Tom flies a kite at a height of 300 ft, and the wind is carrying the height horizontally away at a rate of 25 ft/s. How fast is the angle of elevation changing when y=500ft?

Answer 1

Tom flies a kite at a height of 300ft and the wind is carrying the kite horizontally away at a rate of 25ft/s. We want to determine the rate of change of the angle of elevation when y = 500ft.

It is not clear what y represents. I presume that it represents the length of the string attached to the kite.

Let's consider the diagram below

C

== | ---> u

==/ |

==/ |

==/ |

y ==/ |

==/ | h

==/ |

==/ ) |

==/ ) θ |

==/_________)_______________________ |

A x B

Let Tom and the kite be at point A and point C respectively, as shown in the figure above. Let point B be the point on the ground vertically below the kite.

Let h, x, y, θ, and u and  be the height of the kite, the distance between points A and B, the length of the string, the angle of elevation and the rate at which the wind is carrying away the kite respectively.

u = dx/dt = 25ft/s

tan(θ) = h/x

⇒ sec²θ dθ/dt = -h/x² dx/dt

⇒ dθ/dt = -h/x² cos²θ dx/dt

⇒ dθ/dt = -h/x² x²/y² dx/dt

⇒ dθ/dt = -h/y² dx/dt

It is given that h = 300ft, y = 500ft and u = dx/dt = 25ft/s

∴ dθ/dt = -(300 × 25) / 500² = -3/100 rad/s = -0.03 rad/s

Answer 2