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### Polar Equation of a Ship's Course

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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

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