1. ## Word Problem Calculus!

Carlo is perched precariously on top of a 10-foot ladder leaning against the back wall of an apartment building (spying on an enemy of his) when it starts to slide down the wall at a rate of 4 ft per minute. Carlo's accomplice, Alex, is standing on the groud 6 ft. away from the wall, with the bottom of the ladder between the wall and Alex. How fast is the base of the ladder moving when it hits Alex?
So far... this is by solution.

dy/dt = 4
x = 6

and my equation is (but obviously wrong):
x^2 + y^2 = 10^ / 100.
6^2 + y^2 = 100
y^2 = 100-36
y = 8

2. Originally Posted by anfieldgirl
I posted this in the calculus forum. I'm so sorry for posting again, but i didn't see this forum until afterwards. I'm in urgent need of help. Please please help me out.

So far... this is by solution.

dy/dt = 4
x = 6

and my equation is (but obviously wrong):
x^2 + y^2 = 10^ / 100.
6^2 + y^2 = 100
y^2 = 100-36
y = 8

x^2 + Y^2 =100
diff bth sides wrt. t
we get
2x dx/dt +2y dy/dt =0
dy/dt=-4 (given)
x=6 y=8
find dx/dt.........

3. Thank you so much! Really appreciate the help.

4. Hello, anfieldgirl!

Sounds like "Home Alone" . . .

Carlo is perched precariously on top of a 10-foot ladder leaning
against the back wall of an apartment building (spying on an enemy of his)
when it starts to slide down the wall at a rate of 4 ft per minute.
Carlo's accomplice, Alex, is standing on the ground 6 ft away from the wall.
How fast is the base of the ladder moving when it hits Alex?
Code:
      |
*
|\
| \
|  \
y |   \ 10
|    \
|     \
|      \
* - - - * - - *
x         A

We are given: . $\frac{dy}{dt} \:=\:\text{-}4$ ft/min.

We know that: . $x^2 + y^2 \:=\:10^2$

Differentiate with respect to time: . $2x\frac{dx}{dt} + 2y\frac{dy}{dt} \:=\:0 \quad\Rightarrow\quad \frac{dx}{dt} \:=\:-\frac{y}{x}\,\frac{dy}{dt}$

When $x = 6,\:y = 8$

So we have: . $\frac{dx}{dt} \;=\;-\frac{8}{6}(-4) \;=\;\frac{16}{3}$

The base of the ladder hits Alex at $5\tfrac{1}{3}$ ft/min.