A community for students.
Here's the question you clicked on:
 0 viewing
shann803
 3 years ago
A 14 foot ladder is leaning against a wall. If the top slips down the wall at a rate of 4ft/s, how fast will the foot be moving away from the wall when the top is 11 feet from the ground.
shann803
 3 years ago
A 14 foot ladder is leaning against a wall. If the top slips down the wall at a rate of 4ft/s, how fast will the foot be moving away from the wall when the top is 11 feet from the ground.

This Question is Closed

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1So our first step is to find x. The length of that side. I think we'll end up needing that :o Remember your Pythagorean Theorem for finding that side? :)

shann803
 3 years ago
Best ResponseYou've already chosen the best response.0Ya so x would then be 11 as well because 14^211^2=121 and the square root of that is 11

shann803
 3 years ago
Best ResponseYou've already chosen the best response.0wow lol my bad i mean it's the square root of 75 is x not 11 so like 8.66 is x :) miscalculated

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1So we'll call it ... 5sqrt3 i guess :D in simplified form.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Hmm I'm trying to remember how to do this type of problem XD Lol. I think we can do it like this. \[\large x^2 + y^2 = 14^2\] Taking the derivative from this point. (With respect to t, time).

shann803
 3 years ago
Best ResponseYou've already chosen the best response.0so the derivative of x^2+y^2=14^2 is x/y?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1\[\large 2x\frac{ dx }{ dt }+2y \frac{ dy }{ dt }=0\] Since our derivative is with respect to t, Every time we differentiate a variable that is NOT t, we have to apply the chain rule multiplying by the d/dt term. For example when we differentiate x^2, we get 2x, but a dx/dt will pop out also.

shann803
 3 years ago
Best ResponseYou've already chosen the best response.0yes yes I remember my teacher saying that!

shann803
 3 years ago
Best ResponseYou've already chosen the best response.0so we plug in the dy/dt which is 4 and solve for dx/dt?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Heh :3 Ok cool. So now if we look at our problem. We now have 4 variables! Eeek! But we already KNOW 3 of them from the earlier! Yes plug in dy/dt, and also plug in y and x that we set up earlier :)

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1from the earlier? what is wrong with me _ ugh..

shann803
 3 years ago
Best ResponseYou've already chosen the best response.0so it's 2(5sqrt3)(dx/dt)+2(11)(4) is this right?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Hmm yes good good :) equals 0

shann803
 3 years ago
Best ResponseYou've already chosen the best response.0yes it's right!!1 THANK YOU SOOOO MUCH!! :)

shann803
 3 years ago
Best ResponseYou've already chosen the best response.0I have another question going it's a problem solving like this but in the mathematics section do you think you can help me with that one too please?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1When you do these problems, one way to check your work is this... You should get an answer that MAKES SENSE. If the ladder is sliding down the wall in the y direction at 4 ft/s Then our slide along the x direction should be a similar number.

shann803
 3 years ago
Best ResponseYou've already chosen the best response.0true true, and yes the gravel problem lol :)
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.