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General_Sax
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Beginner "Triangle" Problem
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?
r^2 = x^2 + y^2
dx/dt = 0.3 m/s
dr/dt = ?
dy/dt = 0
x = 2
r = 4
y ~ 3.46
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?
Homework Statement
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?
Homework 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
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?