2 + 6 for x on [2, 8] gives x = 8 − √(10(6 − y)). The inverse is f⁻¹(y) = 8 − √(10(6 − y)), with domain 2.4 ≤ y ≤ 6.
3. The park proposes extending the ramp with a horizontal platform at height 6 m for 8 ≤ x ≤ 11. Explain how this affects the existence of an inverse function for the full extended profile. Suggest a domain restriction that still allows a height-to-distance inverse to be used for the curved part.
Read the worked answer
With the platform, the extended profile takes the value 6 for every x in [8, 11]. A function that is constant on an interval is not one-to-one, because many horizontal distances give the same height. Therefore the full extended profile has no inverse function on [0, 11].
A workable restriction is to keep the inverse only for the curved part, for example restrict the extended function to [0, 8]. On that restricted domain the profile is still strictly increasing, so the height-to-distance inverse f⁻¹(y) = 8 − √(10(6 − y)) remains valid for 2.4 ≤ y ≤ 6. If an inverse is needed on the platform as well, the codomain can be restricted to heights below 6, or a convention can be stated for choosing one point on the platform for height 6.