# Homework Help: An Involved Related Rate Problem

1. Jan 2, 2012

### Wisco

1. The problem statement, all variables and given/known data

A tightrope is stretched 30ft. above the ground between two buildings, which are 50 ft apart. apart. A tightrope walker, walking at a constant rate of 2ft/sec from point A to point B, is illuminated by a spotlight 70ft above point A.

1.) How fast is the shadow of the tightrope walker's feet moving along the ground when she is midway between the buildings?

2.) How far from point A is the tightrope walker when the shadow of her feet reaches the base of the 2nd building

3.) How fast is the shadow of the tightrope walker's feet moving up the 2nd building when she 10 ft from point B.

2. Relevant equations

3. The attempt at a solution

1.) Is this a trick question? Maybe I'm visualizing the problem incorrectly but what reason is there to believe that the shadow is moving at a different rate then the tightrope walker--shouldn't it be 2ft/sec?

2.) The distance between the buildings if 50ft, and the tightrope is 30ft from the ground plus the height of the spotlight making it 100ft in total. Pythagorean theorem gives me a distance from the spotlight of about (√1250) ft. Now that I have this number, and knowing that the triangle formed by the spotlight, walker and point at which the tightrope meets the first building is similar to that formed by the shadow, spotlight and ground, I should be able to solve for an angle and find the distance between the the walker and first building...

...Except that this involves some trigonometric identity that I must not know, and that it's a related rate problem--so the derivative should've been taken, right?

3.) I know that the height from the ground to the building is 30ft and that it's 10ft across to the tightrope so I can solve for the distance from ground to tightrope by Pythagorean theorem. I also know that the rate is 2.0 ft/sec from tightrope to second building and that the height from building to tightrope is 30ft so I can get:

10ft(2.0 ft/sec) + 30ft(0 ft/sec) = (√1000)ft(x) ...but I don't think this is the right equation...?

2. Jan 2, 2012

### SammyS

Staff Emeritus
The tightrope walker will be only part of the way from A to B at the time that her shadow reaches the base of the destination building. Therefore, she walks less than 50ft in the same time it takes the shadow to go the entire 50 ft. So of course she moves at a different rate than her shadow.

You're right about the similar triangles, but this question can be solved with simple proportions, no trigonometry involved.
Hello Wisco. Welcome to PF !

To find the location of the shadow in part 3: Again, this can be done with similar triangles and a proportion.

To find the rates for parts 1 & 3, you will need to use related rates. This will involve derivatives.

3. Jan 2, 2012

### Wisco

Thank You. So if I'm correct, Part 2 is nothing more than (100/50)=(70/x) giving me x=35 correct?