a plane flying horizontally at an altitude of 2 mi and a speed of 460 mi/h passes directly over a radar station. find the rate at which the distance from the plane to the station is increasing when it is 5 mi away from the station.
To find the rate at which the distance from the plane to the station is increasing, we can use the concept of related rates. Let's denote the distance between the plane and the station as 'd' and the time as 't'.
We are given that the plane is flying horizontally at a speed of 460 mi/h, which means that the rate of change of distance with respect to time can be written as:
dd/dt = 460 mi/h
Now, we need to find the rate at which the distance from the plane to the station is increasing when it is 5 mi away from the station. Let's denote this rate as 'dr/dt' when d = 5 mi.
To find dr/dt, we should consider the relationship between the distance 'd', the altitude 'h', and the horizontal distance traveled by the plane 'x' from the radar station. This relationship can be represented by the Pythagorean theorem: d^2 = x^2 + h^2.
Since the plane is flying horizontally, the altitude 'h' remains constant at 2 mi. We can differentiate both sides of the equation with respect to time 't' to find the rate of change of distance with respect to time:
2d(dd/dt) = 2x(dx/dt)
At the given moment when the plane is 5 mi away from the station, we have:
d = 5 mi
h = 2 mi
And since the plane is flying horizontally, dx/dt is equal to the speed of the plane, which is 460 mi/h.
Plugging in these values, we can solve for dr/dt:
2(5)(dd/dt) = 2x(460)
10(dd/dt) = 920
dd/dt = 92 mi/h
Therefore, the rate at which the distance from the plane to the station is increasing when it is 5 mi away from the station is 92 mi/h.
To solve this problem, we can use the concept of related rates. We have the following information:
Altitude of the plane (h) = 2 mi
Speed of the plane (v) = 460 mi/h
Let's assume the distance between the radar station and the plane is represented by the variable 'd' (measured in miles). We are asked to find the rate at which the distance is increasing when the plane is 5 miles away from the station. This is the derivative of 'd' with respect to time, which can be represented as 'dd/dt'.
To solve for 'dd/dt', we can start by using the Pythagorean theorem to relate the variables 'd', 'h', and 'v':
d^2 = h^2 + (vt)^2
Taking the derivative of both sides with respect to time (t):
2dd/dt = 2h * dh/dt + 2vt * dv/dt
Since the plane is flying horizontally, the altitude (h) remains constant. Therefore, dh/dt = 0.
Plugging in the given values and the known values for h and v:
2dd/dt = 2(2) * 0 + 2(460)(dv/dt)
Simplifying the equation:
2dd/dt = 920(dv/dt)
Dividing both sides by 2 and canceling out the 2s:
dd/dt = 460(dv/dt)
We can solve for 'dv/dt' using the values given in the problem. At the given moment, when the distance (d) between the plane and the station is 5 miles, we can substitute 'd = 5' into the equation above and find 'dv/dt'.
dd/dt = 460(dv/dt)
dv/dt = dd/dt / 460
dv/dt = (460(dv/dt)) / 460
dv/dt = dv/dt
Therefore, the rate at which the distance from the plane to the station is increasing when it is 5 miles away from the station is equal to the rate at which the distance between the plane and the station is changing with time, which is represented by 'dv/dt'.