A plane is flying horizontally at an altitude of 0.8 kilometers with a velocity of 485 kilometers per hour when it flies over a radar station. Find the rate at which the distance is changing when the plane is 1.3 kilometers from the station. Round to the nearest tenth.

Drawing a diagram gives:
Identifing $LaTeX: \displaystyle \frac{db}{dt}=485$ , $LaTeX: \displaystyle a=0.8$ , and $LaTeX: \displaystyle c=1.3$ . Since the diagram is a right trinagle we can use the Pythagoren Theorem to get $LaTeX: \displaystyle (0.8)^2 + b^2 = c^2$ . 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 $LaTeX: \displaystyle c = 1.3$ . Using the Pythagoren Theorem gives $LaTeX: \displaystyle b = \sqrt{1.3^2 - 0.8^2}$ . Finally calculating the value of the derivative $LaTeX: \displaystyle \frac{dc}{dt}=\frac{ \sqrt{1.3^2 - 0.8^2} }{ 1.3 }\cdot 485 \approx 382.3$ kilometers per hour.