Tuesday

October 21, 2014

October 21, 2014

Posted by **Lily** on Wednesday, August 17, 2011 at 11:28pm.

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 8:06amUse co-ordinate 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 y-intercept 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 x-axis 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 8:37amJust 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)

x-intercept 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)

**Answer this Question**

**Related Questions**

College Math II - Solve by any method. a^4 – 5a^2 + 4 = 0 I am having problems ...

Math - 3x-4 to y+15 is constant and y=3 when x=2 then when y=12, x equals? My ...

Math - OK, I have had trouble with math, 'C' average, and we are learning ...

Airplane & Radar - (Airplane & Radar) An airplane is flying (horizontally) at ...

Math - Solve for x: x^4-15X^2+14=0 How would I solve this problem? Could someone...

CALCULUS - A plane, flying horizontally at an altitude of 1 km and a speed of ...

calculus - A plane flying horizontally at an altitude of 1 mi and a speed of 500...

Math - I need someone to check and go over my answers to make sure they are ...

Math - I need someone to check and go over my answers to make sure they are ...

College Math II - Solve each problem. Find the quotient and remainder when x^2...