题目

求解一道AP微积分题!
At noon,ship A is 40 nautical miles due east of ship B.Ship A is sailing east at 25 knots and ship B is sailing north at 20 knots.How fast (in knots) is the distance between the ships changing at 4 PM?(Note:1 knot is a speed of 1 nautical mile per hour.)

提问时间：2021-01-13

答案

正午(12点)时,A船位于B船的正东40海里处.A船以25节的速度向正东方航行,船B以20节的速度向正北航行,求在下午4点时两船之间的距离变化率(即速度).
设A船的行程为：S1=40+25t,B船的行程为：S2=20t.
两船的合成距离S=√[(40+25t)^2+(20t)^2].
ds/dt=(1/2)*{1/√[(40+25t)^2+(20t)^2]}*[2(40+25t)*(25)+2(20t)*20].
ds/dt=(1/2)*[1/√(140^2+80^2)]*(2*140*25+40*80).
=(1/2)*(1/161.2)*10200.
=10200/322.4.
≈31.63(节).----即为所求.
【在下午4点时,两船相距ds=31.63*dt=31.63*4=126.5 (海里).】