An airplane is flying towards a radar station at a constant height of 6 km above the ground.

Question

If the distance s between the airplane and the radar station is decreasing at a rate of 400 km per hour when s=10 km, what is the horizontal speed of the plane?       Related Rates of AP Calculus!!! please help!

Answer 1

Hi Gabriele,

 

First, make the approximation that the radar station is at ground level. Then, you have a right triangle:

 

                                 x

                  plane   -----------

                             \           |

                               \         |

                             s   \       |  y

                                   \     |

                                     \   |

                                       \ |

                                         radar

                                         station

 

y is always a constant at y = 6 km.

 

s² = x² + y² by the Pythagorean theorem.

 

Let's now take the time derivative of the Pythagorean relationship.

Using the chain rule, we get:

 

2s ds/dt = 2x dx/dt.

 

Notice we do not have a 2y dy/dt term since y is always constant.

 

Cancel out the 2's in this last equation, and we can write:

 

s ds/dt = x dx/dt

 

We are told in the problem that s decreases at a rate of 400 km/hr when s = 10 km. What is the horizontal speed dx/dt at this time?

 

We can solve the last derivative equation for dx/dt:

 

dx/dt = (s/x) ds/dt

 

But, x = (s² - y²)^1/2

 

So, we can write

 

dx/dt = [s/(s² - y²)^1/2 ]ds/dt

 

s = 10 km, y = 6 km, ds/dt = -400 km/hr, where the negative sign denotes that the hypotenuse of the right triangle drawn above is reducing in size.

 

dx/dt = {(10 km) /[(10 km)² - (6 km)²]^1/2 }(-400 km/hr)

 

 

dx/dt = {(10 km) / [100 km² - 36 km²]^1/2} (-400 km/hr)

 

dx/dt = -500 km/hr