作业帮求解一道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 spee
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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 spee
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.)
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 spee
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).】
楼上解的不对，不是distance between，是changing rate of distance，求的是速度
前半段与楼上解法一致，也是设坐标
之后列写距离s=Sqrt((20t)^2+(40+25t)^2)
其中20t是B离原点的距离，40+25t是A离原点的距离，Sqrt()代表取根号
对此方程中的t求导，可得
v=(1250 t + 40 (4...
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楼上解的不对，不是distance between，是changing rate of distance，求的是速度
前半段与楼上解法一致，也是设坐标
之后列写距离s=Sqrt((20t)^2+(40+25t)^2)
其中20t是B离原点的距离，40+25t是A离原点的距离，Sqrt()代表取根号
对此方程中的t求导，可得
v=(1250 t + 40 (40 + 20 t))/(2 Sqrt[625 t^2 + (40 + 20 t)^2])
带入t=4，v=245/Sqrt(61)
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at noon, mark B as (0,0), then A is (40.0)
at 4pm, B change to (0, 80), A change to (140,0)
distance AB=√(80^2+140^2)=20√65 (nautical mile)