Sunday

March 29, 2015

March 29, 2015

Posted by **Parker** on Thursday, September 24, 2009 at 11:55pm.

- calculus -
**Reiny**, Friday, September 25, 2009 at 12:17amlet t hours be some time past 12:00 noon

After t hours, ship B has gone 20t knots

and ship A has gone 17t knots

let the distance between them be D

I see a right-angled triangle and

D^2 = (20t)^2 + (17t+10)^2

D^2 = 689t^2 + 340t + 100

2D(dD/dt) = 1378t + 340

dD/dt = (689t + 170)/D

at 4:00 pm, t=4 and

D^2 = 12484

D = 111.7318

so at t=4

dD/dt = (689(4) + 170)/111.7318

= 26.19 knots

- calculus -
**Parker**, Friday, September 25, 2009 at 1:56amIt still says NOT CORRECT.

**Answer this Question**

**Related Questions**

Calculus - At noon, ship A is 50 nautical miles due west of ship B. Ship A is ...

calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...

calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...

calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...

calculus - At noon, ship A is 30 nautical miles due west of ship B. Ship A is ...

CALCULUS - At noon, ship A is 40 nautical miles due west of ship B. Ship A is ...

Calculus - At noon, ship A is 20 nautical miles due west of ship B. Ship A is ...

Calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...

Calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...

calculus 1 - At noon, ship A is 20 nautical miles due west of ship B. Ship A is ...