Question 1117084: A woman standing on a cliff is watching a motor boat through a telescope, s the boat approaches the shoreline directly below her. If the telescope is 25 m above the water level and if the boat is approaching the cliff at 20 m/s, what rate is the acute angle made by the telescope with the vertical changing when the boat is 250 m form the shore?
Answer by addingup(3677)   (Show Source):
You can put this solution on YOUR website! Time:
(d/dt)(tanθ) = d/dt(x/250)
(sec^(2)θ)(dθ/dt) = (1/250)(dx/dt)
solve for θ:
tanθ = 250/250
θ = 45° --> π/4 radians
Substitute for θ:
sec^(2)π/4(dθ/dt) = (1/250)(dx/dt)
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let me pause to solve for sec^(2)π/4:
sec^(2)θ = 1/cos^(2)θ
sec^(2)π/4 = 1/(cos^(2)π/4)
sec^(2)π/4 = 1/(sqrt2/2) = 2
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
OK, so I'm going to substitute for sec^(2)π/4 with 2:
(2)(dθ/dt) = (1/250)(dx/dt)
dθ/dt = (1/500)(dx/dt)
Now I'll substitute dx/dt with -20 (negative, because the boat is getting closer):
dθ/dt = (1/500)(-20); where dx/dt = -20
dθ/dt = -1/25 radians/sec at x = 250 feet
So, the rate of change is -1/25 radians/sec, which is about -2.292 degrees when the boat is 250 meters from the shore
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