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hamidic
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
\[\frac{dy}{dt} = 2 cm/\sec\] \[\frac{dx}{dt} = 3 cm/\sec\] \[determine \frac{dA}{dt}\]
sorry wrote that wrong, should be -2cm/sec
i need the area value at any instant in term of "t"
A(t) = (1/2)(20-2t)(15+3t)
\[\frac{d}{dt}A(t)= \frac{d}{dt}\frac{1}{2}*(20-2t)(15+3t)\]
that right ; this problem was at my test and i want to check the second part "wiil the variation stop or not?'
it has to at t=10, because after this point, the area would be negative and negative areas dont exist (sort of)
i mean you might see negative areas when doing derivatives, but for this problem, i dont think negative areas count
but area at any instant =0.5(20-2t)(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
@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