Bird's Path
The solutions discussed here have not been
checked, and no guarantee is made as to whether the arguments
presented are correct.
By Brad Rodgers (P1930) on Friday,
March 9, 2001 - 10:42 pm :
"A bird is flying towards its nest which is 100 feet away due
east. Without considering the wind the bird can fly at a speed of
5 feet per second. There is a north blowing wind, however, at a
speed of 2 feet per second. At all times the bird is flying in a
direction such that he is directly facing the nest. The bird is
not tacking into the wind as a boat would do but is continuously
being blown off course and trying to set a new course. How long
does it take for the bird to reach the nest?"
I have no clue how to solve this. Please help,
Brad
By Brad Rodgers (P1930) on Saturday,
March 10, 2001 - 10:20 pm :
I've found a (weak) scheme that I think might work, but I
don't know how to finish it, or whether it's finishable.
Letting the bird start at (0,0) in the cartesian plane, and
letting the nest be at (100,0), we know that the bird travels at
a rate of 5, and the wind blows at a rate of 2. We can use the
concept of relative motion to say that the nest travels at a
speed of 2 relative to the x axis. Letting the curve we are
writing an equation for be the path of the bird,
dy/dx=(2t-y)/(100-x)
We also know, as the bird always travels at a rate of 5, that t=(distance of
path)/5=D/5 and D=ò(1+(dy/dx)2)1/2 dx
assuming that since D=0 at x=0, we can neglect the constant and
ignore parameters.
We are given
| dy/dx=( |
2
5
|
ò[1+(dy/dx)2]1/2dx -y)/(100-x)
|
But I have no idea how to solve this. Any thoughts?
Thanks,
Brad
By Olof Sisask (P3033) on Saturday,
March 10, 2001 - 11:22 pm :
It's a very nice problem! I'm still working on it though I'm
afraid (in a different way to yours).
Regards,
Olof.
By David Loeffler (P865) on Sunday,
March 11, 2001 - 10:10 pm :
Brad,
Try using radial coordinates with the nest as the origin. Then you can write
down equations for the time derivatives of r and q and mess around a
bit. This should allow you to find the equation of the path the bird follows
(but I'm not sure if that'll help!) I'll have a go and see what comes up.
By David Loeffler (P865) on Sunday,
March 11, 2001 - 10:48 pm :
Time of flight = 500/21 seconds.
Put the nest at the origin and let the bird's initial position be r=100,
q = 0. Then we have dr/dt = 2×sin(q) - 5,
r dq/dt = 2×cos(q). Dividing the first by the second,
1/r dr/dq = tan(q) - 5/2 secq.
Integrating, we have lnr = ln(secq) - 5/2ln(secq+tanq) + C,
or
r = A secq/ (secq+ tanq)5/2. Putting q = 0
gives A = 100
It is now easily checked that the bird reaches the nest when
q = p/2. So the time of flight
= ò0p/2 dt/dqdq = ò0p/2 r/(2 cosq) dq.
Substituting the value of r found above gives a somewhat tedious
integral that can be evaluated by the t-formulae. The answer you
eventually emerge with is 500/21 (I hope).
(What does anyone else think of this? It looks a rather low
answer.)
By David Loeffler (P865) on Sunday,
March 11, 2001 - 10:50 pm :
(Incidentally, I have committed the mathematical sin of
measuring theta clockwise, since it makes the setup a bit
easier.)
By Hal 2001 (P3046) on Monday, March
12, 2001 - 09:55 pm :
David, what is the t-formula?
By Brad Rodgers (P1930) on Monday,
March 12, 2001 - 10:30 pm :
I think that I agree with you, though I'm not sure why dr/dt=so and so, and
dq/dt=such and such.
Also, when you use hte schema for relative motion, you end up
with the shortest amount of time being 100 x sqrt(21)/21 (from
pythagoras), which is clearly larger than your answer for the
time. Am I wrong about what takes the shortest amount of
time?
Brad
By David Loeffler (P865) on Tuesday,
March 13, 2001 - 02:21 pm :
Dear Hal, Brad and anyone else who's reading:
Sorry about that - the t-formulae are a set of formulae used for
evaluating complicated integrals involving trig
funtions.
Essentially the idea is: suppose you are evaluating the integral
òf(sinq, cosq) dq.
Let t = tan(q/2).
Then tanq = 2t/(1-t2 ), from the standard double-angle fomulae.
Also sec2 q = 1 + ( 2t/(1-t2 ) )2 = ( (1 + t2 )/(1-t2 ) )2
(easily checked), so cosq = (1 - t2 )/(1 + t2 ).
And sinq = cosqtanq = 2t/(1+t2 ).
Finally dt/dq = 1/2 sec2 (q/2), so dq = 2 dt /(1 + t2 ).
This will convert any rational function of sinq and cosq
into a rational function in t, which can be evaluated using partial
fractions or any other method that comes to hand.
(In fact in this particular problem, the integral is not a rational function
of sin and cos q, so you will end up with a term in (1 - t2 )1/2.
This can be eliminated in various ways - try setting t = sina.)
The bit about dr/dt etc is a bit harder to explain - I'll need to
go off and draw a diagram for you. Watch this space!
David
By Hal 2001 (P3046) on Tuesday, March
13, 2001 - 02:52 pm :
Thanks David for that explanation of the t-formula.
Hal.
By Kerwin Hui (Kwkh2) on Tuesday, March
13, 2001 - 04:17 pm :
Just to add a note: The 'official' name
for t-formula is the Weierstrass' substitution.
