## Calculus: Early Transcendentals 8th Edition

The ships are moving apart at a rate of $55.4~km/h$
Let $x$ be the east-west distance between the two ships. Then $x = 100~km$ Let $y$ be the north-south distance between the two ships. We can find $y$ at 4:00 pm: $y = (35~km/h)(4~h)+ (25~km/h)(4~h)$ $y = 240~km$ Let $z$ be the distance between the two ships. We can find $z$ at 4:00 pm: $z^2 = x^2+y^2$ $z = \sqrt{x^2+y^2}$ $z = \sqrt{(100~km)^2+(240~km)^2}$ $z = 260~km$ We can differentiate both sides of the equation with respect to $t$: $z^2 = x^2+y^2$ $2z~\frac{dz}{dt} = 2x~\frac{dx}{dt} + 2y~\frac{dy}{dt}$ $\frac{dz}{dt} = \frac{1}{z}~(x~\frac{dx}{dt} + y~\frac{dy}{dt})$ $\frac{dz}{dt} = \frac{1}{260}~[(100)(0) + (240)(25+35)]$ $\frac{dz}{dt} = 55.4~km/h$ The ships are moving apart at a rate of $55.4~km/h$