At noon, ship A is 20 nautical miles due west of ships B. Ship A is sailing west at 18 knots and ship B is sailing north at 20 knots. How fast (in knots) is the distance between the ships changing at 6 PM?
To find the speed at which the distance between the ships is changing at 6 PM, we need to use the concept of related rates.
Step 1: Assign variables to the given quantities:
Let:
- Time be represented by t, in hours.
- Distance between ships A and B be represented by D(t), in nautical miles.
- The x-coordinate of ship A be represented by x(t), in nautical miles.
- The y-coordinate of ship B be represented by y(t), in nautical miles.
Step 2: Set up equations related to the given information:
- At noon (t = 0), ship A is 20 nautical miles due west of ship B:
-> x(0) = 20, and y(0) = 0.
- Ship A is sailing west at 18 knots:
-> dx/dt = -18 knots (negative since it is moving west).
- Ship B is sailing north at 20 knots:
-> dy/dt = 20 knots.
Step 3: Determine the relationships between the variables:
- The distance between the ships can be found using the distance formula:
D(t) = sqrt((x(t) - 0)^2 + (y(t) - 0)^2)
=> D(t) = sqrt(x(t)^2 + y(t)^2)
Step 4: Use the chain rule to find dD/dt in terms of dx/dt and dy/dt:
Using the chain rule, we differentiate D(t) with respect to t:
dD/dt = dD/dx * dx/dt + dD/dy * dy/dt
To calculate dD/dt, we need to determine dD/dx and dD/dy.
- Differentiate D(t) with respect to x:
dD/dx = d(sqrt(x^2 + y^2))/dx
= (1/2) * (x^2 + y^2)^(-1/2) * d(x^2 + y^2)/dx
= (1/2) * (x^2 + y^2)^(-1/2) * 2x
= (x/(sqrt(x^2 + y^2))
- Differentiate D(t) with respect to y:
dD/dy = d(sqrt(x^2 + y^2))/dy
= (1/2) * (x^2 + y^2)^(-1/2) * d(x^2 + y^2)/dy
= (1/2) * (x^2 + y^2)^(-1/2) * 2y
= (y/(sqrt(x^2 + y^2))
Substituting the values of dx/dt, dy/dt, dD/dx, and dD/dy into the equation for dD/dt:
dD/dt = dD/dx * dx/dt + dD/dy * dy/dt
= (x/(sqrt(x^2 + y^2))) * (-18) + (y/(sqrt(x^2 + y^2))) * 20
Step 5: Plug in the values at 6 PM (t = 6) to find the rate at which the distance is changing:
Using the values at 6 PM, we calculate dD/dt:
dD/dt = (x/(sqrt(x^2 + y^2))) * (-18) + (y/(sqrt(x^2 + y^2))) * 20
= (x/(sqrt(x^2 + y^2))) * (-18) + (y/(sqrt(x^2 + y^2))) * 20
= (x/(sqrt(x^2 + y^2))) * (-18) + (y/(sqrt(x^2 + y^2))) * 20
Substituting the coordinates of ship A and ship B (x = -18 * 6 and y = 20 * 6):
dD/dt = ((-18 * 6)/(sqrt((-18 * 6)^2 + (20 * 6)^2))) * (-18) + ((20 * 6)/(sqrt((-18 * 6)^2 + (20 * 6)^2))) * 20
Therefore, to find the speed at which the distance between the ships is changing at 6 PM, we need to calculate the above expression.
To find the speed at which the distance between the ships is changing at 6 PM, we need to use the concept of relative velocity.
Let's first visualize the situation. Ship A is sailing west at a constant speed of 18 knots, and ship B is sailing north at a constant speed of 20 knots. At noon, ship A is 20 nautical miles due west of ship B.
To solve this problem, we can use the Pythagorean theorem and consider the distance between the ships as a function of time.
Let's assume that the distance between the ships at time t is represented by the variable d(t). Since the ships are moving at a constant speed, we can differentiate this function with respect to time (t) to find the rate of change of the distance between the ships.
Let's set up a coordinate system where the origin represents ship B's initial position at noon. At time t, ship A's position will be (-18t, 0) (negative because it's moving west) and ship B's position will be (0, 20t) (positive because it's moving north).
Using the distance formula, we can determine the distance between the two ships at time t:
d(t) = sqrt[(0 - (-18t))^2 + (20t - 0)^2] = sqrt[(18t)^2 + (20t)^2] = sqrt[324t^2 + 400t^2] = sqrt[724t^2]
Now, to find the rate at which the distance between the ships is changing at 6 PM, we need to evaluate the derivative of d(t) with respect to time at t=6.
d'(t) = (1/2)*(724t^2)^(-1/2)*(2*724t) = 724/sqrt(724t^2) = 724/(t*sqrt(724))
To find the speed at 6 PM, let's substitute t=6 into the derivative equation:
d'(6) = 724/(6*sqrt(724)) = 724/(6*sqrt(2896)) = 724/(6*53.803) ≈ 2.675 knots
Therefore, the speed at which the distance between the ships is changing at 6 PM is approximately 2.675 knots.