Two ships leave the same port at noon. Ship A sails north at 12 mph, and ship B sails east at 19 mph. How fast is the distance between them changing at 1 p.m.? (Round your answer to one decimal place.)

d^2=(12t)^2 + (19t)^2 where t is time in hours sailing. In this case, t=1hr

solve for d.

d^2=(12t)^2 + (19t)^2 where t is time in hours sailing. In this case, t=1hr

take the derivative
2d dd/dt=2*144t+2*19^2 t

dd/dt = relative velocity

v= (288 +261)/d
where d=sqrt(12^2+19^2)

To determine how fast the distance between the two ships is changing at 1 p.m., we can use the Pythagorean theorem and the concept of rate of change.

Let's consider the position of the ships at 1 p.m. Ship A will have traveled for 1 hour at a speed of 12 mph, so it will be 12 miles north of the starting point. Ship B will also have traveled for 1 hour at a speed of 19 mph, so it will be 19 miles east of the starting point.

Now, we can visualize the situation on a coordinate plane. Let's assume the starting point is the origin (0,0). Ship A will be located at (0, 12), and Ship B at (19, 0).

To find the distance between the two ships, we can use the distance formula:

Distance = square root of ((x2 - x1)^2 + (y2 - y1)^2)

In this case, x1 = 0, y1 = 12, x2 = 19, and y2 = 0.

Distance = square root of ((19 - 0)^2 + (0 - 12)^2)
Distance = square root of (361 + 144)
Distance = square root of 505
Distance ≈ 22.47 miles

Now, let's differentiate the distance formula with respect to time (t) to find how fast the distance is changing:

d(Distance)/dt = (d/dt) [ square root of ((x2 - x1)^2 + (y2 - y1)^2) ]

Since x1 = 0, y1 = 12, x2 = 19, and y2 = 0, we can substitute these values into the distance formula:

d(Distance)/dt = (d/dt) [ square root of ((19 - 0)^2 + (0 - 12)^2) ]
d(Distance)/dt = (d/dt) [ square root of (361 + 144) ]
d(Distance)/dt = (d/dt) [ square root of 505 ]

To differentiate the square root function, we can use the chain rule. Let's denote the square root of 505 as z, so the equation becomes:

d(Distance)/dt = (d/dz) (square root of z) * (dz/dt)

The derivative of the square root of z with respect to z is 1 / (2 * square root of z).

So, the equation becomes:

d(Distance)/dt = (1 / (2 * square root of z)) * (dz/dt)

Now, we need to find dz/dt. From the given information, we know that Ship A is sailing north at a constant speed of 12 mph, and Ship B is sailing east at a constant speed of 19 mph. Since both ships are continuously moving away from the starting point, the x-coordinate of Ship A is increasing at a rate of 0 mph, and the y-coordinate of Ship B is increasing at a rate of 0 mph. Therefore, dz/dt is 0.

Substituting this value into the equation, we have:

d(Distance)/dt = (1 / (2 * square root of z)) * 0
d(Distance)/dt = 0

Therefore, the distance between the two ships is not changing at 1 p.m. The rate of change is 0 mph.

To find how fast the distance between the two ships is changing, we can use the concept of rate of change.

Let's assume that the position of Ship A at time t is given by the function A(t) in miles, and the position of Ship B at time t is given by the function B(t) in miles.

Since Ship A sails north at 12 mph, the rate of change of its position can be represented as A'(t) = 12 mph. Similarly, since Ship B sails east at 19 mph, the rate of change of its position can be represented as B'(t) = 19 mph.

The distance between the two ships at time t can be found using the distance formula in two dimensions:

D(t) = √((A(t) - B(t))^2 + (A(t) - B(t))^2)

To find how fast the distance is changing, we need to find the derivative of the distance function D(t) with respect to time t.

dD/dt = d/dt √((A(t) - B(t))^2 + (A(t) - B(t))^2)

To simplify the derivative, we can square the equation inside the square root:

dD/dt = d/dt ((A(t) - B(t))^2 + (A(t) - B(t))^2)^1/2

Using the chain rule, the derivative can be calculated as:

dD/dt = 1/2 ((A(t) - B(t))^2 + (A(t) - B(t))^2)^(-1/2) * (2 * (A'(t) - B'(t))(A(t) - B(t)) + 2 * (A(t) - B(t))(A'(t) - B'(t)))

Substituting the known values:

A'(t) = 12 mph
B'(t) = 19 mph

We can now evaluate the derivative at t = 1 hour (since the ships leave at noon and we need to find the rate at 1 p.m.):

dD/dt = 1/2 ((A(1) - B(1))^2 + (A(1) - B(1))^2)^(-1/2) * (2 * (12)(A(1) - B(1)) + 2 * (A(1) - B(1))(19))

To find the values of A(1) and B(1), we need to consider the distance traveled by each ship in one hour.

Ship A travels at a constant speed of 12 mph for 1 hour, so A(1) = 12 miles.

Ship B travels at a constant speed of 19 mph for 1 hour, so B(1) = 19 miles.

Now we can substitute the values:

dD/dt = 1/2 ((12 - 19)^2 + (12 - 19)^2)^(-1/2) * (2 * (12)(12 - 19) + 2 * (12 - 19)(19))

Simplifying further:

dD/dt = 1/2 ((-7)^2 + (-7)^2)^(-1/2) * (2 * (12)(-7) + 2 * (-7)(19))

dD/dt = 1/2 (98 + 98)^(-1/2) * (2 * (-84) + 2 * (-133))

dD/dt = 1/2 (196)^(-1/2) * (-168 - 266)

dD/dt = 1/2 (196)^(-1/2) * (-434)

Now we can calculate the final answer:

dD/dt = -434/2 * (196)^(-1/2)

Using a calculator, we find:

dD/dt ≈ -13.94 mph

Therefore, the distance between the two ships is changing at a rate of approximately -13.94 mph at 1 p.m.