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 them changing at the end of 1 hour?

Answer 1

Answer:

$\dfrac{dz}{dt} =-5.6\ mile/h$

Explanation:

distance between ship A and B = 32 mile

Ship A velocity in south, dx/dt = -16 mph

Ship B is sailing toward east with speed, dy/st = 12 mph

time = 1 hour

rate of change of distance between them = ?

x be the distance travel after t time

X = 32 + x

Let distance between them be z

now, using Pythagoras theorem to calculate distance between ships after 1 hours

z² = x² + y²

z² = (32 + x)² + 12²

z² = (32 - 16)² + 12²

z = √400

z = 20 miles

now, calculation of rate of change of distnace

z² = (32 + x)² + y²

differentiating both side w.r.t. time

$2 z \dfrac{dz}{dt} = 2(32+x)\dfrac{dx}{dt} + 2 y\dfrac{dy}{dt}$

$z \dfrac{dz}{dt} =(32-16)\dfrac{dx}{dt} +y\dfrac{dy}{dt}$

$20\times \dfrac{dz}{dt} =16\times (-16) +12\times 12$

$\dfrac{dz}{dt} =\dfrac{-112}{20}$

$\dfrac{dz}{dt} =-5.6\ mile/h$

hence, the rate is the distance between them changing at the end of 1 hour is equal to $\dfrac{dz}{dt} =-5.6\ mile/h$