# Thread: related rates, maybe??

1. ## related rates, maybe??

A tightrope is stretched 30 feet above the ground between two buildings with are 50 feet apart. A tightrope walker, walking at a constant rate or 2 feet per second from point A to point B, is illuminated by a spotlight 70 feet above point A. (Point A is on the first building, 30 feet from the ground, and point B is on the 2nd building, also 30 feet from the ground)

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

I am so confused. Where do I start? Thanks!!

2. This is a related rates problem ,but it isn't as difficult as it would first appear.

Using similar triangles. Since the rate of the shadow of the walker's feet is what's needed. Just use:

$\frac{70}{25}=\frac{100}{x}$

Solving for x, we find $\frac{250}{7}$

Now, since she is walking at 2 ft/sec, it takes her 25/2 seconds to walk 25 feet.

We divide: $\frac{\frac{250}{7}}{\frac{25}{2}}=\boxed{\frac{20 }{7}}$ ft/sec.

3. Hello, ginax7!

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

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

I am so confused. Where do I start? . . . . Make a sketch!
Code:
    L *
| *
|   *
70 |     *
|       *
|         * T
A * - - - - - o - - - - - - * B
|     x       *           |
30 |               *         | 30
|                 *       |
* - - - - - - - - - o - - *
C - - - - s - - - - E     D
: - - - - - -50 - - - - - :

The tightrope walker is at $T$, having walked $x$ feet from $A$ to $T.$
The spotlight is at $L$, casting the shadow at $E$ on the ground.
Let $s \,= \,CE.$

From similar right triangles, we have: . $\frac{s}{100} \:=\:\frac{x}{70}\quad\Rightarrow\quad s \:=\:\frac{10}{7}x$

Differentiate with respect to time: . $\frac{ds}{dt} \:=\:\frac{10}{7}\,\frac{dx}{dt}$

Since $\frac{dx}{dt} \,= \,2$ ft/sec: . $\frac{ds}{dt} \:=\:\frac{10}{7}(2) \:=\:\frac{20}{7}$ ft/sec.