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- 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!

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- anonymous

- jamiebookeater

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- Astrophysics

|dw:1438840509295:dw|

- Astrophysics

Do you agree with the diagram

- anonymous

i think so... i suck at calculus.

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- Astrophysics

|dw:1438841158267:dw| as it's sliding
so we use the Pythagorean theorem \[x^2+y^2=r^2\]

- anonymous

i got that part

- 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

`how fast is the top of the ladder falling at that instance` this tells you what you're solving for, should be clear

- Astrophysics

|dw:1438841472668:dw| just making sure you know it's a right triangle :P

- anonymous

i just confused myself even further. I just can't figure this out.

- Astrophysics

Ok, so given the equation what are we solving for?

- anonymous

dt ??

- Astrophysics

You are solving for dy/dt as we're looking for how fast the top of the ladder is falling

- Astrophysics

So solve for dy/dt first, as you would for x, don't plug anything just isolate dy/dt.

- anonymous

\[\frac{ dy }{dt }=\frac{ -x \frac{ dx }{ dt } }{ y }\] ??

- Astrophysics

Yeah :) good \[\frac{ dy }{ dt }= -\frac{ x }{ y }\frac{ dx }{ dt }\]

- Astrophysics

So we know x = 9, and r = 15, what is y?

- anonymous

12

- Astrophysics

Good

- 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

so 3ft/s ?

- Astrophysics

- 3ft/s the negative just means the distance from the top of the ladder is decreasing

- anonymous

oh okay i get it now!

- anonymous

thank you so much

- Astrophysics

No problem

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