At noon, ship A is 50 nautical miles due west of ship B. Ship A is sailing west at 16 knots and ship B is sailing north at 16 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.)
the distance z between the ships at time t is
z^2 = (50+16t)^2 + (16t)^2
at t=z, z = 2√9697 = 196.95
2z dz/dt = 2*32(50+16t) + 32t = 1056t+3200
so, at t=7,
dz/dt = 1/2 * (7392+3200)/196.5 = 26.95 knots
To find the speed at which the distance between the two ships is changing at 7 PM, we need to calculate the rate at which the distance between the ships is changing with respect to time. In other words, we need to find the derivative of the distance function with respect to time.
Let's start by setting up a coordinate system. We can set the initial position of ship A as (0, 0), with the x-axis pointing west and the y-axis pointing north. Ship B would then be at (50, 0).
We can define the distance between the two ships as a function of time. Let's call this function D(t), where t represents time in hours.
At noon (t = 0), the distance between the ships is simply the horizontal distance between their positions, which is 50 nautical miles.
Since ship A is sailing west at a constant speed of 16 knots, its x-coordinate can be given by the function x(t) = -16t. Similarly, ship B, sailing north at 16 knots, would have a y-coordinate of y(t) = 16t.
Now we can calculate the distance between the two ships at any given time t using the distance formula:
D(t) = sqrt((x(t) - 50)^2 + y(t)^2)
= sqrt((-16t - 50)^2 + (16t)^2)
To find the speed at which the distance is changing with respect to time at 7 PM (t = 7), we need to compute the derivative of D(t) with respect to t, and then evaluate it at t = 7:
dD(t)/dt = (d/dt)((-16t - 50)^2 + (16t)^2)
= 2(-16t - 50)(-16) + 2(16t)(16)
= -32(-16t - 50) + 32(16t)
= 32(16t) + 32(-16t - 50)
= 32(16t - 16t - 50)
= 32(-50)
= -1600
Therefore, the rate at which the distance between the two ships is changing at 7 PM is -1600 knots. The negative sign indicates that the distance is decreasing.