Answer
The boat is approaching the dock at a rate of $1.008~m/s$
Work Step by Step
Let $x$ be the distance between the boat and the dock:
Then $x = 8~m$
Let $y$ be the vertical height of the pulley:
Then $y = 1~m$
Let $z$ be the distance between the boat and the pulley. We can find $z$ when the boat is 8 meters from the dock:
$z^2 = x^2+y^2$
$z = \sqrt{x^2+y^2}$
$z = \sqrt{(8~m)^2+(1~m)^2}$
$z = \sqrt{65}~m$
We can differentiate both sides of the equation with respect to $t$:
$z^2 = x^2+y^2$
$2z~\frac{dz}{dt} = 2x~\frac{dx}{dt} + 2y~\frac{dy}{dt}$
$2x~\frac{dx}{dt} = 2z~\frac{dz}{dt} - 2y~\frac{dy}{dt}$
$\frac{dx}{dt} = \frac{1}{x}~(z~\frac{dz}{dt} - y~\frac{dy}{dt})$
$\frac{dx}{dt} = \frac{1}{8}~[(\sqrt{65})(-1) - (1)(0)]$
$\frac{dx}{dt} = -1.008~m/s$
The boat is approaching the dock at a rate of $1.008~m/s$
