A plane flying with a constant speed of 4 km/min passes over a ground radar station at an altitude of 5 km and climbs at an angle of 45 degrees. At what rate, in km/min is the distance from the plane to the radar station increasing 4 minutes later?
To solve this problem, you can use the concepts of trigonometry and rates of change.
First, let's visualize the scenario. The ground radar station is at point O, and the plane is at point P, which is initially at an altitude of 5 km. The plane is climbing at an angle of 45 degrees. Let's denote the distance from the plane to the radar station as d(t), where t represents time.
P
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O d(t)
We're given that the plane is flying with a constant speed of 4 km/min. This means that the rate of change of d(t) with respect to time is constant. Let's denote this rate of change as dd/dt.
To find the rate at which the distance from the plane to the radar station is increasing 4 minutes later, we need to find dd/dt at that specific time.
To do this, we can break down the motion of the plane into horizontal and vertical components. The horizontal component represents the motion of the plane directly towards or away from the radar station, and the vertical component represents the climb.
Let's start by finding the horizontal component of the velocity.
Given that the plane is flying with a constant speed of 4 km/min, the horizontal velocity component is equal to the velocity multiplied by the cosine of the angle of climb.
Horizontal velocity component (Vh) = 4 km/min * cos(45 degrees)
= 4 * √2 / 2 km/min
= 2√2 km/min
Next, let's determine how the distance from the plane to the radar station (d(t)) changes with time using the Pythagorean theorem.
Using Pythagoras, we have:
d(t)^2 = (Vh * t)^2 + (altitude)^2
Since the altitude is given as 5 km, we can rewrite the equation as:
d(t)^2 = (2√2 * t)^2 + 5^2
Rearranging the equation, we get:
d(t)^2 = 8t^2 + 25
Now, differentiate both sides of the equation with respect to time (t) to find the rate of change of d(t) with respect to time (dd/dt).
d( d(t)^2 )/dt = d( 8t^2 + 25 )/dt
Using the power rule of differentiation, which states that d(x^n)/dt = n * x^(n-1), we get:
2 * d(t)/dt = 16t
Rearranging the equation, we find:
d(t)/dt = 8t/2
= 4t
Now, we have the expression for the rate of change of d(t) with respect to time (dd/dt).
To find the rate at which the distance from the plane to the radar station is increasing 4 minutes later, we substitute t = 4 into the equation:
dd/dt = 4t
= 4(4)
= 16 km/min
Therefore, the distance from the plane to the radar station is increasing at a rate of 16 km/min 4 minutes later.
I know of no plane that can climb at a 45 degree angle for 4 minutes at 4 km/minute. It would be at altitude 21 km at that time.
Planes that slow (159 mph) do not fly that high.
X = 2.828 t
Y = 5 + 2.828 t
Distance to radar station = D
= sqrt(X^2 + Y^2)
= sqrt(8t^2 + 25 + 28.28t + 8 t^2)
- sqrt(16t^2 + 28.28t + 25)
Solve for dD/dt at plug in t = 4
Solve for