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 24 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.)

answer with explanation please because I got the formula wrong. Thanks

I did almost the same question yesterday ....

After the initial definitions, I would get:

d^2 = (50+16t)^2 + (24t)^2
at 7 pm, t = 7
d^2 = 26244+54468
d = 284.099

2d dd/dt = 2(50+16t)(16) + 2(24t)(24)
dd/dt = ((50+16t)(16) + (24t)(24) )/d

sub in t = 7 and d = 284.099 and find dd/dt

Ahoy there! Allow me, Clown Bot, to assist ye with this seafarin' problem. Now, let's set sail to solve it!

We have ship A sailing west at a speed of 16 knots, and ship B sailing north at a speed of 24 knots. To find how fast the distance between the ships is changing, we can use the classic Pythagorean Theorem.

Now, let's consider the distance between the ships as the hypotenuse (c) of a right triangle, with the distance ship A has sailed west as one side (a), and the distance ship B has sailed north as the other side (b).

At noon, ship A is 50 nautical miles due west of ship B, so we can say that a = 50 nautical miles. Since ship A is sailing west at 16 knots for 7 hours until 7 PM, a would become a = 16 knots * 7 hours = 112 nautical miles.

Now, let's find b. Ship B is sailing north at 24 knots for 7 hours until 7 PM, so b = 24 knots * 7 hours = 168 nautical miles.

Now, we can use the Pythagorean Theorem to find the distance between the ships at 7 PM (let's call it d):

d^2 = a^2 + b^2
d^2 = 112^2 + 168^2
d^2 = 12,544 + 28,224
d^2 = 40,768

Taking the square root of both sides gives us:
d = √40,768
d ≈ 202.4 nautical miles

So, at 7 PM, the distance between the ships is approximately 202.4 nautical miles.

To find how fast the distance between the ships is changing, we need to find the derivative of this distance function with respect to time.

d/dt (d^2) = d/dt (a^2 + b^2)
2d * (dd/dt) = 2a * (da/dt) + 2b * (db/dt)

Now, let's plug in the known values at 7 PM:
2 * 202.4 * (dd/dt) = 2 * 112 * (da/dt) + 2 * 168 * (db/dt)

We're looking for (dd/dt), which represents how fast the distance between the ships is changing at 7 PM. We know the speeds of ship A and ship B, which are da/dt = 16 knots and db/dt = 24 knots.

Substituting the known values:
2 * 202.4 * (dd/dt) = 2 * 112 * 16 + 2 * 168 * 24

Now, let's simplify:
2 * 202.4 * (dd/dt) = 3584 + 8064
2 * 202.4 * (dd/dt) = 11648

Dividing both sides by 2 * 202.4 gives us:
(dd/dt) = 11648 / (2 * 202.4)

Calculating this, we find:
(dd/dt) ≈ 28.83 knots

Yarr! So, at 7 PM, the distance between the ships is changing at approximately 28.83 knots.

To solve this problem, we will use the concept of related rates.

Let's start by defining the variables:
Let x be the distance between ship A and ship B at 7 PM.
Let y be the distance ship A has sailed from its initial position at noon (in nautical miles).
Let t be the time elapsed since noon (in hours).

We are given that ship A is sailing west at 16 knots, so its position at 7 PM can be determined as:
Ship A's position at 7 PM = 50 nautical miles (initial position) + (16 knots * t)

We are also given that ship B is sailing north at 24 knots, so its position at 7 PM can be determined as:
Ship B's position at 7 PM = 0 nautical miles (initial position) + (24 knots * t)

Using the Pythagorean theorem, we can calculate the distance between the two ships at 7 PM:
x^2 = (Ship A's position at 7 PM)^2 + (Ship B's position at 7 PM)^2

Differentiating both sides with respect to time (t):
(2x) dx/dt = 2 * (Ship A's position at 7 PM) * (d(Ship A's position at 7 PM)/dt) + 2 * (Ship B's position at 7 PM) * (d(Ship B's position at 7 PM)/dt)

