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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
Please give the whole question
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
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
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?
but wrong.. lol
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!