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

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