# How fast does the distance between the ships changing at 4 P.M.? At noon, ship A is 150 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 25 km/h. Let ship B begin at the origin and ship A at (-150,0) on a cartesian coordinate plane.

Then the distance between the two ships at time t is given by `d=sqrt(x^2+y^2)` . We want to find the rate of change of the distance after 4 hours.

At 4pm x=-10,y=100. We...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Let ship B begin at the origin and ship A at (-150,0) on a cartesian coordinate plane.

Then the distance between the two ships at time t is given by `d=sqrt(x^2+y^2)` . We want to find the rate of change of the distance after 4 hours.

At 4pm x=-10,y=100. We are given that `(dx)/(dt)=-35` and `(dy)/(dt)=25` .

Then `(dd)/(dt)=d/(dt)[sqrt(x^2+y^2)]` Taking the implicit derivative we get:

`(dd)/(dt)=1/2(x^2+y^2)^(-1/2)[2x(dx)/(dt)+2y(dy)/(dt)]` where `(dd)/(dt)` is the rate of change of the distance function with respect to time. Plugging in the known values we get:

`(dd)/(dt)=1/2((-10)^2+100^2)^(-1/2)[2(-10)(-35)+2(100)(25)]`

`=5700/(2sqrt(10100))`

`~~28.36`

----------------------------------------------------------------

So the rate of change of the distance between the ships at 4 pm is approximately 28.36 km/hr.

---------------------------------------------------------------

Approved by eNotes Editorial Team