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anonymous
 one year ago
I need help again!
A 15ft ladder is leaning against one wall of an alley 9 ft wide. the lade slips, its top sliding down the wall, its foot sliding across the alley and striking the opposite wall at a speed of 4ft/sec. how fast is the top of the ladder falling at that instance?
calculus makes me want to cry!
anonymous
 one year ago
I need help again! A 15ft ladder is leaning against one wall of an alley 9 ft wide. the lade slips, its top sliding down the wall, its foot sliding across the alley and striking the opposite wall at a speed of 4ft/sec. how fast is the top of the ladder falling at that instance? calculus makes me want to cry!

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2dw:1438840509295:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Do you agree with the diagram

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think so... i suck at calculus.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2dw:1438841158267:dw as it's sliding so we use the Pythagorean theorem \[x^2+y^2=r^2\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2\[x^2+y^2=15^2 \implies x^2+y^2=225\] now differentiate \[2x \frac{ dx }{ dt }+2y \frac{ dy }{ dt } = 0\] Finish it off

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2`how fast is the top of the ladder falling at that instance` this tells you what you're solving for, should be clear

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2dw:1438841472668:dw just making sure you know it's a right triangle :P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i just confused myself even further. I just can't figure this out.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Ok, so given the equation what are we solving for?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2You are solving for dy/dt as we're looking for how fast the top of the ladder is falling

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2So solve for dy/dt first, as you would for x, don't plug anything just isolate dy/dt.

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

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Yeah :) good \[\frac{ dy }{ dt }= \frac{ x }{ y }\frac{ dx }{ dt }\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2So we know x = 9, and r = 15, what is y?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2x = 9, y = 12, r = 15, and we know dx/dt = 4 ft/s \[\frac{ dy }{ dt } =  \frac{ 9 }{ 12 }(4) ft/s\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2 3ft/s the negative just means the distance from the top of the ladder is decreasing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh okay i get it now!
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