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

- jarp0120

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- jarp0120

|dw:1443859889821:dw|

- Jhannybean

It doesnt seem right for some reason

- IrishBoy123

can you think of an equation that connects l and b in your drawing?

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## More answers

- jarp0120

|dw:1443860176242:dw|

- jarp0120

A=1/2 bh

- Jhannybean

Using the pythagorean theorem, we can find the rate of change for everything moving

- dan815

oh boy a ladder problem :)

- jarp0120

what is the area if the base?

- jarp0120

is 4

- Jhannybean

You don't necessarily need to know that. You're simply looking for \(\dfrac{dl}{dt}\) when \(x=4\)

- jarp0120

how?

- IrishBoy123

Pythagoreas

- Jhannybean

The area is constantly changing w.r.t time as the ladder moves down the wall

- dan815

This question you are doing right now, will make or break your mathematical career

- Jhannybean

Dan... -_-

- jarp0120

there's no change in y right? the ladder?

- Astrophysics

|dw:1443860459057:dw| \[\frac{ dx }{ dt } = 2\] I thought this

- jarp0120

so i'll just substitute dy/dt and dx/dt?

- jarp0120

to find dl/dt?

- dan815

|dw:1443860606641:dw|

- Jhannybean

Yeah you're right, it `moved away from the wall` at that rate, whoops.

- Astrophysics

:)

- jarp0120

y=2 sqrt of 5?

- dan815

yes that is sqrt(2)

- dan815

sqrt(20)*

- dan815

but solve for dy/dt

- Astrophysics

yes dy/dt is what you need

- Jhannybean

I think i read the question wrong, fml.

- dan815

|dw:1443861020387:dw|

- 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

\[x^2+y^2=36\]
\[2x \frac{ dx }{ dt }+2y \frac{ dy }{ dt }=0\]
\[\frac{ dy }{ dt }=-\frac{ x }{ y }\frac{ dx }{ dt }\]

- 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

wait is this right?
dh/dt =-2sqrt5 / 5
that means the Area now is 6/sqrt5?

- Astrophysics

The rate should be negative because the ladder is decreasing

- Astrophysics

it just defines the direction

- jarp0120

or dy/dt= -2/sqrt5

- 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

And we know \[\frac{ dy }{ dt }=-\frac{ x }{ y }\frac{ dx }{ dt }\] just plugging in the values at this point

- Astrophysics

Yes dy/dt = - 2/sqrt(5)

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