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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: 3 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 $$its$$ 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
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Re: Ladder sliding down wall problem (Implicit Diff)

Quote:
 Originally Posted by mathkid 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
You are correct, it is another implicit differentiation. The ladder, wall, and floor make a right triangle so pythagorean theorem applies. DRAW THE PICTURE, with c as hypotenuse and sides x on floor, y on wall.

$x^2 + y^2= c^2$

$\frac{d}{dt} \ (x^2+ y^2) \= \ \frac{d}{dt} \ (c^2)$

$2x \ \frac{dx}{dt} \ + \ 2y \ \frac{dy}{dt} \= 2c \ \frac{dc}{dt}$

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 100-b^2] now go dA/dT = 1/2[dB/dt(100-b^2)^1/2 + B (1/2) (100-b^2)^-1/2 (you said take out sq root agent) dA/dT = 1/2 [dB/dT (root 100-b^2) - B/root 100-B^2 times dB/dT dA/dT = 1/2[8-4.5] 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: 2,810 Thanks: 53 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 ... $x \ \frac{dx}{dt} \ + \ y \ \frac{dy}{dt} \= \ c \ \frac{dc}{dt}$ 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: 2,810 Thanks: 53 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: 2,810 Thanks: 53 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 = ?
 November 12th, 2012, 08:28 PM #11 Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0 Re: Ladder sliding down wall problem (Implicit Diff) 0 ?
 November 12th, 2012, 08:33 PM #12 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 2,810 Thanks: 53 Re: Ladder sliding down wall problem (Implicit Diff) Bravo! Now plug all the known values into $x \ \frac{dx}{dt} + y \ \frac{dy}{dt}= c \frac{dc}{dt}$ and show me what you get.
 November 12th, 2012, 08:41 PM #13 Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0 Re: Ladder sliding down wall problem (Implicit Diff) I get.. 6(1) + 8 (dont have value for dy/dt ??) = 10(0)
 November 12th, 2012, 08:43 PM #14 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 2,810 Thanks: 53 Re: Ladder sliding down wall problem (Implicit Diff) Yes, very good, you are going to solve for dy/dt, let me see you do that.
 November 12th, 2012, 08:49 PM #15 Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0 Re: Ladder sliding down wall problem (Implicit Diff) 6 + 8dy/dt = 0 8dy/dt = -6 dy/dt= -6/8 dy/dt = -3/4 is this right agent?

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