# Beginner Triangle Problem

1. Jun 4, 2009

### General_Sax

Beginner "Triangle" Problem

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

A ladder 4m long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a speed of 30cm/s, how quickly is the top of the ladder sliding down the wall when the bottom of the ladder is 2m away from the wall?

2. Relevant equations

r^2 = x^2 + y^2

dx/dt = 0.3 m/s

dr/dt = ?

dy/dt = 0

x = 2

r = 4

y ~ 3.46

3. The attempt at a solution

I made two attempt at the question.

#1:

r^2 = x^2 + y^2

2*r*dr/dt = 2*x*dx/dt + 2*y*dy/dt

2*4*dr/dt = 2*2*dx/dt + 2*3.46*0

8*dr/dt = 4*0.3

> dr/dt = 0.15 m/s <

#2:

r^2 = x^2 + (r^2 - x^2)^2

r^2 = x^2 + (r^4 - 2*r^2*x^2 + x^4)

I'm not going to type the intermidiate steps, if you want to know how I arrived at an equation please ask.

dr/dt * [4r^3 - 2r - 4rx^2] = [ 4*x*r^2*dx/dt - 4*x^3*dx/dt - 2*x*dx/dt ]

dr/dt * [184] = [27.6]

> dr/dt = 0.15 m/s <

>.< The answer in the textbook is: sqrt(3)/10 ~ 0.17 m/s

Any suggestions as to where I went wrong?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 4, 2009

### zcd

Re: Beginner "Triangle" Problem

Given that the formula given is r^2 = x^2 + y^2, you can assume that r is the hypotenuse. The question is asking for rate at which the ladder falls down (dy/dt). Even if a ladder slides down a wall, it is unlikely that the length of the ladder changes at all (dr/dt = 0).

3. Jun 4, 2009

### General_Sax

Re: Beginner "Triangle" Problem

Ahhh, yes that makes sense. I was confusing dr/dt with dy/dt. Thanks a bunch.

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