anonymous
  • anonymous
A 10-meter ladder is leaning against the wall of a building. The base of the ladder begins sliding away from the building at a rate of 3 meters per second. How fast is the top of the ladder sliding down the wall when the base of the ladder is 6 meters from the wall?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Astrophysics
  • Astrophysics
Related ratesss
anonymous
  • anonymous
@Astrophysics all yourn
Astrophysics
  • Astrophysics
Haha ok I got this

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More answers

anonymous
  • anonymous
the ladder slides down the wall ...
anonymous
  • anonymous
i would wait to see if @reeses is here before starting
Astrophysics
  • Astrophysics
|dw:1450152840804:dw| we need to find dy/dt
Astrophysics
  • Astrophysics
Notice a right triangle will form |dw:1450153093699:dw|
Astrophysics
  • Astrophysics
\[x^2+y^2=100\] then right?
anonymous
  • anonymous
yeah i actually got this one! it took awhile i was posted to see other methods possibly shorter ones! since these are my weaker points and prefer not to waste too much time on them during a test
Astrophysics
  • Astrophysics
Then we simply apply the chain rule
anonymous
  • anonymous
xdx/dt +ydy/dt = 0
Astrophysics
  • Astrophysics
\[2x \frac{ dx }{ dt }+2y \frac{ dy }{ dt }=0\]
Astrophysics
  • Astrophysics
Then solve for dy/dt :)
Astrophysics
  • Astrophysics
Note you will get a negative number since it's going down so don't think you're wrong!
anonymous
  • anonymous
Alright chill! i got -9/4 meters per second
Astrophysics
  • Astrophysics
What's x and y?
Astrophysics
  • Astrophysics
And what did you get when you solved for dy/dx
anonymous
  • anonymous
okay so x=6 y=8 dx/dt = 3 dy/dt = -9/4
Astrophysics
  • Astrophysics
What's the expression for dy/dt
anonymous
  • anonymous
dy/dt=-((x)(dx/dt))/y
Astrophysics
  • Astrophysics
Right, so \[\frac{ dy }{ dt } = - \frac{ x }{ y }\frac{ dx }{ dt }\]
anonymous
  • anonymous
Yup! and im pretty sure you've answered almost every question i posted here!
Astrophysics
  • Astrophysics
You should be good then :)

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