If a boat attached by a rope being pulled by a person walking along the shore is represented by a tractrix and the equation for the distance moved is given by
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A rope is attached to a boat at water level, and a person on the dock is pulling on the rope at a rate of 50 ft/min50 ft/min. If the person's hands are 16 ft16 ftabove the water level, how fast is the boat approaching when the amount of rope out is 20 ft20 ft?
Here is what I did:
hypotenuse =20 ft=C=20 ft=C
height of person above dock =16 ft=A=16 ft=A
distance from dock to boat =202−162−−−−−−−−√=122−−−√=12=B=202−162=122=12=B
A2+B2=C2A2+B2=C2
A(dA/dt)+B(dB/dt)=C(dC/dt)A(dA/dt)+B(dB/dt)=C(dC/dt)
B(dB/dt)=C(dC/dt)B(dB/dt)=C(dC/dt)
12(dB/dt)=20(50)12(dB/dt)=20(50)
dB/dt=83.3 ft/mindB/dt=83.3 ft/min
Here is what I did:
hypotenuse =20 ft=C=20 ft=C
height of person above dock =16 ft=A=16 ft=A
distance from dock to boat =202−162−−−−−−−−√=122−−−√=12=B=202−162=122=12=B
A2+B2=C2A2+B2=C2
A(dA/dt)+B(dB/dt)=C(dC/dt)A(dA/dt)+B(dB/dt)=C(dC/dt)
B(dB/dt)=C(dC/dt)B(dB/dt)=C(dC/dt)
12(dB/dt)=20(50)12(dB/dt)=20(50)
dB/dt=83.3 ft/mindB/dt=83.3 ft/min
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