At noon, ship A is 140 km west of ship B. Ship A is sailing east at 30 km/h and ship B is sailing north at 20 km/h. How fast is the distance between the ships changing at 4:00 PM?

Guest Oct 23, 2014

#1**+5 **

Using the Pythagorean Theorem, the distance between the ships, (r), at 4PM =

[140 - 4(30)]^{2} + [4(20)]^{2} = r^{2}

[140 - 120]^{2} + [80]^{2} =

[20]^{2} + [80]^{2} =

400 + 6400 =

6800 = r^{2}

√6800km = r

And using

r^{2} = x^{2} + y^{2} where x = 20, y = 80 and r = √6800 and differentiating for time, we have

2r(dr/dt) = 2x(dx/dt) + 2y(dy/dt)

r(dr/dt) = x(dx/dt) + y(dy/dt)

Note that, if we consider x, y and r to be sides of a triangle, then (dx/dt) = -30km/hr and (dy/dt) = 20km/hr

√6800km(dr/dt) = 20km(-30km/hr) + 80km(20km/hr)

√6800km(dr/dt) = -600km^{2}/hr + 1600km^{2}/hr

√6800km(dr/dt) = 1000km^{2}/hr

(dr/dt) = [1000/√6800km]km^{2}/hr ≈ 12.13 km/hr

CPhill Oct 23, 2014

#1**+5 **

Best Answer

Using the Pythagorean Theorem, the distance between the ships, (r), at 4PM =

[140 - 4(30)]^{2} + [4(20)]^{2} = r^{2}

[140 - 120]^{2} + [80]^{2} =

[20]^{2} + [80]^{2} =

400 + 6400 =

6800 = r^{2}

√6800km = r

And using

r^{2} = x^{2} + y^{2} where x = 20, y = 80 and r = √6800 and differentiating for time, we have

2r(dr/dt) = 2x(dx/dt) + 2y(dy/dt)

r(dr/dt) = x(dx/dt) + y(dy/dt)

Note that, if we consider x, y and r to be sides of a triangle, then (dx/dt) = -30km/hr and (dy/dt) = 20km/hr

√6800km(dr/dt) = 20km(-30km/hr) + 80km(20km/hr)

√6800km(dr/dt) = -600km^{2}/hr + 1600km^{2}/hr

√6800km(dr/dt) = 1000km^{2}/hr

(dr/dt) = [1000/√6800km]km^{2}/hr ≈ 12.13 km/hr

CPhill Oct 23, 2014