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Math Help - Application of derivatives, to an aircraft landing.

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    Member Goku's Avatar
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    Application of derivatives, to an aircraft landing.

    An approach path for an aircraft landing is shown in the figure and satisfies the following conditions:

    ( i ) The cruising altitude is when descent starts at a horizontal distance from touchdown at the origin.

    ( ii ) The pilot must maintain a constant horizontal speed throughout descent.

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



    1. Find a cubic polynomialP(x)=ax3+bx2+cx+d that satisfies condition ( i) by imposing suitable conditions on P(x) and P'(x) at the start of descent and at touchdown.

    WE know from the understanding of the graph that (0,0) is 1 point in the graph, so we can conclude that d = 0
    P(x)=3ax2+2bx+c
    I am not sure if we can substitute (0,0) here as well.
    Using P'(0) = 0, we can easily see that c=0.
    So now we currently have P(x)=ax3+bx2
    Now if we take P'(l) = 0 then we get l=0 or l=b3a2
    How do we use this to get rid of the a and b in the equation?
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  2. #2
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    Re: Application of derivatives, to an aircraft landing.

    Quote Originally Posted by Goku View Post
    An approach path for an aircraft landing is shown in the figure and satisfies the following conditions:

    ( i ) The cruising altitude is when descent starts at a horizontal distance from touchdown at the origin.

    ( ii ) The pilot must maintain a constant horizontal speed throughout descent.

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



    1. Find a cubic polynomialP(x)=ax3+bx2+cx+d that satisfies condition ( i) by imposing suitable conditions on P(x) and P'(x) at the start of descent and at touchdown.

    WE know from the understanding of the graph that (0,0) is 1 point in the graph, so we can conclude that d = 0
    P(x)=3ax2+2bx+c
    I am not sure if we can substitute (0,0) here as well.
    Using P'(0) = 0, we can easily see that c=0.
    So now we currently have P(x)=ax3+bx2
    Now if we take P'(l) = 0 then we get l=0 or l=b3a2
    How do we use this to get rid of the a and b in the equation?
    You wrote

    Now if we take P'(l) = 0 then we get l=0 or l=−b3a2

    That's the wrong method: l is a known value and you are looking for values of a or b wrt l:

    P'(l)=3al^2+2bl=0~\implies~b=-\frac32 l \cdot a

    That means:

    P(x)=a x^3- \frac32 l \cdot a x^2 = ax^2 \left(x-\frac32 l \right)
    Thanks from Goku
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