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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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

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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?

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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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