Calculus

At 1:00 p.m. ship A is 25 km due north of ship B. If ship A is sailing west at a rate of 16km/h and ship B is sailing south at 20km/h, find the rate at which the distance between the two ships is changing at 1:30 p.m.

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asked by Becky
  1. draw the triangle (lets W, S)

    label the West leg W km, South leg Skm

    Distance between ship x.

    x= sqrt (W^2+S^2)
    dx/dt= 1/2 *1/sqrt( ) * (2w *dw/dt + 2S ds/dt)

    find dx/dt

    Caculate S, W from 1/2 hr at given speeds.
    you know dw/dt, ds/dt

  2. xa = -16 t
    ya = 25

    xb = 0
    yb = -20 t

    at 1/2 hour
    xa = -8
    ya = 25

    xb = 0
    yb = -10

    z = distance between
    z^2 = (xb-xa)^2 + (yb-ya)^2
    z^2 = (16 t)^2 + (-20 t - 25)^2
    2z dz/dt = 2(16t)(16) + 2(-20t-25)(-20)
    z dz/dt = 256 t +400 t + 500
    z dz/dt = 656 t + 500
    now at 1/2 hour
    z = sqrt(64 + 1225) = 35.9
    so
    35.9 dz/dt = 328+500
    dz/dt = 23.1 km/hr

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    posted by Damon
  3. THANKS FOR ALL THE HELP!

    One question, how do you find xa,ya, xb, and yb?

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    posted by Becky
  4. xa is the x position of the first ship
    At time 0 it is at x = 0
    then it proceeds in the west (negative x) direction at 16 km/hr
    so the x position of A is (0 - 16 t) or just -16 t
    Since it starts out 25 km north (positive y directio) and never goes north or sout, its Y position is always ya = 25
    etc

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    posted by Damon
  5. Oh i shoud've been more specific, what i meant to ask was how do you find xa, ya etc. at 1/2 hours?

    Thanks in advance! :)

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    posted by Becky
  6. OH NEVER MIND! I GOT IT HAHAHA!

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    posted by Becky
  7. Can you use that distance equation every problem or is it modified to fit this question?
    Also, how did you find z at 1/2 hours? (z = sqrt(64 + 1225) = 35.9)?

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    posted by Sydney

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