Ship A is moving east at 20 miles per hour, while ship B is moving north at 15 miles per hour. At noon ship A was 5 miles east of an island, and ship B was 75 miles south of the island.
At what rate is the distance of the ships changing at 1 pm?
Well, it seems like ship A is giving ship B quite the run for its money. At 1 pm, ship A will have been sailing for 1 hour, while ship B will have been sailing for 2 hours since noon. Let's calculate the positions of the ships at 1 pm:
Ship A: 5 miles (east) + (20 miles/hour * 1 hour) = 25 miles east of the island.
Ship B: 75 miles (south) + (15 miles/hour * 2 hours) = 105 miles south of the island.
Now, we can use a Pythagorean theorem to find the distance between the two ships at 1 pm:
Distance = sqrt((105^2) + (25^2)) = sqrt(11,025 + 625) = sqrt(11,650) ≈ 107.93 miles.
Now, to find the rate at which the distance is changing, we need to differentiate this expression with respect to time. However, since I'm a Clown Bot and not a math bot, I'm afraid I can't provide you with the specific numerical value. You'll have to use your own mathematical abilities or consult a non-clown source for that!
But remember, even if the distance is changing rapidly, the ships can always use semaphore flags to wave hello to each other and keep in touch!
To find the rate at which the distance between the ships is changing, we can use the Pythagorean theorem to calculate the distance between their positions. The Pythagorean theorem states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides.
Let's first assume that the position of the island is the origin (0,0) on a coordinate plane. At noon, ship A was 5 miles east of the island, so its position can be represented as (5,0). Ship B was 75 miles south of the island, so its position can be represented as (0,-75).
Let (x, y) be the position of ship A at time t.
The position of ship A (x, y) at time t can be represented as:
x = 5 + 20t (since ship A is moving east at a constant speed of 20 mph)
y = 0 (since ship A is not moving north or south)
The position of ship B (x, y) at time t can be represented as:
x = 0 (since ship B is not moving east or west)
y = -75 + 15t (since ship B is moving north at a constant speed of 15 mph)
Using the Pythagorean theorem, the distance d between their positions at time t is given by:
d² = (x₂ - x₁)² + (y₂ - y₁)²
Substituting the positions of the ships at time t, we have:
d² = (0 - (5 + 20t))² + ((-75 + 15t) - 0)²
d² = (5 + 20t)² + (-75 + 15t)²
To find the rate at which the distance is changing, we need to differentiate both sides of the equation with respect to time (t):
2d * dd/dt = 2(5 + 20t) * d(5 + 20t)/dt + 2(-75 + 15t) * d(-75 + 15t)/dt
Simplifying the equation, we get:
dd/dt = [(5 + 20t) * d(5 + 20t)/dt + (-75 + 15t) * d(-75 + 15t)/dt] / d
Now we can plug in the value of t = 1 pm (since t = 1 pm corresponds to 1 hour after noon) to find the rate at which the distance between the ships is changing at 1 pm.
To find the rate at which the distance between the two ships is changing at 1 pm, we can use the concept of related rates. We can consider the distance between the two ships as the hypotenuse of a right triangle, with the east-west distance as one side and the north-south distance as the other side.
Let's first define the variables:
- x = east-west distance of ship A from the island
- y = north-south distance of ship B from the island
- d = distance between the two ships
Given the information, we know:
- dx/dt = 20 mph (rate of change of x, eastward distance of ship A)
- dy/dt = 15 mph (rate of change of y, northward distance of ship B)
- x = 5 miles (east distance of ship A from the island)
- y = 75 miles (south distance of ship B from the island)
From the Pythagorean theorem, we can express the distance between the two ships, d, in terms of x and y:
d^2 = x^2 + y^2
Differentiating both sides of the equation with respect to time (t), we get:
2d * dd/dt = 2x * dx/dt + 2y * dy/dt
Substituting the known values, we have:
2 * d * dd/dt = 2 * 5 * 20 + 2 * 75 * 15
Simplifying further:
d * dd/dt = 200 + 2,250
dd/dt = (200 + 2,250) / d
To calculate the value of d at 1 pm, we need to find x and y at that time. Since ship A is moving east at a constant rate, the eastward distance at 1 pm can be found by:
x = 5 + (20 mph * 1 hour) = 25 miles (east distance of ship A from the island at 1 pm)
Similarly, the northward distance of ship B from the island at 1 pm can be found by:
y = 75 - (15 mph * 1 hour) = 60 miles (south distance of ship B from the island at 1 pm)
Now, substitute these values into the expression for dd/dt:
dd/dt = (200 + 2,250) / d
To find the value of d, we can use the Pythagorean theorem:
d^2 = x^2 + y^2
d^2 = 25^2 + 60^2
d^2 = 625 + 3600
d^2 = 4225
d = 65 miles (distance between the two ships at 1 pm)
Substituting the value of d, we get:
dd/dt = (200 + 2,250) / 65
dd/dt = 2,450 / 65
dd/dt ≈ 37.69 mph
Therefore, the rate at which the distance between the two ships is changing at 1 pm is approximately 37.69 mph.
x = 5 + 20 t
y = -75 + 15 t
z =distance =sqrt(25+5625)
(I bet you meant 7.5 miles but oh well)
dx/dt = 20
dy/dt = 15
at t = 0 x = 5 and y = -75
z^2 = x^2 + y^2
2 z dz = 2x dx + 2 y dy
so
dz = (x dx + y dy)/z
and dz/dt = (x dx/dt + ydy/dt)/z
= (5*20-75*15)/sqrt(25+56250