A plane flying with a constant speed of 19 km/min passes over a ground radar station at an altitude of 4 km and climbs at an angle of 25 degrees. At what rate is the distance from the plane to the radar station increasing 3 minutes later?
The distance is increasing at
The plane's x- and y- speed is
dx/dt = 19 cos 25°
dy/dt = 19 sin 25°
At time t, the distance z is
z^2 = x^2 + (y+4)^2
so,
2z dz/dt 2x dx/dt + 2(y+4) dy/dt.
Now just figure x,y,z at t=3, and plug in the values.
To find the rate at which the distance from the plane to the radar station is increasing, we can use trigonometry and differentiation.
Let's denote the distance from the plane to the radar station as D, and the horizontal distance from the plane to the radar station as x.
From the given information, we know that the plane is climbing at an angle of 25 degrees. This means that the vertical distance from the plane to the radar station, which we'll call y, is given by:
y = 4 km * tan(25 degrees)
Using the given constant speed of the plane (19 km/min) and the fact that time is measured in minutes, we can express the horizontal distance x as:
x = 19 km/min * t
where t is the time in minutes.
Now, let's derive the relationship between x, y, and D using the Pythagorean theorem:
D^2 = x^2 + y^2
Differentiating both sides of this equation with respect to t, we get:
2D * dD/dt = 2x * dx/dt + 2y * dy/dt
Since we want to find the rate at which the distance D is increasing (dD/dt), we substitute the given values:
x = 19 km/min * (3 minutes) = 57 km
y = 4 km * tan(25 degrees)
Plugging these values into the equation and solving for dD/dt, we find:
2D * dD/dt = 2x * dx/dt + 2y * dy/dt
2D * dD/dt = 2(57 km) * (19 km/min) + 2(4 km * tan(25 degrees)) * (0 km/min)
The term involving dx/dt can be ignored since it represents the horizontal speed of the plane, and we're only interested in the rate of change of the distance.
Simplifying the equation further:
2D * dD/dt = 2(57 km) * (19 km/min)
Now, we can solve for dD/dt:
dD/dt = (2(57 km) * (19 km/min)) / (2D)
Since D is the distance from the plane to the radar station, and this distance remains constant, we substitute the value of D:
dD/dt = (2(57 km) * (19 km/min)) / (2 * D)
Substituting D = sqrt(x^2 + y^2):
dD/dt = (2(57 km) * (19 km/min)) / (2 * sqrt(x^2 + y^2))
Plugging in the known values, we get:
dD/dt = (2(57 km) * (19 km/min)) / (2 * sqrt((57 km)^2 + (4 km * tan(25 degrees))^2))
Evaluating this expression using a calculator, we find that the distance is increasing at approximately 19.286 km/min. Therefore, the distance is increasing at about 19.286 km/min 3 minutes later.
To find the rate at which the distance from the plane to the radar station is increasing, we can consider the position of the plane as it moves.
Let's assume that the radar station is located at the origin of a coordinate system, and the plane initially passes directly over it.
We can define:
- x as the horizontal distance from the radar station to the plane,
- y as the vertical distance from the radar station to the plane, and
- t as the time elapsed since the plane passed over the radar station.
Since the plane is flying with a constant speed of 19 km/min, the speed of the plane does not change as time passes. Therefore, the rate of change of x and y with respect to time is constant.
The rate of change of x is given by the horizontal component of the speed, which is 19 km/min multiplied by the cos(25°) (the angle at which the plane is climbing):
dx/dt = 19 km/min * cos(25°)
Similarly, the rate of change of y is given by the vertical component of the speed, which is 19 km/min multiplied by the sin(25°):
dy/dt = 19 km/min * sin(25°)
To find the rate at which the distance from the plane to the radar station is increasing, we can use the Pythagorean theorem:
r^2 = x^2 + y^2
Taking the derivative of both sides with respect to time:
(d/dt)(r^2) = (d/dt)(x^2 + y^2)
2r(dr/dt) = 2x(dx/dt) + 2y(dy/dt)
Since we are interested in finding the rate at which the distance is increasing, we are looking for dr/dt. We know that r = 4 km (the initial altitude of the plane) and the values for dx/dt and dy/dt.
Substituting all the values we know into the equation:
2(4 km)(dr/dt) = 2(x)(19 km/min * cos(25°)) + 2(y)(19 km/min * sin(25°))
Simplifying the equation:
8 km(dr/dt) = (x)(19 km/min * cos(25°)) + (y)(19 km/min * sin(25°))
Now we need to find the values of x and y, 3 minutes after the plane passed over the radar station.
Since the plane is flying at a constant speed, in 3 minutes it will have covered a distance of 19 km/min * 3 min = 57 km horizontally. Thus, x = 57 km.
To find y, we can use the trigonometric relationship between the angle of climb and the vertical distance. Given that the angle of climb is 25° and the altitude of the plane is 4 km:
y = 4 km * tan(25°)
Now we can substitute the values of x and y back into the equation:
8 km(dr/dt) = (57 km)(19 km/min * cos(25°)) + (4 km * tan(25°))(19 km/min * sin(25°))
Simplifying the equation:
8 km(dr/dt) = (57 km * 19 km/min * cos(25°)) + (4 km * tan(25°) * 19 km/min * sin(25°))
Finally, we can solve for dr/dt by dividing both sides of the equation by 8 km:
dr/dt = ((57 km * 19 km/min * cos(25°)) + (4 km * tan(25°) * 19 km/min * sin(25°))) / 8 km
Plugging these values into a calculator or carrying out the arithmetic will give you the rate at which the distance from the plane to the radar station is increasing 3 minutes later.