A fish rising vertically upward with a uniform velocity of 8m s ^ –...

Let the fish be at a depth $x$ from the surface and let the bird be at height $y$ from the surface. So, with respect to fish

$y ' = \muy + x ⇒ y ' = \dfrac{4y}{3} + x$

Differentiating with respect to time,

$\dfrac{dy '}{dt} = \dfrac{4dy}{3dt} + \dfrac{dx}{dt}$

$⇒ \dfrac{4y}{3} + \dfrac{dx}{dt} = - 12$

$⇒ \dfrac{4}{3} \cdot \dfrac{dy}{dt} = - 12 + 8 = - 4$

$⇒ \dfrac{dy}{dt} = - 3m{s}^{- 1}$