* A. * * An approach path for an aircraft landing is shown in the figure and satisfies the following conditions: *
(i) The cruising altitude is h when descent starts at a horizontal distance L from touch- down at the origin. (Note: in this coordinate system, the direction of flight is from right to left or in the negative –x direction).

(ii) The pilot must maintain a constant horizontal speed v throughout the descent.(Note: speed is the absolute value of velocity. In this instance the velocity will be negative, however, the speed will be a positive number of the same magnitude).

(iii) The absolute value of the vertical acceleration should not exceed a constant k (which 
is much less than the acceleration due to gravity). * -------------------------------------------------

* * 1. Find a cubic polynomial that satisfies condition (i) by imposing suitable conditions on and at the start of descent and touchdown. *
Solution: Assume the cubic polynomial function is P(x)= ax3 + bx2 + cx + d
According to the graph, We know P(0) = 0, P'(0) = 0, P(L) = h, P'(L) = 0
The original equation is P(x)= ax3 + bx2 + cx + d
P(0) = 0, so plug in zero for x. You will find that d = 0.
Simplify the equation to P(x)= ax3 + bx2 + cx
Find the derivative of P(X), which is 3ax2 + 2bx + c
P’(0) = 0, so plug in zero for x. We found that c =0
Simplify the equation to P’(X) = 3ax2 + 2bx
Plug in P'(L) = 0
We get the equation of 3aL2+ 2bL= 0 ①
Plug in P(L) = h
We get the equation of aL3 + bL2 = h ②
Solve for a,b
3aL2+ 2bL= 0 ①aL3 + bL2 = h ②

As a result, P(X) = (-2hL3) x3 + (3h2L2) x2

* * 2. Use conditions (ii) and (iii) to show that: 6hv2L2 ≦ K

Steps 1. Find the first implicit derivative of P(L) dPdt= 3aL2( dxdt)+2b(dxdt) dxdt = -v 2. Find the second implicit derivative of P(x) d2Pdt2 = -6avL(dxdt)-2bv(dxdt) d2Pdt2 = -6avL(-v)-2bv(-v) d2Pdt2 = -6Lav2-2bv2 d2Pdt2≤ k 3. Plug in a and b -6Lav2 -2bv2 ≤ K -6Lv2((-2hL3) - 2(3h2L2)v2 ≤ K 12hLv2L3- 6hv2L2 ≤ K 12hv2L2- 6hv2L2 ≤ K 6hv2L2≤ K

3. Suppose that an airline decides not to allow vertical acceleration of a plane to exceed
 k = 860 mi/h2. If the cruising altitude of a plane is 35,000 ft. and the speed is 300 mi/h, how far away from the airport should the pilot start descent? (Recall that 1mi = 5280 ft. You may need to do some units conversation.)
(1). Given: h=35,000 ft. converted to 6.62878 mi v= 300 mi/h k=860 mi/h2
(2). 6hv2/L2 < k we now need to plug in our values

(3). 6(6.62878)(300) 2/ L2 < 860

(4). 3579541.2 /L2< 860 multiply l2 on both sides and divide by 860 to isolate l2 on the right side.

(5). 33579541.2/860 = 4162.257209

(6). 4162.257209≤L2 L=64.515 miles

(7). When the plane is 64.515 miles away the pilot should start the descent to the airport

4. Graph the approach path if the conditions stated in Problem 3 are satisfied.
Given
a= -2h/l3 b=3h/l2 Height=6.629
Length=64.517
Steps
A= -2(6.629)/(64.517)3 B=3(6.629)/(64.517)2
A=-0.00004937 B=0.004777
Now you plug in A and B to the original equation
Solution
P(x)=-0.00004937x3+ 0.004777x2 x≥0

Graph:

B. Commercial airplanes in the conditions described loosely in part 3 (there are many other variables involved in an actual flight situation) typically begin descent around 100 miles from the airport. What might cause the difference in this distance compared to your answer in part 3?
Wind velocity and direction，Model, size or weight of airplane, cabin depression and pressure , passenger comfort

C. For a much smaller plane, that has a cruising altitude of 10,000 ft., the following rule of thumb is used by some pilots to determine the distance L of beginning descent:
“ Take your groundspeed (mi/min), times 2, times altitude in thousands. This gives the distance in miles at which you should begin descent.” i. Using this rule of thumb, calculate the distance to begin descent if the groundspeed (horizontal velocity) is 200 miles/hour.
The rule of thumb is, L = 0.002vh
Given v=200 mi/hour =103 mi/min, h = 10,000 ft. =12566 mi.
Plug v and h, L = 0.002×103×12566=5396≈0.012626 mi.

ii. What K-value does this method assume? What does this tell you about the differences in how large, commercial aircraft descend compared to smaller aircraft? K equals to 2 ( I guess )
K (vertical acceleration) in small plane is much greater that it in commercial plane.
Commercial aircrafts need to start descent at a long distance from airport, but smaller jet can descant at a close distance from the destination.

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