#### Answer

(a) Initially the east-west distance $x = 150~miles$
Initially the north-south distance $y = 0$
Ship A's speed to the east is $35~km/h$ so $\frac{dx}{dt} = -35~km/h$
Ship B's speed to the north is $25~km/h$ so $\frac{dy}{dt} = 25~km/h$
(b) The unknown is $\frac{dD}{dt}$ where $D$ is the distance between the ships.
(c) We can see a sketch below.
(d) $x^2+y^2 = D^2$
(e) At 4:00 pm, the distance between the two ships is increasing at a rate of $~~21.4~km/h$

#### Work Step by Step

(a) Initially the east-west distance $x = 150~miles$
Initially the north-south distance $y = 0$
Ship A's speed to the east is $35~km/h$ so $\frac{dx}{dt} = -35~km/h$
Ship B's speed to the north is $25~km/h$ so $\frac{dy}{dt} = 25~km/h$
(b) The unknown is $\frac{dD}{dt}$ where $D$ is the distance between the ships.
(c) We can see a sketch below.
(d) $x^2+y^2 = D^2$
(e) We can find $x$ after $4~hours$:
$x = 150-(4)(35) = 10$
We can find $y$ after $4~hours$:
$y = (4)(25) = 100$
We can find $D$ after $4~hours$:
$D^2 = x^2+y^2$
$D = \sqrt{x^2+y^2}$
$D = \sqrt{10^2+100^2}$
$D = 100.5~miles$
We can find $\frac{dD}{dt}$:
$2x~\frac{dx}{dt}+2y~\frac{dy}{dt} = 2D~\frac{dD}{dT}$
$\frac{dD}{dt} = \frac{x~\frac{dx}{dt}+y~\frac{dy}{dt}}{D}$
$\frac{dD}{dt} = \frac{(10)(-35)+(100)(25)}{100.5}$
$\frac{dD}{dt} = 21.4~km/h$
At 4:00 pm, the distance between the two ships is increasing at a rate of $~~21.4~km/h$