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January 19, 2017
Posted by **Becky** on Sunday, November 6, 2011 at 5:58pm.

- Calculus -
**bobpursley**, Sunday, November 6, 2011 at 6:27pmdraw 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 - Calculus -
**Damon**, Sunday, November 6, 2011 at 6:39pmxa = -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 - Calculus -
**Becky**, Sunday, November 6, 2011 at 6:48pmTHANKS FOR ALL THE HELP!

One question, how do you find xa,ya, xb, and yb? - Calculus -
**Damon**, Sunday, November 6, 2011 at 6:54pmxa 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 - Calculus -
**Becky**, Sunday, November 6, 2011 at 7:04pmOh 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! :) - Calculus -
**Becky**, Sunday, November 6, 2011 at 7:06pmOH NEVER MIND! I GOT IT HAHAHA!

- Calculus -
**Sydney**, Sunday, November 6, 2011 at 7:27pmCan 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)?