jarp0120
  • jarp0120
a ladder 6 m long leans against a vertical wall. The lower end of the ladder is moved away from he wall at the rate of 2 m/min. Find the rate of change of the area formed by the wall, the floor and the ladder when the lower end is 4 m from the wall. please help
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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jarp0120
  • jarp0120
|dw:1443859889821:dw|
Jhannybean
  • Jhannybean
It doesnt seem right for some reason
IrishBoy123
  • IrishBoy123
can you think of an equation that connects l and b in your drawing?

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More answers

jarp0120
  • jarp0120
|dw:1443860176242:dw|
jarp0120
  • jarp0120
A=1/2 bh
Jhannybean
  • Jhannybean
Using the pythagorean theorem, we can find the rate of change for everything moving
dan815
  • dan815
oh boy a ladder problem :)
jarp0120
  • jarp0120
what is the area if the base?
jarp0120
  • jarp0120
is 4
Jhannybean
  • Jhannybean
You don't necessarily need to know that. You're simply looking for \(\dfrac{dl}{dt}\) when \(x=4\)
jarp0120
  • jarp0120
how?
IrishBoy123
  • IrishBoy123
Pythagoreas
Jhannybean
  • Jhannybean
The area is constantly changing w.r.t time as the ladder moves down the wall
dan815
  • dan815
This question you are doing right now, will make or break your mathematical career
Jhannybean
  • Jhannybean
Dan... -_-
jarp0120
  • jarp0120
there's no change in y right? the ladder?
Astrophysics
  • Astrophysics
|dw:1443860459057:dw| \[\frac{ dx }{ dt } = 2\] I thought this
jarp0120
  • jarp0120
so i'll just substitute dy/dt and dx/dt?
jarp0120
  • jarp0120
to find dl/dt?
dan815
  • dan815
|dw:1443860606641:dw|
Jhannybean
  • Jhannybean
Yeah you're right, it `moved away from the wall` at that rate, whoops.
Astrophysics
  • Astrophysics
:)
jarp0120
  • jarp0120
y=2 sqrt of 5?
dan815
  • dan815
yes that is sqrt(2)
dan815
  • dan815
sqrt(20)*
dan815
  • dan815
but solve for dy/dt
Astrophysics
  • Astrophysics
yes dy/dt is what you need
Jhannybean
  • Jhannybean
I think i read the question wrong, fml.
dan815
  • dan815
|dw:1443861020387:dw|
Jhannybean
  • Jhannybean
Oh okay, as the ladder moves down the wall, the ladder increases the base of the triangle but shortens the height of the triangle, therefore you're looking for the rate of change of the height
Astrophysics
  • Astrophysics
\[x^2+y^2=36\] \[2x \frac{ dx }{ dt }+2y \frac{ dy }{ dt }=0\] \[\frac{ dy }{ dt }=-\frac{ x }{ y }\frac{ dx }{ dt }\]
Astrophysics
  • Astrophysics
So x here is the distance from the bottom of the ladder to the wall and y is the distance of the top of the ladder to the ground, as the image shows. The second part of the equation we simply use chain rule, \[2x \frac{ dx }{ dt }+2y \frac{ dy }{ dt }=0\] and we just solve for dy/dt then
jarp0120
  • jarp0120
wait is this right? dh/dt =-2sqrt5 / 5 that means the Area now is 6/sqrt5?
Astrophysics
  • Astrophysics
The rate should be negative because the ladder is decreasing
Astrophysics
  • Astrophysics
it just defines the direction
jarp0120
  • jarp0120
or dy/dt= -2/sqrt5
Astrophysics
  • Astrophysics
Well what did Pythagorean theorem give you? x=4, r=6 \[x^2+y^2=r^2 \implies y=\sqrt{36-16} = \sqrt{20}\]
Astrophysics
  • Astrophysics
And we know \[\frac{ dy }{ dt }=-\frac{ x }{ y }\frac{ dx }{ dt }\] just plugging in the values at this point
Astrophysics
  • Astrophysics
Yes dy/dt = - 2/sqrt(5)

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