Fitting a cubic through a tangent constraint
A designer models the cross-section of a skate ramp with a cubic polynomial p(x) = ax³ + bx² + cx + d defined on the interval [0, 2]. The ramp must satisfy four requirements: p(0) = 0, so the ramp meets the ground at the origin; p'(0) = 0, so the entry is smooth (horizontal tangent); p(2) = 3, so the ramp reaches a height of 3 metres at its far edge; and p'(2) = 0, so the exit is also horizontal. The designer claims that these four conditions uniquely determine the cubic. A colleague worries that the four conditions might be inconsistent or that a higher-degree polynomial is needed. You are asked to settle the matter rigorously and then check whether the resulting ramp stays above ground level on the whole interval [0, 2].
1. Determine whether the four conditions uniquely determine a cubic p(x) = ax³ + bx² + cx + d. Solve for the coefficients, or explain why no such cubic exists.
Read the worked answer
Apply each condition to p(x) = ax³ + bx² + cx + d.
- p(0) = 0 gives d = 0.
- p'(x) = 3ax² + 2bx + c, so p'(0) = c = 0.
- p(2) = 8a + 4b = 3.
- p'(2) = 12a + 4b = 0.
Subtracting the last two equations: (8a + 4b) − (12a + 4b) = 3 − 0, so −4a = 3, giving a = −3/4. Then 12(−3/4) + 4b = 0 gives −9 + 4b = 0, so b = 9/4.
Thus p(x) = −(3/4)x³ + (9/4)x² = (3/4)x²(3 − x). The four conditions are consistent and determine the cubic uniquely, because the linear system in (a, b, c, d) is square with a nonzero determinant (the two tangent conditions are independent of the two value conditions). The colleague's worry is unfounded.
2. Show that the ramp stays at or above ground level on [0, 2], and find the maximum height of the ramp on that interval.
Read the worked answer
Factor: p(x) = (3/4)x²(3 − x). On [0, 2], x² ≥ 0 and 3 − x ≥ 3 − 2 = 1 > 0, so p(x) ≥ 0 throughout. The ramp never dips below ground.
To find the maximum, solve p'(x) = 0: p'(x) = −(9/4)x² + (9/2)x = (9/4)x(2 − x). Critical points are x = 0 and x = 2, both endpoints, where p = 0 and p = 3 respectively. There is no interior critical point in (0, 2). So the maximum on [0, 2] is p(2) = 3, attained at the far edge.
Check: the function is increasing on (0, 2) because p'(x) = (9/4)x(2 − x) > 0 there, confirming the ramp rises monotonically from 0 to 3 metres.