Finding the Landing Path of an Airplane Using Cubic Polynomial Functions

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Homework Help Overview

The discussion revolves around determining the landing path of an airplane modeled by a cubic polynomial function. The original poster presents a problem involving the altitude of the airplane, the horizontal distance to the runway, and the conditions that define the polynomial function.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the derivatives of the polynomial at specific points, questioning how these relate to the conditions of the problem. There is an exploration of the implications of having flat slopes at the endpoints of the landing path.

Discussion Status

Some participants have offered insights into the relationships between the derivatives and the coefficients of the polynomial. There is an ongoing exploration of how to utilize the conditions provided to derive the polynomial's form, with no explicit consensus reached on the final approach.

Contextual Notes

Participants note the constraints of the problem, including the requirement that the polynomial must meet specific altitude and slope conditions at the endpoints, which are critical to the discussion.

wisredz
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Hi I have this question that is driving me mad any help is appreciated.

An airplane is flying at altitude H when it begins its decent to an airport runway that is at horizontal ground distance L from the airplane. Assume that the landing path of the airplane is the graph of a cubic polynomial fuction y=ax^3+bx^2+cx+dwhere y(-L)=H and y(0)=0.

(a) What is dy/dx at x=0?
(b) What is dy/dx at x= -L?
(c) Use the values for dy/dx at x=0 and x= -L together with y(0)=0 and y(-L)=H to show that

y(x)=H[2(x/L)^3+3(x/L)^2]

Anyhelp is appreciated. Thanks in advance
 
Last edited:
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Btw, that dy/dx at x=0 equals c and that dy/dx at -L equals 0 is clear to me. My problem is that I cannot use those informations...
 
now at the top and bottom it is flat so dy/dx at x=0 and x=-L should be zero,
 
oh u posted again,
[tex]y= ax^3+bx^2+cx[/tex]
on Solving u will have c=0
Therefore
[tex]y= ax^3+bx^2[/tex]
u can get values of a and b from
dy/dx at x= -L, is zero
and also y(-L) = H
 
God, it is that easy then... It was 2 in the morning here(Turkey) Thanks a lot, I was going out of my mind.
 

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