Two commercial airplanes are flying at an altitude of 40,000 ft along straight-line courses that intersect at right angles. Plane A is approaching the intersection point at a speed of 429 knots (nautical miles per hour; a nautical mile is 2000 yd or 6000 ft.) Plane B is approaching the intersection at 460 knots.
At what rate is the distance between the planes decreasing when Plane A is 2 nautical miles from the intersection point and Plane B is 6 nautical miles from the intersection point?
Answer = knots.
Well, there's an easy way to determine the rate at which the distance between the planes is decreasing.
Let's assume that the position of plane A is (x, y) and the position of plane B is (a, b), where x, y, a, and b are measured in nautical miles. The distance between the two planes is given by the formula:
d = √((x - a)^2 + (y - b)^2)
To determine the rate at which the distance is decreasing, we need to differentiate this formula with respect to time, t. So we'll have:
dd/dt = (∂d/∂x)(dx/dt) + (∂d/∂y)*(dy/dt) + (∂d/∂a)(da/dt) + (∂d/∂b)(db/dt)
Simplifying this equation, we get:
dd/dt = (x - a)(dx/dt) + (y - b)(dy/dt)
Now, we need to plug in the values given in the problem. Plane A is 2 nautical miles from the intersection point, so x = 2. Plane B is 6 nautical miles from the intersection point, so a = 6. The speed of plane A is 429 knots, so dx/dt = 429. The speed of plane B is 460 knots, so da/dt = 460. Since the airplanes are flying at a constant altitude of 40,000 ft (which doesn't change), we can simplify the equation even further:
dd/dt = (x - a)(dx/dt) + (y - b)(dy/dt)
= (2 - 6)(429) + (y - b)(dy/dt)
= -4(429) + (y - b)(dy/dt)
= -1716 + (y - b)(dy/dt)
Now, we need to find the value of dy/dt. Since the altitude is constant and the airplanes are flying perpendicular to each other, the value of dy/dt is 0. So we can substitute that in:
dd/dt = -1716 + (y - b)(dy/dt)
= -1716 + (y - b)(0)
= -1716 + 0
= -1716
So, the rate at which the distance between the planes is decreasing is 1716 knots.
To find the rate at which the distance between the planes is decreasing, we need to use the concept of related rates.
Let's assume that the distance between the planes is represented by the variable "d" and the time is represented by "t".
The rate at which Plane A is approaching the intersection point is given as 429 knots. We can express this as the rate of change of the x-coordinate of Plane A with respect to time (dx/dt).
Similarly, the rate at which Plane B is approaching the intersection point is given as 460 knots. This can be represented as the rate of change of the y-coordinate of Plane B with respect to time (dy/dt).
Since the planes are flying at altitudes of 40,000 ft (which is a constant), we can ignore it for the purpose of this question.
As the planes are flying along straight-line courses that intersect at right angles, we can consider the distance between the planes as the hypotenuse of a right-angle triangle. This distance can be represented by the Pythagorean theorem as:
d^2 = x^2 + y^2
We differentiate both sides of the equation with respect to time t:
d^2/dt = 2x(dx/dt) + 2y(dy/dt)
Given that Plane A is 2 nautical miles (nmi) from the intersection point (x = 2nmi) and Plane B is 6 nautical miles from the intersection point (y = 6nmi), we can substitute these values into the equation:
d^2/dt = 2(2nmi)(dx/dt) + 2(6nmi)(dy/dt)
Now we need to find the value of (dx/dt) and (dy/dt). Since Plane A is approaching the intersection point, its x-coordinate is decreasing, so (dx/dt) is negative. Similarly, since Plane B is approaching the intersection point, its y-coordinate is decreasing, so (dy/dt) is also negative.
Given that Plane A is traveling at 429 knots and Plane B is traveling at 460 knots, we have:
(dx/dt) = -429 knots
(dy/dt) = -460 knots
Substituting these values into the equation, we have:
d^2/dt = 2(2nmi)(-429 knots) + 2(6nmi)(-460 knots)
Simplifying the equation gives:
d^2/dt = -1716nmi(knots) - 5520nmi(knots)
Finally, we take the square root of both sides of the equation to find the rate at which the distance between the planes (d) is decreasing:
d/dt = √( -1716nmi(knots) - 5520nmi(knots) )
Therefore, the rate at which the distance between the planes is decreasing is √( -1716nmi(knots) - 5520nmi(knots) ) knots.
To find the rate at which the distance between the planes is decreasing, we need to apply the concept of related rates. We can think of this problem as a right triangle where the two planes form the legs and the distance between them is the hypotenuse.
Let's denote the distance between the planes as D and the time as t. We are given the following information:
- Plane A is approaching the intersection point at a speed of 429 knots.
- Plane B is approaching the intersection at a speed of 460 knots.
We are interested in finding the rate at which D is changing with respect to time (dD/dt) when Plane A is 2 nautical miles from the intersection point and Plane B is 6 nautical miles from the intersection point.
Here's how we can solve the problem step by step:
Step 1: Identify the relevant information and variables:
- Plane A is 2 nautical miles from the intersection, which means one of the legs of the triangle is 2 nautical miles long.
- Plane B is 6 nautical miles from the intersection, which means the other leg of the triangle is 6 nautical miles long.
- We are looking for the rate at which the distance between the planes (D) is decreasing with respect to time (dD/dt).
Step 2: Set up the equation using the Pythagorean theorem:
According to the Pythagorean theorem, in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
So, we have: D^2 = (2 nautical miles)^2 + (6 nautical miles)^2.
Step 3: Differentiate both sides of the equation with respect to time (t):
d(D^2)/dt = d((2 nautical miles)^2 + (6 nautical miles)^2)/dt.
Step 4: Apply the chain rule and differentiate each term:
d(D^2)/dt = 2(2 nautical miles)(dD/dt) + 2(6 nautical miles)(d(6 nautical miles)/dt).
Step 5: Simplify the equation:
Since d(6 nautical miles)/dt represents the speed at which B is moving towards the intersection, we know it is 460 knots. Therefore, the equation becomes:
2(2 nautical miles)(dD/dt) + 2(6 nautical miles)(460 knots) = 0.
Step 6: Solve for dD/dt (the rate at which the distance is decreasing):
2(2 nautical miles)(dD/dt) = - 2(6 nautical miles)(460 knots).
Simplifying further, we have:
4 nautical miles(dD/dt) = -12 nautical miles(460 knots).
Dividing by 4 nautical miles, we get:
dD/dt = -12 nautical miles(460 knots)/4 nautical miles.
Step 7: Calculate the value:
dD/dt = -12(460 knots)/4 = -1380 knots.
Therefore, the rate at which the distance between the planes is decreasing when Plane A is 2 nautical miles from the intersection point and Plane B is 6 nautical miles from the intersection point is -1380 knots.