Question

A plane flying horizontally at an altitude of 1 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 3 mi away from the station.

Answer 1

First remember that the distance between two points (a, b) and (c, d) is given by the equation:

`d = √((a - c)^2 + (b - d)^2)`

Now let's define the position of the radar as:

(0mi, 0mi)

Then we can write the position of the plane as:

(480mi/h*t, 1mi)

where t is time in hours.

Then we can write the distance equation as:

`d(t) = \sqrt{(480(mi)/(h)*t - 0mi)^2 + (1mi -0mi)^2 } \\\\d(t) = \sqrt{(480(mi)/(h)*t )^2 + (1mi)^2 }`

Now we want to get:

the rate at which the distance from the plane to the station is increasing when it is 3 mi away from the station.

So first we want to find the value of t such that:

d(3) = 3mi

We will look at the positive value of t, because at this point the plane is increasing its distance to the station.

`3mi = \sqrt{(480(mi)/(h)*t )^2 + (1mi)^2 }\\\\(3mi)^2 = (480(mi)/(h)*t )^2 + (1mi)^2\\\\9mi^2 - 1mi^2 = (480(mi)/(h)*t )^2\\\\8mi^2 = (230,400 mi^2/h^2)*t^2\\\\\\\sqrt{(8mi^2)/(230,400 mi^2/h^2) } = t = 0.0059 h`

The rate of change when the plane is 3 mi away from the station is:

d'(0.0059h)

remember that:

d'(t) = dd(t)/dt

We can write:

d(t) = h( g(t) )

such that:

h(x) = √x

g(t) = (480mi/h*t)^2 + (1mi)^2

then:

d'(t) = h'(g(t))*g'(t)

This is:

`d'(t) = (dd(t))/(dt) = (1)/(2)*(2*t*480mi/h)/(√((480mi/h*t)^2 + (1mi)^2) )`

The rate of change at t = 0.0059h is then:

`d'(0.0059h) = (1)/(2)*(2*0.0059h*(480mi/h)^2)/(√((480mi/h*0.0059h)^2 + (1mi)^2) ) =452.6 mi/h^2`