A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
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
anonymous
 one year ago
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

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443859889821:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It doesnt seem right for some reason

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0can you think of an equation that connects l and b in your drawing?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443860176242:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Using the pythagorean theorem, we can find the rate of change for everything moving

dan815
 one year ago
Best ResponseYou've already chosen the best response.1oh boy a ladder problem :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what is the area if the base?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You don't necessarily need to know that. You're simply looking for \(\dfrac{dl}{dt}\) when \(x=4\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The area is constantly changing w.r.t time as the ladder moves down the wall

dan815
 one year ago
Best ResponseYou've already chosen the best response.1This question you are doing right now, will make or break your mathematical career

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there's no change in y right? the ladder?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443860459057:dw \[\frac{ dx }{ dt } = 2\] I thought this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so i'll just substitute dy/dt and dx/dt?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah you're right, it `moved away from the wall` at that rate, whoops.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0yes dy/dt is what you need

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think i read the question wrong, fml.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh 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
 one year ago
Best ResponseYou've already chosen the best response.0\[x^2+y^2=36\] \[2x \frac{ dx }{ dt }+2y \frac{ dy }{ dt }=0\] \[\frac{ dy }{ dt }=\frac{ x }{ y }\frac{ dx }{ dt }\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0So 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait is this right? dh/dt =2sqrt5 / 5 that means the Area now is 6/sqrt5?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0The rate should be negative because the ladder is decreasing

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0it just defines the direction

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Well what did Pythagorean theorem give you? x=4, r=6 \[x^2+y^2=r^2 \implies y=\sqrt{3616} = \sqrt{20}\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0And we know \[\frac{ dy }{ dt }=\frac{ x }{ y }\frac{ dx }{ dt }\] just plugging in the values at this point

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Yes dy/dt =  2/sqrt(5)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.