Rates of Change Sliding Ladder Application

AI Thread Summary
A 5m ladder is sliding down a wall at a rate of 0.5m/s, and the discussion revolves around determining the rate at which the bottom of the ladder moves away from the wall when it is 3m from the wall. Participants emphasize the importance of relating the distances from the wall and floor using the Pythagorean theorem, leading to a right triangle scenario. The conversation highlights the need for differentiation to connect the rates of change of the distances. There is confusion regarding the quadratic nature of the problem, but it is clarified that it is a linear problem requiring a dynamic equation. Ultimately, the focus is on correctly applying calculus to find the desired rate of change.
Dirac
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"a 5m ladder is placed against a wall. the top of the ladder is sliding down the wall at 0.5m per second. at what rate is the bottom of the ladder moving away from the wall when the bottom of the ladder is 3m from the wall. ? "

Dirac.
 
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Dirac said:
"a 5m ladder is placed against a wall. the top of the ladder is sliding down the wall at 0.5m per second. at what rate is the bottom of the ladder moving away from the wall when the bottom of the ladder is 3m from the wall. ? "

Dirac.
let x(t), y(t) be the distance respectively from the wall and floor.
1) relate x and y
2) differentiate the relationship to relate x' and y'
3) find write y(t) in terms of t then find x(t),x'(t),y'(t)
4) use x(ti)=3 to find x'(ti)
 
lurflurf said:
let x(t), y(t) be the distance respectively from the wall and floor.
1) relate x and y
2) differentiate the relationship to relate x' and y'
3) find write y(t) in terms of t then find x(t),x'(t),y'(t)
4) use x(ti)=3 to find x'(ti)

Could you please do a step-by-step solution

Dirac.
 
Dirac said:
Could you please do a step-by-step solution

Dirac.
No.
so the ladder, a piece of floor and a piece of wall form a right triangle.
where the legs are x,y and the hypotenus is 5 can you write an equation relating these quantities?
 
lurflurf said:
No.
so the ladder, a piece of floor and a piece of wall form a right triangle.
where the legs are x,y and the hypotenus is 5 can you write an equation relating these quantities?

Yes, ok but what is x(t)
 
Dirac: TRY! If you can't get it show us what you have done and we'll give suggestions. The important thing is that you try yourself.
 
In a simpler way, if the ladder is moving down at a rate of 0.5m/s which is 5m above the ground, the ladder is slidding at the same time, so what you need to do is find the time it takes for the ladder to fall using v=dx/dt or v=x/t and then using the same equation use that elapsed time with the bottom of the ladder that is 3m away from the wall. does this make sense?
 
HallsofIvy said:
Dirac: TRY! If you can't get it show us what you have done and we'll give suggestions. The important thing is that you try yourself.

Yes, I can do it, but I get two solutions for t from

(x^2)=5t-0.25(t^2)

Dirac.
 
Last edited:
Dirac said:
Yes, ok but what is x(t)
x is the distance from the bottom of the ladder to the wall measured along a line along the floor that is perpendicular to the wall.
y is the distance from the top of the ladder to the floor measured along a line along the wall that is perpendicular to the floor
also I am assuming the wall is perpendicular to the floor
thus we have a right triangle with legs x,y and hypotenus 5
 
  • #10
I don't understand how you have a quadratic, this is a linear problem

I get from your values that the velocity of the sliding is 0.333m/s. But don't quote me!

I f you show me your working perhaps i can help

hhh79bigo
 
  • #11
It is not linear

By Pythagoras'

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

y=5-0.5t

r=5

=>(x^2)=25-(25-5t+0.25(t^2))

=>(x^2)=5t-0.25(t^2)
 
  • #12
Oh ok, done it now.

Dirac.
 
  • #13
Dirac said:
Yes, I can do it, but I get two solutions for t from

(x^2)=5t-0.25(t^2)

Dirac.
x and y should be positive
then there is only one solution
There is also an easier way
we know
x*x'+y*y'=0
so
x'=y*y'/x
we know y'=-.5 x=3 y=4 so x' is easy to find
 
  • #14
Dirac said:
Yes, I can do it, but I get two solutions for t from

(x^2)=5t-0.25(t^2)

Dirac.

And you STILL haven't shown us what you have done! Don't just give us your (wrong) answer. Show us how you got it.

In fact, your answer makes no sense. The problem does not even ASK you to find t! If you let x be the distance from the wall to the foot of the ladder, the problem asks you to find dx/dt.

Now do what lurflurf suggested to begin with: Draw a picture, look at the right triangle in the picture and write an equation relating the parts of the picture. That will be a "static" equation- your picture is kind of like a snapshot of the sliding ladder. To get a "dynamic" equation (a moving picture) differentiate the entire equation with respect to t (even though there is no t in the equation!)

For the same reason, the answer to hhh79bigo's question is "No, that makes no sense at all- you were not asked to find the time the ladder takes to fall to the floor."
 
  • #15
For the same reason, the answer to hhh79bigo's question is "No, that makes no sense at all- you were not asked to find the time the ladder takes to fall to the floor."[/QUOTE]

Forgive me for being ignorant

I apologise I thought that the top of the ladder was 5m of the ground

hhh79bigo
 
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