anonymous
  • anonymous
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schrodinger
  • schrodinger
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anonymous
  • anonymous
At what rate is the area of the triangle formed by the ladder wall and the ground changing at that moment? i already know that dy/dt is -12 cm/sec
baldymcgee6
  • baldymcgee6
Please give the whole question
anonymous
  • anonymous
A 13 ft ladder is leaning against a house when its base starts to slide away. By the time the base is 12 ft from the house, the base is moving at the rate 5 ft/sec

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baldymcgee6
  • baldymcgee6
|dw:1352178476520:dw|
anonymous
  • anonymous
yep, well i already found dy/dt and i was deriving dA/dt but it seemed wrong
anonymous
  • anonymous
i got 1/2 db/dt * h + dh/dt *b
baldymcgee6
  • baldymcgee6
|dw:1352178791974:dw|
baldymcgee6
  • baldymcgee6
|dw:1352179052193:dw|
anonymous
  • anonymous
close, the answer is =59.5....
anonymous
  • anonymous
how did you do that? did you just took the derivative inside and distribute the half sign?
baldymcgee6
  • baldymcgee6
oops o missed something while differentiating the area formula
anonymous
  • anonymous
ok but did you do product rule?
baldymcgee6
  • baldymcgee6
yes
baldymcgee6
  • baldymcgee6
but wrong.. lol
baldymcgee6
  • baldymcgee6
|dw:1352179446697:dw|
anonymous
  • anonymous
do you have to distribute?
baldymcgee6
  • baldymcgee6
no theres nothing to distribute
anonymous
  • anonymous
then why do you have 0.5 multiplied on each portion? isn't 1/2 a constant?
baldymcgee6
  • baldymcgee6
just type that into your calc... and you should get it.
baldymcgee6
  • baldymcgee6
0.5 is a constant that could be factored out.. but there is no point
anonymous
  • anonymous
i did. but if i didn't distribute, i would get the wrong answer
anonymous
  • anonymous
oh no i got it/ lol nvm thanks!
baldymcgee6
  • baldymcgee6
:)

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