A plane flying horizontally at an altitude of 1 mi and a speed of 510 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.)

How do I do this related rates problem?

A plane flying horizontally at an altitude of 1 mi and a speed of 510 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.)

wow that's hard

To solve this problem, we can use the concept of related rates.

Let's denote the distance from the plane to the radar station as "d", and the rate at which this distance is changing as "dd/dt". We want to find the value of dd/dt when the distance (d) is 6 miles.

Given that the plane is flying horizontally and passing directly over the radar station, we can consider the situation as a right triangle with the radar station as the right angle vertex, the distance "d" as the hypotenuse, and the horizontal distance between the plane and the radar station as "x".

Using the Pythagorean theorem, we have:

d^2 = x^2 + 1^2

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

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

Simplifying, we have:

d * dd/dt = x * dx/dt

We are given that the speed of the plane is 510 mi/h, which is equal to dx/dt. Also, when the distance (d) is 6 miles, we need to find dd/dt.

At that moment, we can find x using the Pythagorean theorem:

6^2 = x^2 + 1^2
36 = x^2 + 1
x^2 = 35
x = √35

Now we can substitute the known values into our equation:

6 * dd/dt = √35 * 510

Simplifying further:

dd/dt = (√35 * 510) / 6

Using a calculator, we find:

dd/dt ≈ 244

Therefore, the rate at which the distance from the plane to the radar station is increasing when it is 6 miles away from the station is approximately 244 miles per hour.

To solve this problem, we need to use the concept of related rates. We are given the following information:

- The plane is flying horizontally at a speed of 510 mi/h.
- The plane is at an altitude of 1 mi.
- The distance between the plane and the radar station is changing.

Let's denote the distance between the plane and the radar station as "d" and the rate at which this distance is changing as "dy/dt". We are asked to find the rate at which the distance is increasing when it is 6 mi away from the station, which means we need to find dy/dt when d = 6 mi.

To start solving the problem, let's draw a diagram:

|
P | R
L | A
A | D
N | A
E | T
|
|
|

In this diagram, P represents the plane, R represents the radar station, and the line connecting them represents the distance, d.

Since the plane is flying horizontally, the distance between the plane and the radar station is the hypotenuse of a right triangle. The altitude of the plane is one side of the triangle, and the distance the plane has traveled is the other side of the triangle.

Let's denote the distance traveled by the plane as "x" and the altitude of the plane as "y". We can use the Pythagorean theorem to find the relationship between x, y, and d:

d^2 = x^2 + y^2

Since we are given that the altitude of the plane is 1 mi, we can write:

d^2 = x^2 + 1^2
d^2 = x^2 + 1

To proceed, we need to find an equation that relates x and d. We know that the plane is flying at a speed of 510 mi/h, which means that dx/dt = 510 mi/h.

Differentiating both sides of the equation d^2 = x^2 + 1 with respect to time, we get:

2d(dy/dt) = 2x(dx/dt)

Now we can substitute the values we know:

2(6)(dy/dt) = 2(x)(510)

Simplifying the equation, we have:

12(dy/dt) = 1020x

We still need to find the value of x when d = 6 mi. To do this, we can use the relationship between d and x that we found earlier:

d^2 = x^2 + 1

(6)^2 = x^2 + 1
36 = x^2 + 1
x^2 = 35
x = sqrt(35)

Now we can substitute this value into our equation:

12(dy/dt) = 1020(sqrt(35))

To find dy/dt, we divide both sides of the equation by 12:

dy/dt = 1020(sqrt(35))/12

Evaluating this expression, we get:

dy/dt ≈ 90

Therefore, the rate at which the distance from the plane to the radar station is increasing when it is 6 mi away from the station is approximately 90 mi/h.