At a given moment, a plane passes directly above a radar station at an altitude of 6 km. The plane's speed is 800 km/h. Let ℎ be the distance between the plane and the station through the air.

Question

At a given moment, a plane passes directly above a radar station at an altitude of 6 km. The plane's speed is 800 km/h. Let ℎ be the distance between the plane and the station through the air.

(a) How fast is the distance between the plane and the station changing half an hour later?

(Use decimal notation. Give your answer to three decimal places.)

𝑑ℎ𝑑𝑡≈

(b) How fast is the distance between the plane and the station changing when the plane passes directly above the station?

(Use decimal notation. Give your answer to three decimal places.)

𝑑ℎ𝑑𝑡≈

Answer 1

(a) How fast is the distance between the plane and the station changing half an hour later?

Well, half an hour is 30 minutes, which is plenty of time for the plane to make some distance. However, to calculate the rate of change of the distance, we need to know the plane's horizontal velocity.

Since we only have the plane's speed, we will have to assume that it is moving in a straight line. In that case, we can use the Pythagorean theorem to find the horizontal velocity:

v^2 = 800^2 - h^2,

where v is the horizontal velocity.

Solving for v, we have:

v = sqrt(800^2 - h^2).

Now, let's differentiate this equation with respect to time:

dv/dt = (d/dt)(sqrt(800^2 - h^2)).

Using the chain rule and the fact that dh/dt = 0 (since the plane is not changing altitude), we find:

dv/dt = -h * (1/sqrt(800^2 - h^2)) * (dh/dt).

Substituting dh/dt = 0 and h = 6 km, we get:

dv/dt = -6 * (1/sqrt(800^2 - 6^2)) * 0 = 0.

So, the rate of change of the distance between the plane and the station half an hour later is 0 km/h.

(b) How fast is the distance between the plane and the station changing when the plane passes directly above the station?

When the plane is directly above the station, the distance between them is the altitude of the plane, which is 6 km. At this moment, the plane is not changing its altitude or its horizontal velocity.

Therefore, the rate of change of the distance between the plane and the station when the plane passes directly above the station is 0 km/h.

Remember, always look on the bright side - at least you won't have to calculate any funky derivatives here. Keep smiling!

Answer 2

To solve this problem, we can use the Pythagorean theorem to relate the distance between the plane and the station through the air, h, to the horizontal distance traveled by the plane, x.

The Pythagorean theorem states that:

h^2 + x^2 = d^2

where d is the straight-line distance between the plane and the station.

Given that the plane's speed is 800 km/h, we can express the horizontal distance as:

x = 800t

where t is the time in hours.

(a) To find how fast the distance between the plane and the station is changing half an hour later, we need to differentiate the equation with respect to time (t):

2h * dh/dt + 2x * dx/dt = 2d * dd/dt

Since the plane passes directly above the station, d = h and dd/dt = 0. We also know that dx/dt = 800 km/h.

Plugging in these values, we get:

2h * dh/dt + 2(800t)(800) = 2h * 0

Simplifying:

1600h = 1600t * dh/dt

dh/dt = t * (h/t)

Since we want to find the rate of change half an hour later, we can substitute t = 0.5.

dh/dt = 0.5 * h / 0.5

dh/dt = h km/h

So, the rate at which the distance is changing half an hour later is equal to the height.

(b) To find how fast the distance is changing when the plane passes directly above the station, we need to consider the situation just before the plane reaches that position. At that moment, h = 6 km and x = 0.

Following the same differentiation process as in part (a), we get:

2h * dh/dt + 2(0) * dx/dt = 2h * dd/dt

2h * dh/dt = 0

dh/dt = 0 km/h

So, the rate of change in the distance when the plane passes directly above the station is 0 km/h.

Answer 3

To solve this problem, we can use the concept of related rates. We are given the altitude of the plane, the speed of the plane, and we need to find the rate at which the distance between the plane and the station is changing.

Let's start by establishing a triangle. The altitude of the plane forms one side of the triangle, while the distance between the plane and the station through the air (h) forms another side. The hypotenuse of the triangle represents the actual distance between the plane and the station.

(a) We are asked to find how fast is the distance (h) between the plane and the station changing half an hour later.

Given:
Altitude of the plane (y) = 6 km
Speed of the plane (ds/dt) = 800 km/h

We need to find dH/dt, the rate at which the distance (h) between the plane and the station is changing.

Let's use the Pythagorean theorem to relate the sides of the triangle:
h^2 + y^2 = H^2

Differentiating with respect to time (t) using the chain rule:
2h * (dh/dt) + 2y * (dy/dt) = 2H * (dH/dt)

We are given dy/dt, ds/dt, and y. Substitute these values into the equation:
2h * (dh/dt) + 2(6) * (800) = 2H * (dH/dt)

We need to find dh/dt, which represents the rate at which the distance between the plane and the station through the air is changing. To find this, rearrange the equation and solve for dh/dt:

2h * (dh/dt) = - 2(6) * (800) + 2H * (dH/dt)
dh/dt = [- 2(6) * (800) + 2H * (dH/dt)] / [2h]

Substitute t = 0.5 (half an hour later) and solve for dh/dt:
dh/dt ≈ [- 2(6) * (800) + 2H * (dH/dt)] / [2h]
where H is calculated using the Pythagorean theorem and the given values of y and h.

(b) We need to find how fast the distance between the plane and the station is changing when the plane passes directly above the station.

At this moment, the distance between the plane and the station is equal to the altitude of the plane (h = y), and the rate we are looking for is dh/dt.

Using the equation derived earlier:
dh/dt = [- 2(6) * (800) + 2H * (dH/dt)] / [2h]
Substitute h = y:
dh/dt = [- 2(6) * (800) + 2H * (dH/dt)] / [2y]

Remember to calculate H using the Pythagorean theorem.

Now we have the expressions for both (a) and (b) in terms of H, dH/dt, y, and h, so we need to substitute the given values and calculate the final answers.

Note: The values of H, dH/dt, y, and h were not provided in the question. Please provide these values to obtain the specific answers.

Answer 4

If the plane is x km from the point overhead, then

h^2 = 6 + x^2
h dh/dt = x dx/dt

Now you can answer the questions