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

In summary, the 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".
  • #1
McFluffy
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1

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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  • #2
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:)
 
  • #3
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.
 
Last edited:
  • #4
So what is ##
dy\over dx## at (-L, h) in your solution :rolleyes: ?
 
  • #5
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
 
Last edited:
  • #6
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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  • #7
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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  • #8
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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Related to Pick a,b,c,d for y=ax^3+bx^2+cx+d that models path of plane.

1. What does the equation represent?

The equation y=ax^3+bx^2+cx+d represents the path of a plane in three-dimensional space. The variables a, b, c, and d represent the coefficients that determine the shape and direction of the path.

2. How does the equation model the path of a plane?

The equation models the path of a plane by using the variables x and y to represent the horizontal and vertical positions of the plane at any given time. The coefficients a, b, c, and d determine the shape and direction of the path, allowing for different flight paths to be modeled.

3. What does the value of a determine in the equation?

The value of a determines the rate of change of the plane's vertical position. A positive value of a will result in the plane ascending, while a negative value of a will result in the plane descending.

4. How can the equation be used to predict the path of a plane?

By plugging in different values for x into the equation, the corresponding values of y can be calculated, giving the coordinates of the plane at different points in time. This allows for the prediction of the path of the plane and can be used for flight planning and navigation.

5. What are the limitations of this model?

This model assumes that the plane is following a smooth path and does not account for external factors such as wind, air resistance, or changes in altitude. It also does not take into account the plane's speed or changes in direction, making it less accurate for longer flights.

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