Pick a,b,c,d for y=ax^3+bx^2+cx+d that models path of plane.

McFluffy
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Homework Statement


A plane starts its descent from height ##y =h## at ##x = -L## to land at ##(0,0)##. Choose ##a, b, c, d## so its landing path ##y =ax^3 + bx^2 + cx + d## is "smooth". With ##\frac{\mathrm {d}x}{\mathrm {d}t} = V =##constant, find ##\frac{\mathrm {d}y}{\mathrm {d}t}## and ##\frac{\mathrm {d}^2y}{\mathrm {d}t^2}## at ##x =0## and ##x = -L##. (To keep ##\frac{\mathrm {d}^2y}{\mathrm {d}t^2}## small, a coast-to-coast plane starts down ##L > 100## miles from the airport.)

Homework Equations


Let ##y=ax^3 + bx^2 + cx + d## be the vertical distance of the plane as a function of its horizontal distance, ##x##.
Since ##\frac{\mathrm {d}x}{\mathrm {d}t} = V =##constant, let ##x=Vt-L## with the constant ##-L## because I want the plane to be at ##x=-L## when ##t=0##.

The Attempt at a Solution


First thing I did was to interpret what it means for the landing path of the plane to be "smooth". I interpreted this as the plane intersecting ##(0,0)## after being in mid-air which would mean that ##y=ax^3 + bx^2 + cx + d## is tangent to ##(0,0)##.

Since ##(0,0)## is on ##y=ax^3 + bx^2 + cx + d##, this implies ##d=0##, and since it's tangent to ##(0,0)##, this implies that the tangent line that's tangent to ##y=ax^3 + bx^2 + cx## would be ##y=0##, which would mean ##c=0##. So far, we have that ##y=ax^3 + bx^2##.

To find what ##y## would be as a function of time, ##t##, substitute ##x=Vt-L## into ##y## to get ##y=a(Vt-L)^3 + b(Vt-L)^2##. We know that at ##t=0##, ##y=h## so ##y=a(-L)^3 + b(-L)^2=-aL^3 + bL^2=h##.

Since the plane will stay on the ground after the time when it has landed, this would mean that ##\frac{\mathrm {d}y}{\mathrm {d}t}## at ##t=##time when plane has landed is ##0##. We first find when the plane has landed which correspond to solving for ##t## for ##x=Vt-L=0## which gives ##t=\frac{L}{V}##.

With this, ##\frac{\mathrm {d}y}{\mathrm {d}t}## at ##t=\frac{L}{V}## is ##0##. I tried finding the derivative, ##\frac{\mathrm {d}y}{\mathrm {d}t}## and setting it to ##t=\frac{L}{V}## but ended up with an identity ##0=0## so I tried finding ##\frac{\mathrm {d}^2y}{\mathrm {d}t^2}##
at the same ##t## ( because ##\frac{\mathrm {d}y}{\mathrm {d}t}## at ##t=\frac{L}{V}## is ##0##, this implies ##\frac{\mathrm {d}^2y}{\mathrm {d}t^2}## at ##t=\frac{L}{V}## is ##0##) and found ##b=0##(too many stuff to type out what the derivative, ##\frac{\mathrm {d}^2y}{\mathrm {d}t^2}## is, same thing thing with ##\frac{\mathrm {d}y}{\mathrm {d}t}## and have limited time, sorry.)

So ##y=ax^3 + bx^2 + cx + d=ax^3## and from ##-aL^3 + bL^2=h##, we get ##a=-\frac{h}{L^3}## which means ##y=-\frac{h}{L^3}x^3## with ##L>100##. And from this, we can compute what ##\frac{\mathrm {d}^2y}{\mathrm {d}t^2}## and ##\frac{\mathrm {d}y}{\mathrm {d}t}## and at ##x =0## and ##x = -L##.

Graphing the end result equation, it seems correct so is my solution correct?
 
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McFluffy said:
have limited time
You can save yourself some time with:
  • path has to go through (-L, h)
  • path has to go through (0,0)
  • at both these points ##dy\over dx## = 0

(in fact,this elimiates time altogether :smile:)
 
BvU said:
You can save yourself some time with:
  • path has to go through (-L, h)
  • path has to go through (0,0)
  • at both these points ##dy\over dx## = 0

(in fact,this elimiates time altogether :smile:)

The solution I typed however didn't assume ##dy\over dx## =0 at ##(-L, h)##. I'm only considering ##y =-\frac{h}{L^3}x^3## over the interval ##[-L, 0]## for the path of the plane.
 
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So what is ##
dy\over dx## at (-L, h) in your solution :rolleyes: ?
 
BvU said:
So what is ##
dy\over dx## at (-L, h) in your solution :rolleyes: ?

Since I'm considering only the ##[-L,0]## interval, I would say that ##dy\over dx##(the limit is also one-sided) is negative for that point because the plane is going down to land. I think you're suggesting that the path of the plane before ##x=-L## is a straight horizontal line, then it starts going down. o_O Here's the graph of what I'm thinking of, I picked the constants, h and L as 75 and 200, respectively. https://www.desmos.com/calculator/9du7nducy8
 
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McFluffy said:
I think you're suggesting that the path of the plane before x=−Lx=−Lx=-L is a straight horizontal line,
for the exercise, yes. In real life: as good as.

What airline are you flying for ? Kamikaze & co ? I sure would avoid it like ... The acceleration at -L would be lethal !
But the bodies would land very smooth indeed -- if only the plane would surive (-L,0) (it would not)
 
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You're absolutely right. I forgot the passengers in the plane. I fixed the solution and the end result equation will be ##y=\frac{2h}{L^3}x^3+\frac{3h}{L^2}x^2## and with this, the passenger will land safe and sound. :D https://www.desmos.com/calculator/5ah7z7cs11
 
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Bingo. At least: I'm personally convinced this is the answer the exercise writer wants. (I.e.: no guarantee :rolleyes: )

The knowledgeable student is probably confused: he thinks a linear approach path is desired, with a gentle transition at 'start of approach' and also at 'touchdown, with a downward linear path -3 degrees (under the guidance of papi :smile:).
 
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