A boat is being pulled toward a dock by a rope from the bow through a ring on a dock ...

Question

A boat is being pulled toward a dock by a rope from the bow through a ring on a dock 12 ft above the bow. the rope is haled at a rate of 3 ft/sec.

a) how fast is the boat approaching the dock when 20 ft of rope is out

b) at what rate is the angle theta (the angle from the rope to the water) changing then?

I'm just double checking my work.

I got dx/dt is equal to 3.75 ft/s

and dΘ/dt = -72/640 rad/s

Is this correct?

Answer 1

Your work is correct. Your answer for part b is not simplified, which is insignificant depending on your teacher.

a.)

Known:

1. b = 12 ft (triangle height)
2. c = 20 ft (triangle hypotenuse)
3. ^dc/_dt = ^3ft/_sec

Solving for ^da/_dt ... (change in triangle length with respect to t)

Set up Pythagorean theorem

a² + b² = c²

(^da/_dt)² + b² = (^dc/_dt)² Take the derivative of both sides with respect to t

(^dc/_dt)² - (^da/_dt)² = b²

^d/_dt[(^dc/_dt)² - (^da/_dt)²] = ^d/_dt[12²] Apply chain rule

2c(t)c'(t) - 2a(t)a'(t) = 0

2c(t)c'(t) = 2a(t)a'(t) Divide both sides by 2 and rearrange to solve for a'(t)

a'(t) = ^c(t)c'(t)/_a(t) Plugin known variables, a = √(20² - 12²) = 16

a'(t) = ²⁰⁽³⁾/₁₆

a'(t) = ^da/_dt = 3.75ft/sec

b.)

Solving for ^dθ/_dt ...

Set up trigonometric equation

sinθ = ^opposite/_hypotenuse = ^b/_c

^d/_dt(sinθ) = ^d/_dt(¹²/_c) Left: (sinθ)' = cosθ; Right: apply chain rule

cosθ•^dθ/_dt = -12c^-2•^dc/_dt Rearrange to solve for ^dθ/_dt

^dθ/_dt = (-12c^-2•^dc/_dt)/cosθ Plugin known variables, cosθ = ^adjacent/_hypotenuse = 16/20 = 4/5

^dθ/_dt = (-12(20)^-2(3))/(⁴/₅)

^dθ/_dt = -⁹/₈₀rad/sec = -0.1125rad/sec