At noon, ship A is 150Km west of ship B. Ship A is sailing east at 35km/h and ship B is sailing north at 25km/h. How fast is the distance between the ships changing at 4:00PM?

Printable View

- March 18th 2007, 05:59 PMcamherokidDifferentiation
At noon, ship A is 150Km west of ship B. Ship A is sailing east at 35km/h and ship B is sailing north at 25km/h. How fast is the distance between the ships changing at 4:00PM?

- March 18th 2007, 07:39 PMJhevon
Very Important: the first thing we want to do is draw a diagram so we can see what's going on. Fill in everything on the diagram that we know, don't know, and want to know. (See the diagram I have attached before reading the solution).

Let x be the distance between ship A and the initial position of ship B (B1).

Let y be the distance ship B has traveled.

Let z be the distance between the two ships.

After 4 hours:

A travels 140 km (35km/h *4 hrs). So the distance between A and B1 is 10 km. It is travelling at a speed of 35 km/h so dx/dt = - 35 km/h (note, the rate is negative since the distance is decreasing).

B travels 100 km from its initial position.

By pythagoras then, after 4 hours, z^2 = 10^2 + 100^2

=> z = sqrt(100 + 10000) = 100.50 km

Now to solve the problem, we again use Pythagoras' Theorem:

z^2 = x^2 + y^2 ..........now differentiate implicitely

=> 2z dz/dt = 2x dx/dt + 2y dy/dt

=> dz/dt = (2x dx/dt + 2y dy/dt)/2z

after 4 hours: x = 10, y = 100, dx/dt = -35, dy/dt = 25, z = 100.50

=> dz/dt = [2(10)(-35) + 2(100)(25)]/2(100.5)

=> dz/dt = 21.39 km/h ....the rate at which the distance between the two ships are changing - March 18th 2007, 07:40 PMSoroban
Hello, camherokid!

Quote:

At noon, ship A is 150Km west of ship B.

Ship A is sailing east at 35km/h and ship B is sailing north at 25km/h.

How fast is the distance between the ships changing at 4:00pm?

Code:`|`

* B

x * |

* | 25t

* |

* - - - - * - - - - - * Q

P 35t A 150-35t

Ship A starts at point P, 150 km west of point Q.

. . In*t*hours, it has moved*35t*km east to point A.

. . Hence: .AQ = 150 - 35t

Ship B starts at point Q.

. . In*t*hours, it has moved*25t*km north to point B.

Let x = AB

Then: .x² .= .(25t)² + (150 - 35t)²

. . and we have: .x² .= .1850t² - 10,500t + 22,500 .**[1]**

Differentiate with respect to time: .2x(dx/dt) .= .3700t - 10,500

. . and we have: .dx/dt .= .(1850t - 5250)/x .**[2]**

When t = 4, we have: .[1] .x² .= .1850·4² - 10,500·4 + 22,500 .= .10,100

. . Hence: .x .= .√10,100 .= .10√101

Substitute into [2]: .dx/dt .= .(1850·4 - 5350)/(10√101) .= .21.39329959

Therefore, the ships are separating at about 21.4 km/hr.

- March 18th 2007, 07:41 PMJhevon