### Where should a pilot start descent?

```Where should a pilot start
descent?
By: Alex, Chelsea, Gio, Ian, Jake, and Tessa
The Problem
An approach path for an
aircraft landing in shown in the
figure and satisfies the
following conditions…
Conditions
1) The cruising altitude is h when descent starts at a
horizontal distance l from touch-down at the origin
1) The pilot must maintain a constant horizontal
speed v through the descent
1) The absolute value of the vertical acceleration
should not exceed a constant k (which is much less
than the acceleration due to gravity)
Question #1
Find a cubic polynomial p(x)= ax3+bx2+cx+d that
satisfies condition 1 , by imposing suitable
conditions on p(x) and p’(x) at the start of
descent and at touchdown.
Objective
Find constants a, b, c, and d that justify an
equation for the plane’s flight path.
Question #1 Work
Rates
Distance: p(x)= ax3+bx2+cx+d
Velocity: p’(x)= 3ax2+bx+c
Acceleration: p’’(x)= 6ax+2b
12. Slide 12
13. Slide 13
We found…
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d=0, by plugging 0 into the distance equation
c = 0, by setting the velocity equation (the tangent line)
a in terms of b and L
Knowing that p(L)=h, we plugged a back into the distance equation to find b in
terms of h and L
We had an equation for a in terms of b and L, so we plugged our value for b
back in to find a in terms of h and L
Solution
p(x) = -2hx3/L3 + 3hx2/L2
Question #2
Use the conditions 2 and 3 to show that
6hv2/L2 < k
Given
• a = -2h/L2
• b = 3h/L3
• p(x) = ax3+bx2
• x=L
Question #2 Work
Steps
1. Find the first implicit derivative of p(L)
dp/dt = 3aL2(dx/dt)+2b(dx/dt)
note: dx/dt = -v
2.
Find the second implicit derivative of p(x)
d2p/dt2 = -6avL(dx/dt)-2bv(dx/dt)
d2p/dt2 = -6avL(-v)-2bv(-v)
d2p/dt2 = -6Lav2-2bv2
note: d2p/dt2 < k
3.
Plug in a and b
-6Lav2 -2bv2 < K
-6Lv2(-2h/L3) - 2(3h/L2)v2 < K
12hLv2/L3 – 6hv2/L2 < K
12hv2/L2 - 6hv2/L2 < K
6hv2/L2 < k
Question #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. how far
away from the airport should the pilot start
descent?
Objective
Find L, which is the point at which the pilot
should start the descent
Question #3 Work
• Given: h=35,000 ft converted to 6.62878 mi
v= 300 mi/h k=860 mi/h2
• 6hv2/L2 < k we now need to plug in our values
• 6(6.62878)(300) 2/ L2 < 860
• 3579541.2 /L2< 860 multiply l2 on both sides
and divide by 860 to isolate l2 on the right
side.
• 33579541.2/860 = 4162.257209
• √4162.257209 < √L2
L=64.515 miles
• When the plane is 64.515 miles away the
pilot should start the descent to the airport
Question #4
Graph the approach path if the conditions stated in
Problem 3 are satisfied.
• In order to graph we use the original equation
which was ax3 + bx2 + cx+d. Because in question
one we found that c and d equaled 0, the
equation is ax3 + bx2.
• In problem 2 we found what a and b equaled in
terms of h and l, now that we have what h equals
and l equals from number 3, we plug it in for a
and b.
Question #4 Work
Given
a= -2h/l3
b=3h/l2
Height=6.629
Length=64.517
Steps
A= -2(6.629)/(64.517)3
A=-.00004937
B=3(6.629)/(64.517)2
B=.004777
Now you plug in A and B to the original equation
Solution
p(x)=-.00004937x3+ .004777x2
Window xmin=0, xmax=64.5(length), ymin=0, ymax=6.6 (height)
Walk away…
P(0) = 0
\
P’(0) = 0
P’(L) = 0
P(L) = h
Question
5. Question
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