At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing south

Question

At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing south at 23 knots and ship B is sailing north at 25 knots. How fast (in knots) the distance between the ships changing at 7 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.) Let y = the distance ship B has traveled since noon. Let x = the distance ship A has traveled since noon. Let z = the direct distance between ship A and ship B. In this problem you arc given two rates. What arc they? Express your answers in the form dx/dt, dy/dt, or dz/dt = a number. Enter your answers in the order of the variables shown; that is, dx/dt first, dy/dt, etc. next. What rate are you trying to find? Write an equation relating the variables. Note: In order for WeBWorK to check your answer you will need to write your equation so that it has no denominators. For example, an equation of the form 2/x = 6/y should be entered as 6x=2y or y = 3x or even y - 3x = 0. Use the chain rule to differentiate this equation and then solve for the unknown rate, leaving your answer in equation form. Substitute the given information into this equation and find the unknown rate. Express your answer in the form dx/dt, dy/dt, or dz/dt = a number.

Explanation / Answer

x = 23*7 = 161

y = 25*7 = 175

z^2 = 10^2 + (161+175)^2, so z = sqrt(112996) = 2 sqrt(28249)

dz/dt = (161+175)(23+25)/2 sqrt(28249) = 7*48^2/(2 sqrt(28249)) = 8064/sqrt(28249)

8064/sqrt(28249) is approximately 47.9787556088062. Note how this value is clos to 23+25=48; that is because the distance they are apart east-west has less and less effect on the total distance as they move further apart north-south.

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