At noon, ship A is 20 nautical miles due west of ship B. Ship A is sailing west at 22 knots and ship B is sailing north at 18 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.)
QUESTION 3
A boat sails 30 miles to the east from a point P, then it changes direction and sails to the south. If this boat is sailing at a constant speed of 10 miles/hr, at what rate is its distance from the point P increasing
a) 2 hours after it leaves the point P
b) 7 hours after it leaves the point P
To find how fast the distance between the ships is changing at 4 PM, we need to calculate the rate at which the distance is changing, which is the derivative of the distance with respect to time.
Let's consider the position of each ship as a function of time. Let the position of ship A at time t be (xA(t), yA(t)) and the position of ship B be (xB(t), yB(t)).
Given:
At noon, ship A is 20 nautical miles due west of ship B.
This means that at t = 0, xA(0) = xB(0) + 20.
Ship A is sailing west at 22 knots, which means its x-coordinate is changing at a rate of -22 knots.
So, the velocity of ship A is dxA/dt = -22.
Ship B is sailing north at 18 knots, which means its y-coordinate is changing at a rate of 18 knots.
So, the velocity of ship B is dyB/dt = 18.
We want to find the rate at which the distance between the ships is changing at 4 PM, which is the derivative of the distance with respect to time: d(distance)/dt.
The distance between the ships at any time t can be calculated using the distance formula:
distance = sqrt((xA - xB)^2 + (yA - yB)^2)
To differentiate this distance function, we need to apply the chain rule. The derivative is:
d(distance)/dt = (d(distance)/dxA) * (dxA/dt) + (d(distance)/dxB) * (dxB/dt) + (d(distance)/dyA) * (dyA/dt) + (d(distance)/dyB) * (dyB/dt)
Let's calculate each partial derivative:
(d(distance)/dxA) = (xA - xB) / distance // Partial derivative with respect to xA
(d(distance)/dxB) = -(xA - xB) / distance // Partial derivative with respect to xB
(d(distance)/dyA) = (yA - yB) / distance // Partial derivative with respect to yA
(d(distance)/dyB) = -(yA - yB) / distance // Partial derivative with respect to yB
Now, let's plug in the given values:
At 4 PM, t = 4.
Velocity of ship A, dxA/dt = -22 knots.
Velocity of ship B, dyB/dt = 18 knots.
We also need to find the values of xA and yB at 4 PM.
Since ship A is 20 nautical miles due west of ship B at noon, it will have moved 20 nautical miles westward by 4 PM.
So, xA(4) = xB(4) + 20.
Finally, we can substitute all these values into the derivative equation to calculate d(distance)/dt at 4 PM.
I hope this explanation helps you understand how to approach the problem and how to find the solution step by step.