A plane is flying horizontally at an altitude of 1.7 kilometers with a velocity of 455 kilometers per hour when it flies over a radar station. Find the rate at which the distance is changing when the plane is 4.6 kilometers from the station. Round to the nearest tenth.

Drawing a diagram gives:
Identifing $LaTeX: \displaystyle \frac{db}{dt}=455$ , $LaTeX: \displaystyle a=1.7$ , and $LaTeX: \displaystyle c=4.6$ . Since the diagram is a right trinagle we can use the Pythagoren Theorem to get $LaTeX: \displaystyle (1.7)^2 + b^2 = c^2$ . 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 $LaTeX: \displaystyle c = 4.6$ . Using the Pythagoren Theorem gives $LaTeX: \displaystyle b = \sqrt{4.6^2 - 1.7^2}$ . Finally calculating the value of the derivative $LaTeX: \displaystyle \frac{dc}{dt}=\frac{ \sqrt{4.6^2 - 1.7^2} }{ 4.6 }\cdot 455 \approx 422.8$ kilometers per hour.