Related Rates Problemin Calculus: At noon, ship A is 180 km west of ship B. Ship A is sailing south at 20 km/h and ship B is sailing north at 40 km/h. How fast is the distance between the ships

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Related Rates Problemin Calculus: At noon, ship A is 180 km west of ship B. Ship A is sailing south at 20 km/h and ship B is sailing north at 40 km/h. How fast is the distance between the ships

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oops . how fast is the distance between the ships changing at 4:00pm?
ok, so you know the distance formula & how to find derivatives right?

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yes
ok, in 4 hourse, ship b will have traveled 160 miles due north, and ship a will have traveled 80 miles due south, there are still 180 miles between them east to west, right?
oh yeah okayy
z^2=180^2 + (x+y)^2..z=diagonal distance between ship a & ship b, x = ship a's distance south & y = shipb's distance north...
implicit diff. w/respect to t...you should knon the distances for x, y, z after 4 hrs, dx/dt = 20 kmh, dy/dt = 40 kmh, so just sub & solve.
wait where do i use the implicit differentiation? on the equation: 180^2 + (x + y)^2? because i'm still kind of confused
the whole thing...so you get... 2z dz/dt = 2(x+y) (dx/dt + dy/dt) *chain rule & 180 is constant so its der. is 0. then just sub & solve.
okay i think i might get it now
from z^2 = 180^2 + (x + Y)^2 (this was from pythag. theorem)
kay thanks for the help
just another question. is x = 20 and y = 40 and z = 180?
oh wait nvm thats wrong i think
sorry nat, but no x, will be 80 (the distance that ship a has sailed south after 4 hrs), y will be 160 (the distance that ship b has sailed north in 4 hrs), and z will be 300 (z^2 = 180^2 +(x+y)^2)...those are your distances. 20 and 40 are your rates (km/h), so dx/dt=20 and dy/dt = 40.
you begin with z^2 = 180^2 + (x+y)^2 (pythag. theorem*distance) and RELATE rates to it using implicit diff., helps to draw it sometimes as well.
ohhh okay that makes more sense

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