anonymous
  • anonymous
I need help again! A 15-ft 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!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Astrophysics
  • Astrophysics
|dw:1438840509295:dw|
Astrophysics
  • Astrophysics
Do you agree with the diagram
anonymous
  • anonymous
i think so... i suck at calculus.

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More answers

Astrophysics
  • Astrophysics
|dw:1438841158267:dw| as it's sliding so we use the Pythagorean theorem \[x^2+y^2=r^2\]
anonymous
  • anonymous
i got that part
Astrophysics
  • Astrophysics
\[x^2+y^2=15^2 \implies x^2+y^2=225\] now differentiate \[2x \frac{ dx }{ dt }+2y \frac{ dy }{ dt } = 0\] Finish it off
Astrophysics
  • Astrophysics
`how fast is the top of the ladder falling at that instance` this tells you what you're solving for, should be clear
Astrophysics
  • Astrophysics
|dw:1438841472668:dw| just making sure you know it's a right triangle :P
anonymous
  • anonymous
i just confused myself even further. I just can't figure this out.
Astrophysics
  • Astrophysics
Ok, so given the equation what are we solving for?
anonymous
  • anonymous
dt ??
Astrophysics
  • Astrophysics
You are solving for dy/dt as we're looking for how fast the top of the ladder is falling
Astrophysics
  • Astrophysics
So solve for dy/dt first, as you would for x, don't plug anything just isolate dy/dt.
anonymous
  • anonymous
\[\frac{ dy }{dt }=\frac{ -x \frac{ dx }{ dt } }{ y }\] ??
Astrophysics
  • Astrophysics
Yeah :) good \[\frac{ dy }{ dt }= -\frac{ x }{ y }\frac{ dx }{ dt }\]
Astrophysics
  • Astrophysics
So we know x = 9, and r = 15, what is y?
anonymous
  • anonymous
12
Astrophysics
  • Astrophysics
Good
Astrophysics
  • Astrophysics
x = 9, y = 12, r = 15, and we know dx/dt = 4 ft/s \[\frac{ dy }{ dt } = - \frac{ 9 }{ 12 }(4) ft/s\]
anonymous
  • anonymous
so 3ft/s ?
Astrophysics
  • Astrophysics
- 3ft/s the negative just means the distance from the top of the ladder is decreasing
anonymous
  • anonymous
oh okay i get it now!
anonymous
  • anonymous
thank you so much
Astrophysics
  • Astrophysics
No problem

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