nal Prolet Sample Dram: Final Project-Sample Drawi1WeeworK : winter2017 1 + /web

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nal Prolet Sample Dram: Final Project-Sample Drawi1WeeworK : winter2017 1 + /webworklatech 01718-Math220-Crow oow/section,44/3/Tuser: hsg003&effectivelser;= hsg003&key;: 1 LOUISIANA TECH webwork / winter201718-math220-crow / section_4-4/3 Section 4-4: Problem 3 Previous Problem Problem List Next Problem (1 point) At noon, ship A is 40 nautical miles due west of ship B. Ship A is sailing west at 23 knots and ship 8 is sailing north at 25 knots. How fast (in knots) is the distance between the ships changing at 6 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.) Preview My Answers Submit Answers You have attempted this problem 0 times You have unlimited attempts remaining Email instructor Page peperated at 01/062018 at 04 07pm cST PO 2 13 The WeBWork Project

Explanation / Answer

So we use the Pythagorean Theorm:
D^2 = A^2 + B^2

Differentiate with respect to time: we get

2D (dD/dt) = 2A (dA/dt) + 2B (dB/dt)

finding all variables.

A = time(speed) + original distance = 6(23) + 40 = 178
B = 6(25) + 0 = 150
D = (A^2 + B^2)

=(178^2+150^2)

D = 232.775
dA/dt = 23
dB/dt = 25
dD/dt = how fast the distance is changing

Plug in all your variables and solving for (dD/dt):

2D (dD/dt) = 2A (dA/dt) + 2B (dB/dt)
2(232.775) (dD/dt) = 2(178)(23) + 2(150)(25)
465.55 (dD/dt) = 15688
dD/dt = 15688 /465.55
dD/dt = 33.7 knots

At 6pm the distance between the ships is changing at the speed of 33.7 knots

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