# Application of derivatives

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 polynomial$P(x)= ax^3 + bx^2 + cx + d$ 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

$P'(x)=3ax^2+2bx+c$

I am not sure if we can substitute (0,0) here as well.

Please give any hints, I am not looking for a solution.

UPDATE

Using P'(0) = 0, we can easily see that c=0.

So now we currently have $P(x) = ax^3 + bx^2$

Now if we take P'(l) = 0 then we get l=0 or $l=\frac{-b}{3a^2}$

How do we use this to get rid of the a and b in the equation?

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I think you want $P(\ell)=h$, $P'(\ell)=0$, and $P'(0)=0$. – Gerry Myerson Feb 27 '13 at 6:05
And $P(0)=0$, of course... And $|P'(x)||\dot{x}| \leq V_{\text{max}}$ for all $x \in [0,l]$. The latter involves finding the $x' \in (0,l)$ for which $p''(x') = 0$ and then checking $|p'(x')||\dot{x}| \leq V_{\text{max}}$. – copper.hat Feb 27 '13 at 6:46
Edited question, not sure on how to calculate for a and b. – Dreamer78692 Feb 27 '13 at 6:49
You haven't used $P(\ell)=h$. – Gerry Myerson Feb 27 '13 at 7:50

First of all create an independent variable (time) since the question has velocity and acceleration constraints, such that $x=x(t)$. For $x(t)$ we can derive the eqution by following conditions.

• The pilot must maintain a constant horizontal speed throughout descent.

$$\frac{dx(t)}{dt}=\alpha\quad \alpha \ is \ some \ constant$$

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

$$x(t_0)=l\ and\ x(t_1)=0$$

By using above conditions we can derive the equation for $x(t)$ (I skip derivation since it is straightforward) to be

$$x(t)=l*\frac{t_1-t}{t_1-t_0}$$

Now we can start with the solution of $y=y(x(t))$.

• At time $t_1$ plane must be in the origin. [Note that $x(t_1)=0$]

$$y(x(t_1))=a*x(t_1)^3+b*x(t_1)^2+c*x(t_1)+d=0\Rightarrow d=0$$

• The cruising altitude is when descent starts at a horizontal distance from touchdown at the origin. [Until descent starts plane was parallel to x-axis and y-velocity was zero]

$$y(l)=a*l^3+b*l^2+c*l=h$$

$$\frac{dy}{dx}_{x=l}=3a*l^2+2b*l+c=0$$

$$\frac{dy}{dx}_{x=0}=0\Rightarrow c=0\quad Assumption:\ touchdown\ means\ "beeing\ paralel\ to\ x-axis"$$

By solving above system we find a and b to be

$$a=\frac{-2h}{l^3}\quad b=\frac{3h}{l^2}$$

We have to check the last solution that the absolute value of the vertical acceleration should not exceed a constant [means that it must have an upper bound]. We can derive the acceleration by chain rule of second order such as

$$\frac{d^2y}{dt^2}=\frac{d^2y}{dx^2}{\bigg(\frac{dx}{dt}\bigg)}^2+\frac{dy}{dx}\frac{d^2x}{dt^2}$$

Since horizontal velocity is constant as given in the question we conclude that $\frac{d^2x}{dt^2}=0$. The only term for vertical acceleration remains as below

$$\frac{d^2y}{dt^2}=\frac{d^2y}{dx^2}{\bigg(\frac{dx}{dt}\bigg)}^2=\frac{6h}{(t_1-t_0)^2}-\frac{12h(t_1-t)}{(t_1-t_0)^3}$$

Since $t_0\le t\le t_1$ the above equation never goes to infinity within solution domain unless $t_1=t_0$. To satisfy the last condition we must impose a condition that $t>0$.

PS: By making different assumptions for boundary conditions you can come up with different solutions.

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