A plane is flying horizontally at an altitude of 0.9 kilometers with a velocity of 475 kilometers per hour when it flies over a radar station. Find the rate at which the distance is changing when the plane is 3.1 kilometers from the station. Round to the nearest tenth.

Drawing a diagram gives:
Identifing $LaTeX: \displaystyle \frac{db}{dt}=475$ , $LaTeX: \displaystyle a=0.9$ , and $LaTeX: \displaystyle c=3.1$ . Since the diagram is a right trinagle we can use the Pythagoren Theorem to get $LaTeX: \displaystyle (0.9)^2 + b^2 = c^2$ . Take the derivative with respect to time gives $LaTeX: \displaystyle 0 + 2b\frac{db}{dt} = 2c\frac{dc}{dt}$ . Solving for $LaTeX: \displaystyle \frac{dc}{dt}$ gives $LaTeX: \displaystyle \frac{dc}{dt} = \frac{b}{c}\frac{db}{dt}$ To find $LaTeX: \displaystyle \frac{dc}{dt}$ we need to calculate $LaTeX: \displaystyle b$ when $LaTeX: \displaystyle c = 3.1$ . Using the Pythagoren Theorem gives $LaTeX: \displaystyle b = \sqrt{3.1^2 - 0.9^2}$ . Finally calculating the value of the derivative $LaTeX: \displaystyle \frac{dc}{dt}=\frac{ \sqrt{3.1^2 - 0.9^2} }{ 3.1 }\cdot 475 \approx 454.5$ kilometers per hour.