A plane flying horizontally at an altitude of 3 mi and a speed of 480 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 6 mi away from the station. (Round to the nearest whole number.)
Why did the plane go to a therapist? Because it had issues with its altitude! Now let's solve this problem, shall we?
We can use the Pythagorean theorem to relate the distance between the plane and the radar station, the altitude of the plane, and the horizontal distance between the plane and the radar station.
Let's assume that the distance between the plane and the radar station is represented by 'x'. We know that the altitude of the plane remains constant at 3 mi. Therefore, the horizontal distance between the plane and the radar station is √(x^2 - 3^2).
To find the rate at which the distance from the plane to the station is increasing, we need to differentiate this expression with respect to time. Let's call the rate at which the distance is increasing 'dx/dt'.
Using the chain rule of differentiation, we can differentiate √(x^2 - 3^2) with respect to time 't'.
d/dt [√(x^2 - 3^2)] = d/dx [√(x^2 - 3^2)] * dx/dt
The derivative of √(x^2 - 3^2) with respect to x is (1/2) * (x^2 - 3^2)^(-1/2) * 2x, which simplifies to x/(√(x^2 - 3^2)).
Plugging in the value of 'x' as 6 mi, we get ((6)/(√(6^2 - 3^2))) * dx/dt.
Since we want to find dx/dt when x = 6 mi, let's plug in x = 6 mi into the equation:
((6)/(√(6^2 - 3^2))) * dx/dt = dx/dt
Now, we know dx/dt = 480 mi/h (as given in the problem).
Plugging in these values, we get:
((6)/(√(6^2 - 3^2))) * (480) = 480
Therefore, the rate at which the distance from the plane to the station is increasing when it is 6 mi away from the station is approximately 480 mi/h.
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 consider a right triangle formed by the plane, the radar station, and the line connecting them. The altitude of the plane is one side of the triangle, and the distance from the plane to the radar station is the hypotenuse of the triangle. We need to find the rate at which the hypotenuse is increasing when it is 6 mi away from the station.
We can use the Pythagorean theorem to relate the altitude, the distance from the plane to the radar station, and the hypotenuse. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Let's denote the distance from the plane to the radar station as x, the altitude of the plane as y, and the hypotenuse as z. From the problem statement, we have x = 6 mi and y = 3 mi. We want to find dz/dt, the rate at which z is changing (i.e., the rate at which the distance from the plane to the radar station is increasing).
Now, let's differentiate both sides of the Pythagorean theorem with respect to time (t) using implicit differentiation:
2z * (dz/dt) = 2x * (dx/dt) + 2y * (dy/dt)
Simplifying the equation, we have:
z * (dz/dt) = x * (dx/dt) + y * (dy/dt)
Since the plane is flying horizontally, its altitude y is not changing, so dy/dt = 0. The problem also states that the plane is flying at a speed of 480 mi/h, so dx/dt = 480 mi/h.
Substituting the known values into the equation:
z * (dz/dt) = 6 mi * (480 mi/h) + 3 mi * 0
z * (dz/dt) = 2880 mi²/h
Finally, we can solve for dz/dt by dividing both sides of the equation by z:
dz/dt = (2880 mi²/h) / (z)
Since we want to find the rate when the plane is 6 mi away from the station, we substitute x = 6 mi into the equation:
dz/dt = (2880 mi²/h) / (6 mi)
Simplifying, we get:
dz/dt = 480 mi/h
Therefore, the rate at which the distance from the plane to the station is increasing when it is 6 mi away from the station is approximately 480 miles per hour.
64
Did you make a sketch and got a right-angled triangle?
let the horizontal distance between the plane and the station be x miles
let the distance between the plane and the station be y miles
given: dx/dt = 480 mph
find: dy/dt when y = 6
y^2 = x^2 + 3^2
when y = 6
36 = x^2 + 9
x = √27
2y dy/dt = 2x dx/dt
dy/dt = (x/y)dx/dt
= √27(480)/6
= 415.7 mph or 416 mph