Subject: ap calculus for high school

I am a parent, Amy. I have a problem which I am trying to assist my child in solving:

A tightrope is stretched 30 feet above the ground between the Jay and the Tee buildings, which are 50 feet apart. A tightrope walker, walking at a constant rate of 2 feet per second from point A to point B, is illuminated by a spotlight 70 feet above point A. | |

(a) | how fast is the shadow of the tightrope walker's feet moving along the ground when she is midway between the buildings? (Indicate units of measure) |

(b) | how far from point A is the tightrope walker when the shadow of her feet reaches the base of the Tee building? (Indicate units of measure) |

(c) | how fast is the shadow of the tightrope walker's feet moving up the wall of the Tee building when she is 10 feet from point B? (Indicate units of measure) |

Thank you for any help you can offer.

Hi Amy

I drew a diagram of the situation and labeled it. You have W increasing at the rate of 2 feet per second, ie W'=2 feet per second. Triangles DCA and DEF are similar so W/70=X/(70+30). Thus 70X=100W. Differentiating with respect to time gives 70X'=100W'=100(2) and hence X'=20/7 feet per second.
Notice that this result does not depend on the value of X so the shadow is moving along the ground at a constant rate of 20/7 feet per second. When the tightrope walker is far enough that the point E is on the wall of the building say Y feet up from the ground (you need to redraw the diagram), triangles DCA and CBE are similar. This should allow you to find the rate at which E is moving up the wall.
Cheers |