Optimization

At 1:00 PM ship A is 30 miles due south of ship B and is sailing north at a rate of 15mph. If ship B is sailing due west at a rate of 10mph, at what time will the distance between the two ships be minimal? will the come within 18 miles of each other?

The answer is 2:23:05 and yes,
i just need help with the setup

The positions of A and B are parametric functions of time.

Take the position of ship A at 1:00 pm be (0,0), the
xa(t) = 0
ya(t) = 10t

xb(t) = -15t
yb(t) = 30

Distance between the two ships
D(t) = √((xb(t)-xa(t))²+(yb(t)-ya(t))²)

Substitute xa,xb,ya,yb in the expression for D(t) and differentiate with respect to t.

For the minimum distance, D'(t0) = 0.
Verify that D(t0) is a minimum by confirming that D"(t0)>0, or by comparing values of D'(t0-) and D'(t0+) with D'(t0).

Evaluate D(t0) for the remainder of the question.

To find the time at which the distance between the two ships will be minimal, we can use the concept of relative motion.

Let's first establish a coordinate system. Assume that ship B is located at the origin (0,0), and ship A is located at (0, -30) at 1:00 PM.

Since ship A is sailing north at a rate of 15 mph, its position after time t in hours can be represented as (0, -30 + 15t).

Similarly, ship B is sailing west at a rate of 10 mph, so its position after time t can be represented as (-10t, 0).

The distance formula between two points (x1, y1) and (x2, y2) is given by:
d = √((x2 - x1)^2 + (y2 - y1)^2)

Substituting the positions of ship A and ship B into the distance formula, we get:
d = √((-10t - 0)^2 + (-30 + 15t)^2)
= √(100t^2 + 900 - 900t + 225t^2)
= √(325t^2 - 900t + 900)

To find the time at which the distance is minimal, we need to find the minimum value of the distance function. We can do this by finding the minimum value of the quadratic function 325t^2 - 900t + 900.

The minimum value of a quadratic function occurs at its vertex, which can be found using the formula:
t = -b / (2a)

For our quadratic function, a = 325 and b = -900. Substituting the values into the formula, we get:
t = -(-900) / (2 * 325)
= 900 / 650
= 1.38

Converting 1.38 hours to minutes:
1.38 hours * 60 minutes/hour = 82.8 minutes

Therefore, the time at which the distance between the two ships will be minimal is approximately 1 hour and 23 minutes after 1:00 PM, which is 2:23 PM.

To determine if the two ships will come within 18 miles of each other, we can substitute the value of t (1.38 hours) into the distance formula and check if the resulting distance is less than 18 miles.

d = √(325(1.38)^2 - 900(1.38) + 900)
= √(325(1.9044) - 1242 + 900)
= √(615.89 - 1242 + 900)
= √(273.89)
≈ 16.55 miles

Since the resulting distance is less than 18 miles, the two ships will come within 18 miles of each other.

To find the time at which the distance between the two ships is minimal, we need to set up a mathematical equation based on their positions and velocities.

Let's start by breaking down the given information:

- Ship A is initially 30 miles due south of ship B.
- Ship A is sailing north at a rate of 15 mph (miles per hour).
- Ship B is sailing due west at a rate of 10 mph.

We can model the positions of the ships at any given time with a coordinate system, where the starting position of ship A is (0, -30) and ship B is (0, 0).

Let's denote the time as "t". Ship A's position after time "t" can be represented as (0, -30 + 15t) and ship B's position as (-10t, 0). The distance between the two ships, D, can be calculated using the distance formula:

D = √((x2 - x1)^2 + (y2 - y1)^2)

Substituting the positions of the ships:

D = √((-10t - 0)^2 + (0 - (-30 + 15t))^2)
D = √(100t^2 + (30 - 15t)^2)
D = √(100t^2 + 900 - 900t + 225t^2)
D = √(325t^2 - 900t + 900)

To find the time at which D is minimal, we need to find its derivative with respect to time and set it equal to zero:

dD/dt = 0

Differentiating the equation:

dD/dt = (650t - 900) / (2√(325t^2 - 900t + 900))

Setting this derivative equal to zero:

650t - 900 = 0

Solving for "t":

t = 900 / 650 = 1.38461 hours

To convert this into minutes and seconds, multiply by 60:

1.38461 x 60 = 83.0769 minutes

Therefore, the time at which the distance between the two ships is minimal is approximately 1 hour and 23 minutes.

To determine if they come within 18 miles of each other, substitute the calculated time back into the distance equation:

D = √(325(1.38461)^2 - 900(1.38461) + 900)
D = √(815.38461 - 1246.15385 + 900)
D = √(-530.76824)

As the distance is a negative value under the square root, it indicates that the ships do not come within 18 miles of each other.