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anonymous
 3 years ago
a right angled triangle with length of the edges of the right angle "20" cm and"15"cm ;the first decrease by rate 2cm/sec . the second increase by 3cm/sec. find at any moment the rate of the change of the area and when the change stop
anonymous
 3 years ago
a right angled triangle with length of the edges of the right angle "20" cm and"15"cm ;the first decrease by rate 2cm/sec . the second increase by 3cm/sec. find at any moment the rate of the change of the area and when the change stop

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{dy}{dt} = 2 cm/\sec\] \[\frac{dx}{dt} = 3 cm/\sec\] \[determine \frac{dA}{dt}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry wrote that wrong, should be 2cm/sec

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i need the area value at any instant in term of "t"

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0A(t) = (1/2)(202t)(15+3t)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{d}{dt}A(t)= \frac{d}{dt}\frac{1}{2}*(202t)(15+3t)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that right ; this problem was at my test and i want to check the second part "wiil the variation stop or not?'

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it has to at t=10, because after this point, the area would be negative and negative areas dont exist (sort of)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i mean you might see negative areas when doing derivatives, but for this problem, i dont think negative areas count

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but area at any instant =0.5(202t)(15+3t) ;then area=\[3t ^{2}+15t+150\] then ;the rate of change of area at any instant=6t+15; at rate stops 6t+15=0;then the variation stops at (15/6) not at 10 ;if i have any mistake tell me please

y2o2
 3 years ago
Best ResponseYou've already chosen the best response.0@completeidiot He asked when will the variation stop, so the idea is about letting the derivative(dA/dt) = 0 , not the the area [A(t)]= 0
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