Posted by Lily on Wednesday, August 17, 2011 at 11:28pm.
Can someone please go over the steps to solve this problem? I know I need to find where the tugboat intersects the radar but I'm not sure exactly how to do this.
A radar buoy detects any boats within a radius of 12 miles. A tugboat starts at a location 18 miles SOUTH and 10 miles EAST of the radar buoy. The tugboat travels at a constant speed of 15 mph. The tugboat travels on a straight line toward the NORTHERNMOST point of the radar region. When the tugboat is directly EAST of the buoy, it turns and travels DUE NORTH until it exits the radar region.
How long (in hours) is the tugboat in the radar region?

Math  Reiny, Thursday, August 18, 2011 at 8:06am
Use coordinate geometry.
let the radar region be represented by the circle
x^2 + y^2 = 144
and the initial position P(10,18) , heading to (0,12)
so find the equation of that line:
slope = (12+18)/(0+10) = 3 , with a yintercept of 12
so first path : y = 3x + 12
where does it cross the circle:
x^2 + (3x+12)^2 = 144
x^2 + 9x^2 + 72x + 144 = 144
10x^2 + 72x = 0
x(10x + 72) = 0
x = 0 or x = 7.2
we know that the x=0 is our point (0,12) so
if x = 7.2
y = 3(7.2) + 12 = 9.6
So the intersection of his path with the radar region is (7.2, 9.6)
I crosses the xaxis at x = 4 , (let y=0 in y = 3x+12 )
So first leg is from (7.2 , 9.6) to (4,0) , then it heads north
so let x = 4 in the circle equation:
16 + y^2 = 144
y = √128 , which is the distance it travels northbound within the region.
Finally find the distance between (7.2, 9.6) and (4,0) , add on √128 for the total distance.
Since Time = distance/rate
divide your total distance be 15 mph to get the total time.
Good luck.

correction  Math  Reiny, Thursday, August 18, 2011 at 8:37am
Just realized that I read the EAST as WEST .
however, because of the symmetry of the diagram, the final answer would be the same.
P should be (10,18)
equation of first path should by y = 3x+12
crosses the circle at (7.2, 9.6)
xintercept is x = 4
first leg from (7.2, 9.6) to (4,0)
(because of the squaring in the distance formula, the results do not change)
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