At t=0, ship A is 12 miles due north of ship B. Ship A travels 12 miles/hour due south, while ship B travels 8 miles/hour due east.

a. Write a function for the distance between the two ships.
b. At what time are the two ships closest?

At a time of t after t=0 , the southward distance covered by A is 12t miles, and the distance covered by B is 8t

Let the distance between them be d
I see a right - angled triangle.
a) d^2 = (8t)^2 + (12-12t)^2
b)
dd/dt = 2(8t)(8) + 2(12-12t)(-12)
= 0 for a min d
128t -288 + 288t = 0
416t = 288
t = 288/416 = 9/13 = .6923

they will be closest at 9/13 hours
= .6923 hrs
= 41.5 minutes

check:
at t = .6923 , d = 6.65640
take a value of t slightly higher and lower than .3923
t = .69 , d = 6.65648 , a bit farther than at t = .6923
t = .70 , d = 6.657 , a bit farther than at t = .6923
My answer is correct

Thank you so much!

I appreciate it!

To find the distance between the two ships at any given time, we can use the Pythagorean theorem. The Pythagorean theorem states that for any right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Let's assume that t represents the time in hours that has passed since t=0. We can break down the motion of each ship as follows:

Ship A:
- Initial position: 12 miles due north of ship B
- Velocity: 12 miles/hour due south

Ship B:
- Initial position: Same as ship A
- Velocity: 8 miles/hour due east

To find the distance at any given time t, we need to find the coordinates of each ship. Let's assume that the initial position of ship B is (0, 0), since no coordinates have been given. The initial position of ship A will then be (0, 12).

Using the velocities, we can find the coordinates of each ship after time t as follows:

Ship A coordinates: (0, 12 - 12t)
Ship B coordinates: (8t, 0)

Now we can calculate the distance between the two ships using the Pythagorean theorem:

Distance^2 = (x2 - x1)^2 + (y2 - y1)^2
Distance^2 = (8t - 0)^2 + (0 - (12 - 12t))^2
Distance^2 = (8t)^2 + (12 - 12t)^2
Distance^2 = 64t^2 + 144 - 288t + 144t^2
Distance^2 = 208t^2 - 288t + 144

a. The function for the distance between the two ships is Distance(t) = sqrt(208t^2 - 288t + 144).

To find the time at which the two ships are closest, we need to find the minimum of this distance function. We can do this by finding the critical points of the function (where the derivative is equal to zero) and then checking for local minimums.

Taking the derivative of the distance function with respect to t:

Distance'(t) = 416t - 288

Setting Distance'(t) = 0 and solving for t:

416t - 288 = 0
416t = 288
t = 288/416
t = 0.6923

b. The two ships are closest at t = 0.6923 hours.