Thursday

October 23, 2014

October 23, 2014

Posted by **JEN** on Monday, July 15, 2013 at 9:04am.

- math -
**Steve**, Monday, July 15, 2013 at 10:22amWell, since the ship is equidistant from A and B, it lies on the perpendicular bisector of AB, so if A is at (0,0) and B is at (6,0), the ship is at (3,4).

If the ship is at (x,y), the distances u from A and v from B are

u^2 = x^2 + y^2

v^2 = (x-6)^2 + y^2

2u du/dt = 2x dx/dt + 2y dy/dt

2v dv/dt = 2(x-6) dx/dt + 2y dy/dt

When x is at (3,4)

3 dx/dt + 4 dy/dt = 140

-3 dx/dt + 4 dv/dt = 20

dx/dt = 20

dy/dt = 20

Looks like it's moving NE at 20√2 km/h

**Answer this Question**

**Related Questions**

Thinking Mathematically - Geometry - A rocket ascends vertically after being ...

Thinking Mathematically - Geometry - A rocket ascends vertically after being ...

calculus - A drone aeroplane is flying horizontally to a constant height of 4000...

Trig - From two different tracking stations, a weather balloon is spotted from ...

Calculus - 1.At noon of a certain day,ship A is ^0 km due north of ship B. If A ...

Phyics - A ship is first seen on a radar screen to be 13 km east of the radar ...

Calculus - At 8:00am, ship A is 120km due east of ship B. Ship A is moving north...

Math - To chart the movement of a polar bear, scientists attached a radio ...

calculus - shortly after taking off, a plane is climbing at an angle of 30° and ...

Calculus - At noon of a certain day,ship A is 60 km due north of ship B. If A ...