Asked by Jermaine
                A jet flying due north at 16km/min passes 2 km directly above a plan flying due east at 8km/min. How quickly are they separating when the plane is 32 km from the crossover?
What I have done (I am not sure if it is correct.):
r^2=x^2+y^2+z^2
Since the plane has gone 32 meters, 32km/8km/min=4seconds
Therefore for the jet, 16km/min(4seconds)=64km.
So, 32km=x value, 64km=y value and z would be 2 because that is how far they are from each other?
Solve for r^2=x^2+y^2+z^2
= 32^2+64^2+2^2
=5124
r=71.58
So, then,
r^2=x^2+y^2+z^2
2r d(r)/d(t)= 2x dx/dt + 2y dy/dt + 0 (since 2^2 will be a constant)
Now, we subb in the values.
71.58 dr/dt = 32 (8) + 64 (16)
dr/dt = 32 (8) + 64 (16)/71.58
=17.8km/min
Is this done correctly, or did I go wrong somewhere?
            
            
        What I have done (I am not sure if it is correct.):
r^2=x^2+y^2+z^2
Since the plane has gone 32 meters, 32km/8km/min=4seconds
Therefore for the jet, 16km/min(4seconds)=64km.
So, 32km=x value, 64km=y value and z would be 2 because that is how far they are from each other?
Solve for r^2=x^2+y^2+z^2
= 32^2+64^2+2^2
=5124
r=71.58
So, then,
r^2=x^2+y^2+z^2
2r d(r)/d(t)= 2x dx/dt + 2y dy/dt + 0 (since 2^2 will be a constant)
Now, we subb in the values.
71.58 dr/dt = 32 (8) + 64 (16)
dr/dt = 32 (8) + 64 (16)/71.58
=17.8km/min
Is this done correctly, or did I go wrong somewhere?
Answers
                    Answered by
            Steve
            
    Aside from all the carelessness with the units, the math look  good.
    
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