Now, let's plug in the values we know:
dx/dt (the rate at which the distance is changing) is what we want to find.
x = distance between the ships at 7 PM.
(Ship A's position at 7 PM) = 50 + (16 * 7) (since 7 hours have passed since noon)
(Ship B's position at 7 PM) = 0 + (24 * 7) (since 7 hours have passed since noon)
(d(Ship A's position at 7 PM)/dt) = 16 knots (since ship A's velocity is constant)
(d(Ship B's position at 7 PM)/dt) = 24 knots (since ship B's velocity is constant)

Now, substitute the values into the equation:
(2x) dx/dt = 2 * (50 + (16 * 7)) * 16 + 2 * (24 * 7) * 24

Simplifying the equation:
2x dx/dt = 2 * (50 + 112) * 16 + 2 * 24 * 24
2x dx/dt = 2 * 162 * 16 + 2 * 24 * 24
2x dx/dt = 5184 + 1152
2x dx/dt = 6336

We can now isolate dx/dt:
dx/dt = 6336 / (2x)

Substituting x (the distance between the ships at 7 PM) as 50 + (16 * 7):
dx/dt = 6336 / (2 * (50 + (16 * 7)))

Simplifying further:
dx/dt = 6336 / (2 * (50 + 112))
dx/dt = 6336 / (2 * 162)
dx/dt ≈ 19.56 knots

Therefore, the rate at which the distance between the ships is changing at 7 PM is approximately 19.56 knots.

To find the rate at which the distance between the ships is changing at 7 PM, we need to make use of the concept of related rates. Let's break down the problem and solve it step by step:

Step 1: Understand the situation
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 24 knots.

Step 2: Set up a coordinate system
Let's assign coordinates to the ships. We'll place ship A at the origin (0, 0) and ship B at the point (-50, 0).

Step 3: Determine the positions of the ships at 7 PM
Since ship A has been sailing west for 7 hours at a speed of 16 knots, it would have traveled a distance of 7 * 16 = 112 nautical miles. Therefore, the position of ship A at 7 PM is (112, 0).
Since ship B has been sailing north for 7 hours at a speed of 24 knots, it would have traveled a distance of 7 * 24 = 168 nautical miles. Therefore, the position of ship B at 7 PM is (-50, 168).

Step 4: Calculate the distance between the ships at 7 PM
Using the distance formula, we can determine the distance between the ships at 7 PM:
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Distance = sqrt((112 - (-50))^2 + (0 - 168)^2)
Distance = sqrt(162^2 + (-168)^2)
Distance = sqrt(26244 + 28224)
Distance = sqrt(54468)
Distance ≈ 233.4 nautical miles

Step 5: Find the rate at which the distance is changing at 7 PM
To find the rate at which the distance is changing, we need to use the concept of related rates. We already know the rates at which ship A and ship B are moving. Let's call the rate at which the distance is changing dD/dt.
Using the chain rule, we have:
dD/dt = (∂D/∂x)*(dx/dt) + (∂D/∂y)*(dy/dt)
where D is the distance, x and y are the coordinates, and dx/dt and dy/dt are the rates of change of x and y, respectively.

Let's break it down:
∂D/∂x is the partial derivative of the distance function with respect to x at (112, 0) = 112 / sqrt(54244) approximately 112 / 233.4
∂D/∂y is the partial derivative of the distance function with respect to y at (0, 168) = 168 / sqrt(54244) approximately 168 / 233.4
dx/dt is the rate at which ship A is moving, which we know is 16 knots.
dy/dt is the rate at which ship B is moving, which we know is 24 knots.

Now let's substitute the values:
dD/dt = (112 / 233.4) * 16 + (168 / 233.4) * 24
dD/dt ≈ 9.63 + 17.31
dD/dt ≈ 26.94 knots

Therefore, the rate at which the distance between the ships is changing at 7 PM is approximately 26.94 knots.