A plane is flying horizontally at an altitude of 1.5 kilometers with a velocity of 395 kilometers per hour when it flies over a radar station. Find the rate at which the distance is changing when the plane is 3.9 kilometers from the station. Round to the nearest tenth.

Drawing a diagram gives:
Identifing $LaTeX: \displaystyle \frac{db}{dt}=395$ , $LaTeX: \displaystyle a=1.5$ , and $LaTeX: \displaystyle c=3.9$ . Since the diagram is a right trinagle we can use the Pythagoren Theorem to get $LaTeX: \displaystyle (1.5)^2 + b^2 = c^2$ . 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 $LaTeX: \displaystyle c = 3.9$ . Using the Pythagoren Theorem gives $LaTeX: \displaystyle b = \sqrt{3.9^2 - 1.5^2}$ . Finally calculating the value of the derivative $LaTeX: \displaystyle \frac{dc}{dt}=\frac{ \sqrt{3.9^2 - 1.5^2} }{ 3.9 }\cdot 395 \approx 364.6$ kilometers per hour.