A plane flies horizontally with a speed Of 600km/h at an altitude of 10km and pasess directly over th town of quinton .find the rate at which the distance from the plane to quinton is increasing when its 20km away from quinton ?

Not sure why the oblique distance needs to be calculated, but that's what the question asks.

The solution is similar to the cone problem.

dx/dt=600 km h-1
h=10 km
x=20 km
L=sqrt(x²+h²)
dL/dx=x/sqrt(x²+h²)
dL/dt
=(dL/dx) * (dx/dt)
=(x/sqrt(x²+h²)) * (600 km h-1)
=600x/sqrt(x²+h²) km-h-1
=537 km-h-1

To find the rate at which the distance from the plane to Quinton is increasing when it is 20 km away from Quinton, we can use the concept of similar triangles and the Pythagorean theorem.

Let's denote the distance from the plane to Quinton as d, and the horizontal distance from Quinton to the point directly below the plane as x. We are given that the plane is flying horizontally with a speed of 600 km/h, so the rate of change of x with respect to time (dx/dt) is equal to 600 km/h.

Using the Pythagorean theorem, we can express d in terms of x and the altitude of the plane:

d^2 = x^2 + (10 km)^2

Differentiating both sides of the equation with respect to time (t), we get:

2d * (dd/dt) = 2x * (dx/dt)

Since we want to find the rate at which d is increasing, we can solve for (dd/dt):

dd/dt = (x * dx/dt) / d

Now, we are given that the plane is 20 km away from Quinton, so x = 20 km.

Substituting this value into the equation, we have:

dd/dt = (20 km * 600 km/h) / d

To find the rate, we need to determine the value of d at this instant when x = 20 km. To do this, we can use the right triangle formed by the plane's altitude and the distance between the plane and Quinton. We can apply the Pythagorean theorem:

d^2 = x^2 + (10 km)^2
d^2 = (20 km)^2 + (10 km)^2
d^2 = 400 km^2 + 100 km^2
d^2 = 500 km^2

Taking the square root of both sides, we get:

d ≈ 22.36 km

Now, substituting this value into the equation for dd/dt, we have:

dd/dt = (20 km * 600 km/h) / 22.36 km
dd/dt ≈ 847.46 km/h

Therefore, the rate at which the distance from the plane to Quinton is increasing when the plane is 20 km away from Quinton is approximately 847.46 km/h.