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# math

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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?
Oct 23, 2014

#1
+94526
+5

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

[140 - 4(30)]2 + [4(20)]2 = r2

[140 - 120]2 + [80]2 =

[20]2 + [80]2  =

400 + 6400 =

6800 = r2

√6800km = r

And using

r2 = x2 + y2   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) = -600km2/hr + 1600km2/hr

√6800km(dr/dt) = 1000km2/hr

(dr/dt) = [1000/√6800km]km2/hr ≈ 12.13 km/hr

Oct 23, 2014

#1
+94526
+5

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

[140 - 4(30)]2 + [4(20)]2 = r2

[140 - 120]2 + [80]2 =

[20]2 + [80]2  =

400 + 6400 =

6800 = r2

√6800km = r

And using

r2 = x2 + y2   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) = -600km2/hr + 1600km2/hr

√6800km(dr/dt) = 1000km2/hr

(dr/dt) = [1000/√6800km]km2/hr ≈ 12.13 km/hr

CPhill Oct 23, 2014