Question

A woman standing on a cliff is watching a motor boat though a telescope as the boat approaches the shoreline directly below her. If the telescope is 250 feet above the water and if the boat is approaching at 20 feet per second, at what rate is the angle of the telescope changing when the boat is 250 feet from shore

Answer 1

Answer:

Dα/dt   = 0.079 degree/sec

Step-by-step explanation:

From problem statement, is easy to see, that if point A is ubicated at the top of the telescope,  the shoreline is directly below the woman ( point B), and the point where the boat is, which is at distance x from shoreline  is point C. These three point  shape a right triangle with angle α (the angle of the telescope).

So we have

tan α  =  x/250    

Differentiating both sides of the equation we get

D (tan α)/dt    =  ( 1/250)* Dx/dt

sec² α Dα/dt  =  ( 1/250)* Dx/dt  

we already know that    Dx/dt   = 20 feet/sec

sec² α Dα/dt  =  20/250   ⇒     sec² α Dα/dt  = 0.08  

Dα/dt   =  0.08 / sec² α

Then

tan α  =  20/250  = 0,08        α   = arctan 0.08      α  ≈ 5⁰

Dα/dt   =  0.08/ sec² α  

From tables we get    cos  5⁰  =  0.9961  then

1/ 0.9961  = 1.003

sec α  = 1.003      and     sec²  α   =   1.0078

Dα/dt   =  0.08/ sec² α    ⇒   Dα/dt   =  0.08/1.0078

Dα/dt   = 0.079 degree/sec