A plane (A) heading south at 120 mph and a plane (B) heading west at 600 mph are rlying tword the same point at the same altitude. Plane A is 100 miles from the point where the flight patterns intersect and plane B is 550 miles from there.
A) Find the parametric equation that model the motion of both planes
B) Find a formula for the distance between the planes as a function of time.
C) What is the min distance between the planes? When are the planes closest?
Thanks!!
A) To find the parametric equation that models the motion of both planes, we can use the distance formula.
Let t be the time in hours.
For plane A, which is heading south at 120 mph, we can use the equation:
xA = 0
yA = -120t + 100
Since plane A is not moving horizontally, its x-coordinate remains constant at 0. The y-coordinate changes at a rate of -120t, as the plane is moving south at 120 mph. The initial y-coordinate is 100 since the plane is 100 miles from the point where the flight patterns intersect.
For plane B, which is heading west at 600 mph, we can use the equation:
xB = -600t + 550
yB = 0
Similarly, plane B is not moving vertically, so its y-coordinate remains constant at 0. The x-coordinate changes at a rate of -600t, as the plane is moving west at 600 mph. The initial x-coordinate is 550 since the plane is 550 miles from the point where the flight patterns intersect.
B) To find a formula for the distance between the planes as a function of time, we can use the distance formula:
distance = √((xB - xA)^2 + (yB - yA)^2)
Substituting the equations for xA, yA, xB, and yB, we get:
distance = √((-600t + 550 - 0)^2 + (0 - (-120t + 100))^2)
distance = √((-600t + 550)^2 + (120t - 100)^2)
Thus, the formula for the distance between the planes as a function of time is:
distance = √((-600t + 550)^2 + (120t - 100)^2)
C) To find the minimum distance and when the planes are closest, we need to find the time when the distance between the planes is minimized.
To do this, we can find the derivative of the distance function with respect to t and set it equal to zero:
(distance)' = 0
Simplifying the expression and solving for t will give us the time when the distance between the planes is minimized.
Taking the derivative of the distance function and setting it equal to zero:
d(distance)/dt = -120(120t - 100) + 600(-600t + 550) = 0
Simplifying the expression:
-14400t + 12000 - 360000t + 330000 = 0
-376400t + 342000 = 0
t = 342000 / 376400
t ≈ 0.9083
Therefore, at approximately t = 0.9083 hours, the planes are closest, and the minimum distance between them can be found by substituting this value of t into the distance function:
distance = √((-600(0.9083) + 550)^2 + (120(0.9083) - 100)^2)
distance ≈ √((-544.98)^2 + (-7.84)^2)
distance ≈ √(297039.28 + 61.4656)
distance ≈ √297100.7456
distance ≈ 545.17
Thus, the minimum distance between the planes is approximately 545.17 miles, and it occurs at approximately t = 0.9083 hours.
To find the parametric equations that model the motion of both planes, let's consider the position vectors for each plane at a given time t.
Let P(t) be the position vector of plane A at time t, and Q(t) be the position vector of plane B at time t.
A) Parametric equations for plane A:
Since plane A is heading south at a constant speed of 120 mph and is initially 100 miles from the intersection point, its position vector can be defined as:
P(t) = <0, -120t + 100>
Parametric equations for plane B:
Since plane B is heading west at a constant speed of 600 mph and is initially 550 miles from the intersection point, its position vector can be defined as:
Q(t) = <-600t + 550, 0>
Here, the x-component of each vector represents the east-west position, and the y-component represents the north-south position.
B) To find the formula for the distance between the planes as a function of time, we can use the distance formula:
d(t) = |P(t) - Q(t)|
Substituting the values of P(t) and Q(t), we get:
d(t) = |<0, -120t + 100> - <-600t + 550, 0>|
= |<-600t + 550, -120t + 100>|
= sqrt((-600t + 550)^2 + (-120t + 100)^2)
C) To find the minimum distance between the planes and when they are closest, we'll need to find the derivative of d(t) with respect to t and set it equal to zero.
First, let's simplify the equation for d(t):
d(t) = sqrt((-600t + 550)^2 + (-120t + 100)^2)
= sqrt(360000t^2 - 1320000t + 1320000)
Next, let's find the derivative of d(t):
d'(t) = (1/2) * (360000t^2 - 1320000t + 1320000)^(-1/2) * (720000t - 1320000)
Setting d'(t) equal to zero and solving for t:
720000t - 1320000 = 0
t = 11/6
So, at t = 11/6, the planes are closest. To find the minimum distance, substitute t = 11/6 into the equation for d(t):
d(11/6) = sqrt((360000 * (11/6)^2) - 1320000 * (11/6) + 1320000)