Question

Ship A is 32 miles north of ship B and is sailing due south at 16 mph. Ship B is sailing due east at 12 mph. At what rate is the distance between the ships changing an hour later? Are they getting closer together or farther apart?

Answer 1

Answer:

-5.6 mi/h

Explanation:

$\frac{da}{dt}$ = Velocity of ship A = -16 mi/h (negative as the distance is decreasing)

Distance ship A travels in 1 hour = 16×1 = 16 mi

a = Distance remaining = 32-16 = 16 mi

$\frac{db}{dt}$ = Velocity of ship B = 12 mi/h

b = Distance ship B travels in 1 hour = 18×1 = 12 mi

c = Distance between A and B after 1 hour = √(a²+b²) = √(16²+12²) = 20 mi

From Pythagoras theorem

a²+b² = c²

Now, differentiating with respect to time

$2a\frac{da}{dt}+2b\frac{db}{dt}=2c\frac{dc}{dt}\\\Rightarrow a\frac{da}{dt}+b\frac{db}{dt}=c\frac{dc}{dt}\\\Rightarrow \frac{dc}{dt}=\frac{a\frac{da}{dt}+b\frac{db}{dt}}{c}\\\Rightarrow \frac{dc}{dt}=\frac{18\times 54+24\times 72}{90}\\\Rightarrow \frac{dc}{dt}=-5.6\ mi/h$

∴ Rate at which distance between the ships is decreasing one hour later is 5.6 mi/h