Answer:
First remember that the distance between two points (a, b) and (c, d) is given by the equation:
[tex]d = \sqrt{(a - c)^2 + (b - d)^2}[/tex]
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:
[tex]d(t) = \sqrt{(480\frac{mi}{h}*t - 0mi)^2 + (1mi -0mi)^2 } \\\\d(t) = \sqrt{(480\frac{mi}{h}*t )^2 + (1mi)^2 }[/tex]
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.
[tex]3mi = \sqrt{(480\frac{mi}{h}*t )^2 + (1mi)^2 }\\\\(3mi)^2 = (480\frac{mi}{h}*t )^2 + (1mi)^2\\\\9mi^2 - 1mi^2 = (480\frac{mi}{h}*t )^2\\\\8mi^2 = (230,400 mi^2/h^2)*t^2\\\\\\\sqrt{\frac{8mi^2}{230,400 mi^2/h^2} } = t = 0.0059 h[/tex]
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:
[tex]d'(t) = \frac{dd(t)}{dt} = \frac{1}{2}*\frac{2*t*480mi/h}{\sqrt{(480mi/h*t)^2 + (1mi)^2} }[/tex]
The rate of change at t = 0.0059h is then:
[tex]d'(0.0059h) = \frac{1}{2}*\frac{2*0.0059h*(480mi/h)^2}{\sqrt{(480mi/h*0.0059h)^2 + (1mi)^2} } =452.6 mi/h^2[/tex]
