Pad 9:46 PM 81%--+ a webwork.daltonstate.net Dalton State College WeBWorK sp18-M

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Pad 9:46 PM 81%--+ a webwork.daltonstate.net Dalton State College WeBWorK sp18-Math2253 Section 4.1:9 Logged in as afisher6. MATHEMATICAL ASSOCIATION OF AMERICA Log Out C WeBWorK Kwebwork sp18-math2253 section_4.1/9 MAIN MENU Courses Section 4.1: Problem 9 Homework Sets Section Previous Problem Problem List Next Problem Probl (1 point) At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 17 knots and ship B is sailing north at 17 knots. How fast (in knots) is the distance between the ships changing at 4 PM? User Settings Grades The distance is changing at knots Problems (Note: 1 knot is a speed of 1 nautical mile per hour.) Problem1 Preview My Answers Submit Answers Problem 2 You have attempted this problem 2 times Your overall recorded score is 0% You have unlimited attempts remaining Problem3 Problem4 Problem 5 Email instructor Problem6 Problem7 Problem8 Problem9 Problem 10 v Page generated at 04/03/2018 at 09:45pm EDT WeBWork 1996-2017 theme: math4 | ww_version: WeBWork-2.13 pg_version PG-2.13 The WeBWorK Project

Explanation / Answer

First we find an equation to represent both ships and the distance between each. A is moving 25knots west and B is moving 17knots north. D will be the distance between the two.
Drawing it, you'll notice that it creates a right triangle. So we use the Pythagorean Theorm:
D^2 = A^2 + B^2

Differentiate in relation to time:

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

Now we must find all of our variables.

A = time(speed) + original distance = 4(17) + 10 = 78
B = 4(17) + 0 = 68
D = ?(A^2 + B^2) = 103.48
dA/dt = 17
dB/dt = 17
dD/dt = how fast the distance is changing

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

2D (dD/dt) = 2A (dA/dt) + 2B (dB/dt)
2(103.48) (dD/dt) = 2(78)(17) + 2(68)(17)
103.48 (dD/dt) = 2482
dD/dt = 2482/103.48
dD/dt = 23.98 knots = 24knots

At 4pm the distance between the ships is changing at the speed of 24knots.

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