At noon, ship A is 30 nautical miles due west of ship B. Ship A is sailing west at 21 knots and ship B is sailing north at 15 knots. How fast (in knots) is the distance between the ships changing at 4 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)
this is a cal problem.
I did this problem an hour or so ago.
http://www.jiskha.com/display.cgi?id=1310260104
i think u did it wrong
I got the same answer as Damon.
Damon is right.
To solve this problem, we can use the concept of rates and apply the Pythagorean theorem to find the distance between the ships.
Step 1: Find the position of each ship at 4 PM.
Since ship A is sailing west at a constant rate of 21 knots for 4 hours, it will be 21 * 4 = 84 nautical miles west of its initial position at noon. So, its position at 4 PM will be 30 - 84 = -54 nautical miles.
Ship B is sailing north at a constant rate of 15 knots for 4 hours, so it will be 15 * 4 = 60 nautical miles north of its initial position at noon.
Step 2: Calculate the distance between the ships at 4 PM using the Pythagorean theorem.
The distance between the ships is the hypotenuse of a right-angled triangle formed by the positions of the ships. We can use the Pythagorean theorem, which states that the square of the hypotenuse (c) is equal to the sum of the squares of the other two sides (a and b).
Let's consider the position of ship B as the origin (0, 0) on a coordinate plane. Ship A will be at (-54, 0).
The distance between the ships at 4 PM is given by:
distance = sqrt((-54)^2 + 60^2)
distance = sqrt(2916 + 3600)
distance = sqrt(6516)
distance ≈ 80.77 nautical miles
Step 3: Find how fast the distance between the ships is changing at 4 PM.
To find how fast the distance between the ships is changing at 4 PM, we need to determine the rate of change of the distance with respect to time. In other words, we need to find the derivative of the distance function at 4 PM.
Let d(t) be the distance between the ships at time t, where t represents the number of hours after noon. Using the Pythagorean theorem, we have:
d(t) = sqrt((30 - 21t)^2 + (15t)^2)
We can differentiate this function with respect to t to find the derivative:
d'(t) = (1/2) * (30 - 21t)^(-1/2) * (-21) + (1/2) * (15t)^(-1/2) * 30
d'(t) = -10.5 * (30 - 21t)^(-1/2) + 7.5 * (15t)^(-1/2)
To find the rate of change of the distance at 4 PM, we substitute t = 4 into the derivative:
d'(4) = -10.5 * (30 - 21(4))^(-1/2) + 7.5 * (15(4))^(-1/2)
d'(4) = -105 * (30 - 84)^(-1/2) + 30 * 15^(-1/2)
d'(4) = -105 * (-54)^(-1/2) + 30 * 15^(-1/2)
d'(4) ≈ -105 * (-54)^(-1/2) + 30 * 0.36
d'(4) ≈ -105 * (-54)^(-1/2) + 10.8
d'(4) ≈ -105 * (-54)^(-1/2) + 10.8
Therefore, the speed at which the distance between the ships is changing at 4 PM is approximately -105 * (-54)^(-1/2) + 10.8 knots.