Pick a,b,c,d for y=ax^3+bx^2+cx+d that models path of plane.

[McFluffy]

Homework Statement

A plane starts its descent from height $y =h$ at $x = -L$ to land at $(0,0)$. Choose $a, b, c, d$ so its landing path $y =ax^3 + bx^2 + cx + d$ is "smooth". With $\frac{\mathrm {d}x}{\mathrm {d}t} = V =$constant, find $\frac{\mathrm {d}y}{\mathrm {d}t}$ and $\frac{\mathrm {d}^2y}{\mathrm {d}t^2}$ at $x =0$ and $x = -L$. (To keep $\frac{\mathrm {d}^2y}{\mathrm {d}t^2}$ small, a coast-to-coast plane starts down $L > 100$ miles from the airport.)

Homework Equations

Let $y=ax^3 + bx^2 + cx + d$ be the vertical distance of the plane as a function of its horizontal distance, $x$.
Since $\frac{\mathrm {d}x}{\mathrm {d}t} = V =$constant, let $x=Vt-L$ with the constant $-L$ because I want the plane to be at $x=-L$ when $t=0$.

The Attempt at a Solution

First thing I did was to interpret what it means for the landing path of the plane to be "smooth". I interpreted this as the plane intersecting $(0,0)$ after being in mid-air which would mean that $y=ax^3 + bx^2 + cx + d$ is tangent to $(0,0)$.

Since $(0,0)$ is on $y=ax^3 + bx^2 + cx + d$, this implies $d=0$, and since it's tangent to $(0,0)$, this implies that the tangent line that's tangent to $y=ax^3 + bx^2 + cx$ would be $y=0$, which would mean $c=0$. So far, we have that $y=ax^3 + bx^2$.

To find what $y$ would be as a function of time, $t$, substitute $x=Vt-L$ into $y$ to get $y=a(Vt-L)^3 + b(Vt-L)^2$. We know that at $t=0$, $y=h$ so $y=a(-L)^3 + b(-L)^2=-aL^3 + bL^2=h$.

Since the plane will stay on the ground after the time when it has landed, this would mean that $\frac{\mathrm {d}y}{\mathrm {d}t}$ at $t=$time when plane has landed is $0$. We first find when the plane has landed which correspond to solving for $t$ for $x=Vt-L=0$ which gives $t=\frac{L}{V}$.

With this, $\frac{\mathrm {d}y}{\mathrm {d}t}$ at $t=\frac{L}{V}$ is $0$. I tried finding the derivative, $\frac{\mathrm {d}y}{\mathrm {d}t}$ and setting it to $t=\frac{L}{V}$ but ended up with an identity $0=0$ so I tried finding $\frac{\mathrm {d}^2y}{\mathrm {d}t^2}$
at the same $t$ ( because $\frac{\mathrm {d}y}{\mathrm {d}t}$ at $t=\frac{L}{V}$ is $0$, this implies $\frac{\mathrm {d}^2y}{\mathrm {d}t^2}$ at $t=\frac{L}{V}$ is $0$) and found $b=0$(too many stuff to type out what the derivative, $\frac{\mathrm {d}^2y}{\mathrm {d}t^2}$ is, same thing thing with $\frac{\mathrm {d}y}{\mathrm {d}t}$ and have limited time, sorry.)

So $y=ax^3 + bx^2 + cx + d=ax^3$ and from $-aL^3 + bL^2=h$, we get $a=-\frac{h}{L^3}$ which means $y=-\frac{h}{L^3}x^3$ with $L>100$. And from this, we can compute what $\frac{\mathrm {d}^2y}{\mathrm {d}t^2}$ and $\frac{\mathrm {d}y}{\mathrm {d}t}$ and at $x =0$ and $x = -L$.

Graphing the end result equation, it seems correct so is my solution correct?

 

[BvU]

have limited time

You can save yourself some time with:

• path has to go through (-L, h)
• path has to go through (0,0)
• at both these points $dy\over dx$ = 0

(in fact,this elimiates time altogether )

 

[McFluffy]

You can save yourself some time with:

• path has to go through (-L, h)
• path has to go through (0,0)
• at both these points $dy\over dx$ = 0

(in fact,this elimiates time altogether )

The solution I typed however didn't assume $dy\over dx$ =0 at $(-L, h)$. I'm only considering $y =-\frac{h}{L^3}x^3$ over the interval $[-L, 0]$ for the path of the plane.

 

[BvU]

So what is $dy\over dx$ at (-L, h) in your solution ?

 

[McFluffy]

So what is $dy\over dx$ at (-L, h) in your solution ?

Since I'm considering only the $[-L,0]$ interval, I would say that $dy\over dx$(the limit is also one-sided) is negative for that point because the plane is going down to land. I think you're suggesting that the path of the plane before $x=-L$ is a straight horizontal line, then it starts going down. Here's the graph of what I'm thinking of, I picked the constants, h and L as 75 and 200, respectively. https://www.desmos.com/calculator/9du7nducy8

 

[BvU]

I think you're suggesting that the path of the plane before x=−Lx=−Lx=-L is a straight horizontal line,

for the exercise, yes. In real life: as good as.

What airline are you flying for ? Kamikaze & co ? I sure would avoid it like ... The acceleration at -L would be lethal !
But the bodies would land very smooth indeed -- if only the plane would surive (-L,0) (it would not)

 

[McFluffy]

You're absolutely right. I forgot the passengers in the plane. I fixed the solution and the end result equation will be $y=\frac{2h}{L^3}x^3+\frac{3h}{L^2}x^2$ and with this, the passenger will land safe and sound. :D https://www.desmos.com/calculator/5ah7z7cs11

 

[BvU]

Bingo. At least: I'm personally convinced this is the answer the exercise writer wants. (I.e.: no guarantee )

The knowledgeable student is probably confused: he thinks a linear approach path is desired, with a gentle transition at 'start of approach' and also at 'touchdown, with a downward linear path -3 degrees (under the guidance of papi ).