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Salsk061 [2.6K]

3 years ago

6

A tourist at scenic Point Loma, California uses a telescope to track a boat approaching the shore. If the boat moves at a rate o

f 10 meters per second, and the lens of the telescope is 35 meters above water level, how fast is the angle of depression of the telescope (θ) changing when the boat is 210 meters from shore? Round any intermediate calculations to no less than six decimal places, and round your final answer to four decimal places.

1 answer:

Taya2010 [7]3 years ago

8 0

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

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