# Application of derivatives, to an aircraft landing.

• February 26th 2013, 09:56 PM
Goku
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?
• February 26th 2013, 11:12 PM
earboth
Re: Application of derivatives, to an aircraft landing.
Quote:

Originally Posted by Goku
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:

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