Question

at noon, ship a is 40 nautical miles due west of ship b. ship a is sailing west at 22 knots and ship b is sailing north at 17 knots. how fast (in knots) is the distance between the ships changing at 7 pm? (note: 1 knot is a speed of 1 nautical mile per hour.)

Explanation:

Set up coordinate system and variables

Let the initial position of Ship B at noon be the origin \((0,0)\).
Since Ship A is \(40\) nautical miles due west of Ship B at noon, its initial position is \((-40, 0)\).
Let \(x(t)\) be the distance of Ship A west of the origin at time \(t\) (in hours after noon).
Let \(y(t)\) be the distance of Ship B north of the origin at time \(t\).
Since Ship A sails west at \(22\text{ knots}\):

$$x(t) = 40 + 22t \implies \frac{dx}{dt} = 22$$

Since Ship B sails north at \(17\text{ knots}\):

$$y(t) = 17t \implies \frac{dy}{dt} = 17$$

Let \(s(t)\) be the distance between the two ships:

$$s^2 = x^2 + y^2$$

Evaluate values at 7 PM

At \(7\text{ PM}\), \(t = 7\) hours:

$$x(7) = 40 + 22(7) = 40 + 154 = 194$$

$$y(7) = 17(7) = 119$$

$$s = \sqrt{194^2 + 119^2} = \sqrt{37636 + 14161} = \sqrt{51797}$$

Differentiate and solve for ds/dt

Differentiating \(s^2 = x^2 + y^2\) with respect to \(t\):

$$2s \frac{ds}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \implies \frac{ds}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{s}$$

Substitute the values at \(t = 7\):

$$\frac{ds}{dt} = \frac{194(22) + 119(17)}{\sqrt{51797}} = \frac{4268 + 2023}{\sqrt{51797}} = \frac{6291}{\sqrt{51797}} \approx 27.64$$

Answer:

At noon, ship A is 40 nautical miles due west of ship B. Ship A is sailing west at 22 knots and ship B is sailing north at 17 knots. How fast (in knots) is the distance between the ships changing at 7 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.) <blank>\(\frac{6291}{\sqrt{51797}}\)</blank> knots