At noon, ship A is 50 nautical miles due west of ship B. Ship A is sailing west at 19 knots and ship B is sailing north at 24 knots. How fast (in knots) is the distance between the ships changing at 3 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)
after 3hrs the distance travelled by A increases by 19*3=57 nautical miles.Distance by A equals 50+57=117nm distance by B equals 24*3=72nm.distance between them is sqrt(117^2+72^2) equals 137.4nm
To find the rate of increase, note that the distance x after t hours is given by
x^2 = (50+19t)^2 + (24t)^2
At 3 pm, t=3, and we have
x^2 = 117^2 + 72^2 = 16633
x = 137.4
Now we have to find the derivative:
2x dx/dt = 38(50+19t)+48(24t) = 1874t + 1900
So, at t=3, we have
2(137.4) dx/dt = 1874(3)+1900
dx/dt = 7522/274.8 = 27.4 knots
To find the rate at which the distance between the ships is changing, we need to compute the derivative of the distance function with respect to time.
Let's first set up a coordinate system. Let the origin be the initial position of ship B, and let the x-axis be east-west and the y-axis be north-south.
At noon, ship A is 50 nautical miles due west of ship B, so its position can be represented as (−50, 0). Ship A is sailing at a speed of 19 knots due west, so its position at any given time can be represented as (−50 − 19t, 0), where t represents time in hours since noon.
Ship B is sailing north at a speed of 24 knots, so its position at any given time can be represented as (0, 24t).
The distance between the ships can be calculated using the distance formula:
d = sqrt((x2 − x1)² + (y2 − y1)²),
where (x1, y1) and (x2, y2) represent the positions of the ships at a given time.
Substituting the positions of ship A and ship B at any given time, we have:
d = sqrt((0 − (−50 − 19t))² + (24t − 0)²)
= sqrt((50 + 19t)² + (24t)²)
= sqrt(2500 + 950t + 361t² + 576t²)
= sqrt(2500 + 1526t + 937t²).
Now, to find how the distance is changing at 3 PM, we need to find the derivative of the distance function with respect to time and evaluate it at t = 3.
Taking the derivative of d with respect to t, we have:
dd/dt = (2t + 1526)/(2√(2500 + 1526t + 937t²)).
Evaluating this derivative at t = 3:
dd/dt = (2(3) + 1526)/(2√(2500 + 1526(3) + 937(3)²))
= (6 + 1526)/(2√(2500 + 4578 + 2817))
= 1532/(2√9895).
Simplifying further:
dd/dt = 766/√9895.
Thus, at 3 PM, the distance between the ships is changing at a rate of 766/√9895 knots.