Beginner ski slope and drainage pipe
A snow park is shaping a beginner slope whose centreline height above the base station is modelled by h(x)=x^3/8 - 3x^2/4 +4 for 0≤x≤6, where x is horizontal distance in metres and h is height in metres. The designers require the slope to descend from the start, pass through a local minimum, and then rise gently to the exit so that skiers can reach the lift without pushing. A horizontal drainage pipe is planned at height 1.0 m above the base station; it must remain below the snow surface along the whole run. The maintenance team also needs to know the steepest descent, because that section determines where anti-slip matting is installed, and the average height of the centreline above the base station for earthwork calculations. Treat the model as exact on the closed interval and ignore the width of the slope.
1. Find h'(x), determine and classify all critical points on [0,6], and identify the point of steepest descent.
Read the worked answer
Differentiate:
h'(x)=3x^2/8 - 3x/2 = (3x/8)(x-4).
Thus h'(x)=0 at x=0 and x=4. Also h''(x)=3x/4 - 3/2, so h''(0)=-1.5<0 and h''(4)=1.5>0. Therefore x=0 is a local maximum and x=4 is a local minimum. The endpoint values are h(0)=4, h(4)=0 and h(6)=4, so the minimum height on the whole slope is 0 m at x=4, and the slope descends on (0,4) and rises on (4,6).
The steepest descent occurs where h' is most negative. Since h''(x)=0 gives x=2, this is the only candidate for an extremum of the slope. There h'(2)=3(2)(-2)/8=-1.5. Hence the steepest descent is -1.5, at x=2 m, where the slope drops 1.5 m per horizontal metre.
2. Compute the average height of the centreline above the base station over 0≤x≤6.
Read the worked answer
The average height is
(1/(6-0))∫_0^6 h(x)dx.
An antiderivative is
H(x)=x^4/32 - x^3/4 +4x.
Evaluate at the endpoints:
H(6)=1296/32 -216/4 +24 =40.5-54+24=10.5,
H(0)=0.
So the integral is 10.5 m^2, and the average height is 10.5/6=1.75 m. This is a mean over horizontal distance, not over slope length; it is the value needed for a rectangular earthwork estimate of the same horizontal extent.
3. The drainage pipe is horizontal at height 1.0 m. Decide whether it can remain below the snow surface for the whole run. If not, estimate the interval where it would be exposed.
Read the worked answer
The pipe remains below the surface exactly where h(x)>1.0 for every x in [0,6]. But h(4)=0, so at x=4 the snow surface is at height 0 m while the pipe is at height 1.0 m. Therefore the pipe is above the surface there and cannot remain buried along the whole run.
To find the exposed interval, solve h(x)=1:
x^3/8 -3x^2/4 +4 =1,
multiply by 8:
x^3 -6x^2 +24=0.
Numerically, the roots are approximately x≈2.69 and x≈5.06. Between these roots h(x)<1.0, so the pipe would be at or above the surface; outside this interval the surface is higher than 1.0 m. The pipe would therefore need to be lowered, rerouted, or replaced by a sloping pipe that stays below the surface over the approximately 2.4 m-wide central section.