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).
- 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?
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.
$$\frac{dx(t)}{dt}=\alpha\quad \alpha \ is \ some \ constant$$
$$x(t_0)=l\ and\ x(t_1)=0$$
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 $y=y(x(t))$.
$$y(x(t_1))=a*x(t_1)^3+b*x(t_1)^2+c*x(t_1)+d=0\Rightarrow d=0$$
$$y(l)=a*l^3+b*l^2+c*l=h$$
$$\frac{dy}{dx}_{x=l}=3a*l^2+2b*l+c=0$$
$$\frac{dy}{dx}_{x=0}=0\Rightarrow c=0\quad Assumption:\ touchdown\ means\ "beeing\ paralel\ to\ x-axis"$$
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.