9514 1404 393
Answer:
0.4424°/s
Step-by-step explanation:
We can get an answer to the desired accuracy by approximating the rate of change over a small time interval.
0.05 seconds before the time of interest, the boat will be ...
210 +10(0.05) = 210.5
meters from shore. The angle of depression from the observer is ...
tan(θ) = (35 m)/(210.5 m) ⇒ θ = arctan(35/210.5) ≈ 9.440251°
Then 0.05 seconds after the time of interest, the angle of depression will be ...
θ = arctan(35/(210-0.05(10))) = arctan(35/209.5) ≈ 9.484495°
The rate of change of angle of depression over that 0.1-second period is ...
(9.484495° -9.440251°)/(0.1 s) ≈ 0.4424°/s
_____
<em>Additional comment</em>
If you write the function relating angle θ and time t, you would have ...
θ = arctan(35/(210-10t))
The derivative of this with respect to time is ...
θ' = 1/(1 +(35/(210-10t))²)×(350/(210 -10t)²) = 350/((210 -10t)² +35²)
At t=0, this is 350/(210² +35²) = 2/259 . . . . . radians per second
Then in degrees per second, the rate of change of angle is ...
(2/259)(180/π) = 360/(259π) °/s ≈ 0.4424°/s