Kerwin
By Hal 2001 (P3046) on Tuesday, March
13, 2001 - 04:48 pm :
Kerwin, how do you pronounce Weierstrass?
By Dan Goodman (Dfmg2) on Tuesday,
March 13, 2001 - 05:04 pm :
Hal, I think it's pronounced
vye-er-strass
By David Loeffler (P865) on Tuesday,
March 13, 2001 - 10:49 pm :
Consider this diagram.
Here B=bird, N=nest, and r = N B. The component along N B of the wind
speed is 2 sinq, so the rate of change of length N B is
2 sinq- 5. That is, dr/dt = 2 sinq- 5.
As for the second, the component of wind speed perpendicular to N B is
2 cosqft/s. Thus, every dt seconds, the bird has moved
2 cosqdt feet in this direction. This subtends an added angle
of 2 cosqdt / r at N (plus various negligibly small
quantities), so dq/dt = 2 cosq/ r. Thus
r dq/dt = 2 cosq.
I probably haven't explained that very well; so if there is
anything you're not sure about, say so.)
David
By Brad Rodgers (P1930) on Wednesday,
March 14, 2001 - 11:08 pm :
In what way is the idea for finding a geodesic flawed though.
Here's a picture of the relative motion concept for shortest
path-
Using pythagoras, we get the least amount of time is
100/sqrt(21).
By Kerwin Hui (Kwkh2) on Thursday,
March 15, 2001 - 04:53 pm :
Brad,
The direction in which the bird flies is NOT constant.
Kerwin
By Anonymous on Thursday, March 15,
2001 - 04:55 pm :
Guess that depends on how one defines 'direction'.
:-)
By Brad Rodgers (P1930) on Thursday,
March 15, 2001 - 07:11 pm :
But if the direction in which he travels is constant, it would
seem as though he would travel in a shorter path (this doesn't
deal with this specific problem). but this is apparently not
true. Why?
By Brad Rodgers (P1930) on Wednesday,
March 21, 2001 - 03:04 am :
I may be wrong on this, but I think I've found a way to
evaluate the integral using relative motion. First, use the same
logic in my first example, except invert the position of the bird
and the postion of the nest around, so that the bird starts
at(100,0) and the nest starts at (0,0) being pushed at a speed of
2 upwards. Then we get
dy/dx = (y-2t)/x
Using t=1/5òx100(1+[dy/dx]2)1/2 dx, and substituting,
we can solve the equation to get (check this though)[also
substitute initial conditions to get the 1/252 ]
dy/dx = tan2 (-4/(25x)+1/252 )
Using the formula for time, we get that t roughly = 20.411 This doesn't make a
ton more sense than the other answer, but have a look. David, on your method,
are you assuming that the arc forming the angle becomes a straight line as
dq goes to 0? Can we do this?
Thanks,
Brad
By Olof Sisask (P3033) on Sunday, March
25, 2001 - 02:45 pm :
Hi,
What's the reasoning behind the (1+[dy/dx]2
)1/2 in the integral?
Thanks,
Olof.
By Brad Rodgers (P1930) on Sunday,
March 25, 2001 - 08:43 pm :
The integral òab (1+[dy/dx]2 )1/2 dx is used to find the length of
a curve. As the birds time in flight (neglecting wind) is length/5, and
because using relative motion allows us to neglect wind on the bird, we are
given time=1 /5 òab (1+[dy/dx]2 )1/2 dx for y=y(x) being the
equation of the birds path when considering the whole occurance relative to the
wind.
By the way, we can prove that formula gives the length of a curve
by first noting ds2 =dy2 +dx2
for s= length of curve, then solving for ds/dx, then changing
this to integral form.
Sorry if that was jumbled,
Brad
By Olof Sisask (P3033) on Tuesday,
March 27, 2001 - 04:32 pm :
I see, thanks!
/Olof.
By Brad Rodgers (P1930) on Sunday,
April 1, 2001 - 10:14 pm :
I don't know if anyone's still following this thread, but in
case anyone is, I've worked out an answer using my method.
Using the formula for t, and
dy/dx=(y-2t)/x, then solving we get
dy/dx=sinh(2 /5 *ln(x)+C)
as dy/dx=0 at x=100, we get C=-2 /5
ln(100), put this back into the equation for dy/dx, then
simplifying, we get
dy/dx=[(x/100)2/5 -(x/100)-2/5 ]/2
Finding y and using intitial condition x=100 at y=0, I think
we'll end up with
y=(101/5 /28)x7/5 -(5*104/5
/6)x3/5 +1000/21
Now, at x=0, y=2t, So
2t=1000/21, or
t=500/21 roughly 23.81 which is greater than 100/sqrt(21).
Brad
By Brad Rodgers (P1930) on Sunday,
April 1, 2001 - 10:42 pm :
Which means I guess I must've been doing something wrong
before when I thought 500/21 was less than 100/sqrt(21). Which
means that all of you probably noticed that and just didn't have
the heart to tell me I was being an idiot :)
Brad
By Arun Iyer (P4587) on Wednesday, June
13, 2001 - 09:14 pm :
The given problem is similar to the problem given below,
Two particles are initially located at points A and B a distance
d apart. They start moving at time t=0 such that the velocity
(vector) u of B is along the horizontal direction and the
(vector) v of A is continually aimed at B. At t=0,
u is perpendicular to v . When will the particles
meet?
The answer to this question as I evaluate is
T = vd/(v2 - u2
This should probably satisfy your question as well. Please
correct me if I am wrong.
Arun