A fun Math problem

[Cogman]

I posted a message over at vbforums.com and It looks like I stumped them. I submit the question to you. Again this is not home work as I have graduated HS and Calculus, this is just for fun. Im going to post the answer on that forum in a little while.

 

[Cogman]

Ohh, and for those of you that dont want to go there, here is the problem

I Just thought this up, so I don't have the answer to it.

An airplane traveling at a constant speed of 150 km/h receives report that a tropical storm is heading its way. As the Pilot hears the report, he feels the wind start to pick up. The wind is coming in at a 90* angle from the airport and is accelerating at a constant speed of 20km/h/h. The airplane is 50 km from the airport. What is the rate at which the airplane will have to change its direction (in degrees) in order to keep a straight path to the airport. Will the airplane make it?

Again I just thought this up, so I don't even know if our Pilot will make it. (but Now I can start to work on it.) Enjoy.

[Spell checked, and wind acel changed to faster speed]

 

[silverpig]

I'm lazy but just set up the triangle, set your x, y, and 0(theta). You've got dx/dt and d2y/dt2. You want d0/dt.

 

[silverpig]

I'm lazy but just set up the triangle, set your x, y, and 0(theta). You've got dx/dt and d2y/dt2. You want d0/dt.

 

[TuxDave]

I'm too lazy too, but here are the equations.

x: Plane's horizontal velocity
y: Plane's vertical velocity

x^2 + y^2 = 150 : Given in the problem.

t = 50/x : Time to reach the airport

ddown = y*t=y*(50/x) .....(d = v*t)
dup = 0.5*20*t^2=10*(50/x)^2 ......(d = 0.5*a*t^2)

ddown = dup in order for it to end up at the proper up/down position

Solve.

y = 500/x
x^2+y^2 = 150

Pick non-negative solutions for both values if possible.

 

[blahblah99]

I'm too lazy too but this is just a vector problem.

 

[DrPizza]

I'm bored enough to do it, but don't have a pen or pencil next to me.

 

[DrPizza]

Wow, excellent problem! I like it. Working in paint to post the solution.

 

[DrPizza]

(of course, Silverpig's method is correct)
Here's the explanation... and link to picture to follow if I AT will let me post it before JRunning me outta town..
solution.jpg

magnitude of horizontal velocity = wind velocity. 150 km/hr is constant. Speed toward airport is, of course, decreasing, but doesn't matter for the problem.

Here's the equations:
tan(theta) = horiz/150
take the derivative of both sides with respect to t...
sec^2(theta)*(d_theta/dt) = 1/150 (d_horiz/dt)
we know that d_horiz/dt = 20 km/h/h
d_theta/dt (what you're looking for) = 20/150 cos^2(theta)

just need a substitution for cos^2(theta)
cos^2(theta) = (speed toward airport)^2 / 150^2
speed toward airport (pythag thm) = sqrt(150^2 - horiz^2)

so, cos^2(theta) = [150^2 - horiz^2] / 150^2
t=0 is the starting time,
horiz(t) = 20t (t in hours, I'd prefer minutes, but don't want 1/3 or .333 in here)

thus, cos^2(theta) = [150^2 - 400t^2] / 150^2

And, finally, d_theta/dt = 20/150 * ^^^^
which simplified =


20[150^2 - 400 t^2] / 150^3


I didn't write this on paper, thus I'm quite prone to errors while typing (especially since I can't get a reply box to open today and have to do all this in a tiny little quick reply box)
Since, without paper and pen, I've lost the "flow" that I'm more used to, I make no guarantees of this solution. But, I will guarantee if it was on paper, I wouldn't have made any mistakes that I may have made above.


Nonetheless, I propose this as a somewhat more challenging follow-up question:

What is the maximum distance from the airport that the plane could travel in this manner and still reach the airport (excluding landing, of course)

 

[Cogman]

Well then, It looks like I could have a job writing these things .

Just so you know, What got me thinking of the problem was another problem we had in calculus. (yes, when I get bored I start thinking about various things including math) The problem, as I remember it, wanted you to find the rate of change of the angle theta as an airplane was traveling towards a fixed object.

The vbforum solution I posted is just wrong, I took the intergral of the wind aceleration to get the velocity, then last night I cant define the v = 20km/h*v + V. that it had to be take with respect to t. So I came here to find out if someone did it right (which I see the Dr. Pizza has, assuming math was done correctly (no I did not check it)).

 

[TuxDave]

oh.. my bad, i misread the problem. I thought you wanted to know what angle the plane should turn such that he'll hit the airport.

 

[DrPizza]

Here's one that I thought of while driving yesterday (long boring drive with 7 teenaged boys... I needed something to distract my mind from the distractions)

Similar situation:
The plane travels at 150 km/hr
Plane is located some distance due North from the airport.
Wind travels from due East to due West.

What is the maximum distance from the airport that the plane can be, 1. If it has to fly in a straight line, and 2. If it doesn't have to fly in a straight line.

The solution to #1 isn't that difficult, just an integral

The solution to #2 is even easier than #1.

 

[jagec]

Originally posted by: DrPizza
The solution to #2 is even easier than #1.

No kidding, the plane can be any distance from the airport because it can just circle until the storm passes

I know that's not what you were thinking, don't worry.

 

[TuxDave]

Originally posted by: DrPizza
Here's one that I thought of while driving yesterday (long boring drive with 7 teenaged boys... I needed something to distract my mind from the distractions)

Similar situation:
The plane travels at 150 km/hr
Plane is located some distance due North from the airport.
Wind travels from due East to due West.

What is the maximum distance from the airport that the plane can be, 1. If it has to fly in a straight line, and 2. If it doesn't have to fly in a straight line.

The solution to #1 isn't that difficult, just an integral

The solution to #2 is even easier than #1.

#1) If I had to set up the equations.

Wind is accelerating at a rate of X km/hr/hr
Plane can fly at maximum Y km/hr

Time before the plane is screwed = Y/X hrs

Maximum distance the plane can fly in a straight line =

INT t = 0 to Y/X of {vvertical * dt}
INT t = 0 to Y/X of {sqrt[150^2-(X*t)^2] dt}

#2) I guess it would be like throwing a baseball problem no? Angle the plane at 45 degrees and do the normal math.

 

[DrPizza]

The actual solution to #2: The plane could reach any airport in the world. The earth is round...

 

[TuxDave]

Originally posted by: DrPizza
The actual solution to #2: The plane could reach any airport in the world. The earth is round...

I was tricked....