# calculus ab At noon, ship A is 150 km west of ship B. Ship A is sailing east at 35 km/hr and ship B is sailing north at 25 km/hr. How fast is the distance changing between them at 4:00 PM? o

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The initial distance between A and B is 150 km.

Four hour has gone by, we know that A travels to east at 35km/hr. Hence, for 4hrs ship A travels -35 * 4 = - 140km. (the rate is negative since, ship A goes nearer and nearer to B).

This will affect the horizontal distance 150km, 150km - 140km = 10km. Let set that distance to be "x".

Now, we know that B travels to north at 25km/hr. Hence, for 4hrs ship B travels 25*4 = 100km. Let set that distance to be "y".

After 4 hrs, our triangle will be:

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-      -   y = 100

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x = 10

Let set the hypotenuse to be r. We can solve for the measure of r using Pythagorean Theorem:

x^2 + y^2 = r^2

===>  10^2 + 100^2 = r^2

===>  100 + 10000 = r^2

===> 10100 = r^2

Getting the square root of both sides we will get,

r = 100.499

Now, we differentiate implicitly the x^2 + y^2 = r^2 with respect to t (time).

===>  2xdx/dt + 2y/dt = 2rdr/dt

Dividing both sides by 2:

===> xdx/dt + ydy/dt = rdr/dt

We can now solve for the dr/dt.

Plug-in x = 10, dx/dt = -35, y = 100, dy/dt = 25 and r = 100.499.

===> (10)(-35) + (100)(25) = 100.499(dr/dt)

===> -350 + 2500 = 100.499dr/dt

===> 2150 = 100.499dr/dt

Dividing both sides by 100.499 to isolate the dr/dt on right side,

2150/100.499 = (100.499dr/dt)/100.499

===>  21.39 = dr/dt

or dr/dt = 21.39 km/hr

Approved by eNotes Editorial Team Label the position of ship B at noon as P.

At 4:00pm ship A is 10km from P (it started 150km from P and has travelled 35km/h for 4hr or 140km.) Ship B is 100km from P (it has travelled 25km/h for 4 hours or 100km.)

Ship A's speed is 35km/h, so `(dA)/(dt)=-35` (negative since the distance from A to P is decreasing as t increases). Also ship B's speed is 25km/h so `(dB)/(dt)=25` (positive as the distance between B and P increases as t increases.)

The distance between A and B is `d=sqrt(A^2+B^2)` where A is the distance fro A to P, and B is the distance from B to P.

Then the rate at which the distance between the ships is changing is:

`(dd)/(dt)=1/2(A^2+B^2)^(-1/2)(2A(dA)/(dt)+2B(dB)/(dt))`

Substituting the known values we get:

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

`=4300/(2sqrt(10100))`

`~~21.39`

Thus the distance between the ships is increasing at approxiamtely 21.39km/h

Approved by eNotes Editorial Team