Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Polar Equation of a Ship's Course


Date: 05/30/2000 at 05:19:37
From: mark
Subject: polar coordinates - problem solving

The question is:

A ship is following a counterclockwise course along a curve whose 
polar equation is r = 6 cos(angle) km. A radar station at the origin, 
O, first places the ship at the position (3root3, pi root6). Eight 
minutes later the ship is observed at a distance 3root2 km from point 
O. Assuming constant speed is maintained, how long after the second 
observation is the ship expected to arrive at point O?

I can't even get started on this one.

Thanks.


Date: 05/30/2000 at 14:52:55
From: Doctor Rob
Subject: Re: polar coordinates - problem solving

The formula for the arc-length of a curve is:

      s = INTEGRAL sqrt([dr]^2 + r^2*[d(theta)]^2)

with appropriate limits. You know that r = 6*cos(theta), so that

     dr = 6*sin(theta)*d(theta)

so the arc-length integral then becomes

      s = INTEGRAL sqrt(36*sin^2[theta]+36*cos^2[theta]) d(theta)
        = INTEGRAL 6 d(theta)

Now when the ship is 3*sqrt(2) km from the point O, the angle is given 
by the equation:

     3*sqrt(2) = 6*cos(theta)
     sqrt(2)/2 = cos(theta)
         theta = Pi/4

There are other solutions, but they can be rejected by the facts that 
the ship is going counterclockwise, and doesn't reach O until after 
the second observation. Thus after 8 minutes, the ship is at 
Q:(3*sqrt[2],Pi/4), and at the start, it is at P:(3*sqrt[3],Pi/6).

Now you have enough information to compute the distance traveled 
during those 8 minutes: it is the arc-length from P to Q, that is, 
from theta = Pi/6 to theta = Pi/4, which is the integral evaluated 
between those limits.

Once you know the distance and time, you can find the rate of speed, 
which is given to be constant. Next you need to compute the arc-length 
from Q to O:(0,Pi/2). This is again the integral evaluated between the 
appropriate values of theta.

Finally, divide by the rate of speed of the ship to find the length of 
time it took to reach O from the point (3*sqrt[2],Pi/4).

There is another approach, which uses no calculus at all. Instead, it 
uses the fact that the curve r = 6*cos(theta) is a circle through O 
and with center on the polar axis. The ship travels along the circle 
with constant speed, so the central angle also changes at a constant 
rate.  A theorem from plane geometry says that if you have an angle 
with vertex on a circle, it cuts an arc whose central angle is double 
the size of the original angle. Now the time used is proportional to 
arc-length traveled (because the speed is constant), which is 
proportional to the central angle traversed (since the radius is 
constant), which is 2*theta, which is proportional to theta. Now from 
P to Q theta differs by Pi/12, and from Q to O theta differs by Pi/4, 
whose ratio is 3. Thus the times traveled also are in a 3-to-1 ratio. 
That means that the time to go from Q to O is 3*8 = 24 minutes.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Calculus

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/