Rate of change of distance between two ships

Since the ships sail south and west from the same starting point, the paths form a right angle, so the distance $z$ between the ships satisfies $z^2 = x^2 + y^2$, where $x = 20t$ and $y = 15t$. Differentiating both sides with respect to $t$ gives $2z\frac{dz}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}$, which simplifies to $z\frac{dz}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt}$.

At $t = 3$ hours, $x = 20(3) = 60$ miles and $y = 15(3) = 45$ miles, so $z = \sqrt{60^2+45^2} = \sqrt{3600+2025} = \sqrt{5625} = 75$ miles. Substituting $x=60$, $y=45$, $z=75$, $\frac{dx}{dt}=20$, and $\frac{dy}{dt}=15$ into $z\frac{dz}{dt} = x\frac{dx}{dt}+y\frac{dy}{dt}$ gives $75\frac{dz}{dt} = 60(20)+45(15) = 1200+675 = 1875$, so $\frac{dz}{dt} = 25$ mi/h.

The separation rate found in part a was 25 miles per hour, while Ship B travels at 20 miles per hour. Since 25 is greater than 20, the distance between the ships increases faster than Ship B itself travels. This makes sense because the separation rate combines the effects of both ships moving away from the port simultaneously. The answer is faster.