# Math Help - Differentiation

1. ## Differentiation

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?

2. Originally Posted by camherokid
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?
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

3. Hello, camherokid!

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: . .= .(25t)² + (150 - 35t)²
. . and we have: . .= .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] . .= .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.

4. Originally Posted by Soroban
Hello, camherokid!

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: . .= .(25t)² + (150 - 35t)²
. . and we have: . .= .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] . .= .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.
Sorry again Soroban, i had no idea you were working on this problem. I checked the "Who's Online" Forum to see if anyone was working on it and saw no one. It's good to verify that I'm right though, so your work is not in vain.