The plane horizontal altitude is 1 mi and speed is 450 mi/h.
The plane distance from the station is 3 mi.
Draw the figure,
P is the plane's position
R is the radar station's position
V is the point located vertically of the radar station at the plane's height.
h is the plane's height
d is the distance between the plane and the radar station
x is the distance between the plane and the V point
Since the plane flies horizontally, we can conclude that PVR is a right triangle. Therefore, the pythagoreas theorem is used to find x,
[tex]d=\sqrt{h^2+x^2}[/tex]Substitute the values,
[tex]\begin{gathered} x=\sqrt{3^2-1^2} \\ x=\sqrt{9-1} \\ x=\sqrt{8} \\ x=2\sqrt{2} \end{gathered}[/tex]When d=3 mi, and, since the plane flies horizontally, we know that h=1mi regardless of the situation. We are looking for
[tex]\frac{dd}{dt}=d[/tex][tex]\begin{gathered} d^2=h^2+x^2 \\ 2d\frac{dd}{dt}=2h\frac{dh}{dt}+2x\frac{dx}{dt} \\ d\frac{dd}{dt}=x\frac{dx}{dt} \end{gathered}[/tex]Substitute the values.
[tex]\begin{gathered} \frac{dd}{dt}=\frac{x}{d}\frac{dx}{dt} \\ \frac{dd}{dt}=\frac{2\sqrt{2}}{3}\times450 \\ \frac{dd}{dt}=150\times2\times\sqrt{2} \\ \frac{dd}{dt}=300\sqrt{2} \end{gathered}[/tex]Hence the rate at which the distance from the plane to the station is increasing when it has a total distance of 3 mi away from the station is
[tex]424.6mi\text{ /h}[/tex]