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November 12th, 2012, 02:03 PM  #1 
Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0  Ladder sliding down wall problem (Implicit Diff)
A 10 ft ladder rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 ft/sec, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 feet from the wall? Mark, Agent...somebody would like help setting this one up. I am sure this is another Implicit application. Until I get comfortable with these... I am going to keep bugging out every time one pops up 
November 12th, 2012, 02:32 PM  #2 
Senior Member Joined: Sep 2007 Posts: 2,409 Thanks: 5  Re: Ladder sliding down wall problem (Implicit Diff)
I am concerned that you keep asking the same type of questions and don't seem to have learned anything from the responses you get. Draw a picture pf the ladder, ground and house. Do you see a right triangle? Label the two legs of the triangle and use the Pythagorean theorem to write a formula involving those two lengths. Differentiate both sides to get an equation involving the rates of change of those. You are told the distance from the house to the base of the ladder and its rate of change. You can use those two equations, the Pythagorean theorem and its derivative, to find the height of the ladder on the house and [tex]its[/tex] rate of change.

November 12th, 2012, 06:13 PM  #3 
Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0  Re: Ladder sliding down wall problem (Implicit Diff)
ok

November 12th, 2012, 07:27 PM  #4  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 2,959 Thanks: 111  Re: Ladder sliding down wall problem (Implicit Diff) Quote:
divide out 2 then plug in everything you know and we'll continue from there... you should be able to get x, dx/dt, y, c, and dc/dt using the info given that means dy/dt can be solved for.  
November 12th, 2012, 07:47 PM  #5 
Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0  Re: Ladder sliding down wall problem (Implicit Diff)
so is this right ..? b^2 + h^2 = 10^2 h^2 = 100 b^2 h = root 100  b^2 now I know from the past that Area = 1/2 base times height db/dt = 1 b = 6 so A = 1/2 [ b (root 100b^2] now go dA/dT = 1/2[dB/dt(100b^2)^1/2 + B (1/2) (100b^2)^1/2 (you said take out sq root agent) dA/dT = 1/2 [dB/dT (root 100b^2)  B/root 100B^2 times dB/dT dA/dT = 1/2[84.5] dA/dT = 1.75 ft^2/sec ? 
November 12th, 2012, 08:02 PM  #6 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 2,959 Thanks: 111  Re: Ladder sliding down wall problem (Implicit Diff)
You're not looking for dA/dt, so area formula is not usefull. You are looking for ROC of side against the wall. y^2 = 10^2  6^2 y^2 = 64 y = 8 so, so far you have x = 6 , y = 8 , c = 10 , dx/dt = 1 if you find dc/dt then you will have 5 of the 6 unknowns in ... what is the ROC of hypotenuse, dc/dt = ? 
November 12th, 2012, 08:15 PM  #7 
Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0  Re: Ladder sliding down wall problem (Implicit Diff)
can u give me a hint?

November 12th, 2012, 08:20 PM  #8 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 2,959 Thanks: 111  Re: Ladder sliding down wall problem (Implicit Diff)
as the ladder slides down, does the length of the hypotenuse change?

November 12th, 2012, 08:24 PM  #9 
Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0  Re: Ladder sliding down wall problem (Implicit Diff)
since the ladder is the hypotanuse I don't see how it's lengh can change it seems to me the lengh of the ladder is fixed

November 12th, 2012, 08:27 PM  #10 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 2,959 Thanks: 111  Re: Ladder sliding down wall problem (Implicit Diff)
correct! very good, now, if c is the hypotenuse AND IT'S CONSTANT AT 10, dc/dt = ? 

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