- #1

McFluffy

- 37

- 1

## 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?