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in the expression y = 63t e^(-64t) the variation of y and time (t) gives a critically damped type of oscillation. In the question i have already differentiated it to dy/dt = -62e^(-62t)63t and need to find the maximum value of y and at which time it occurs then draw a graph showing the variation of y and t with t ranging from 0 to 1/32. thanks.
If y=63te^(-64t) then dy/dt=y'(t)=63e^(-64t)-(63*64)te^(-64t).
y'(t)=0 -> 63e^(-64t)-(63*64)te^(-64t)=0 ->
63e^(-64t)=(63*64)te^(-64t) -> 1=64t -> t=1/64. To find if this is
a max or min we will derive 2nd time :
y''(t)=(-64*63)e^[-64t]-(63*64)e^[-64t]+(64*63*64)te^[-64t].
y''(t)=-8064e^[-64t]+258048te^[-64t].
y''(t)=8064 { -1+32t } e^[-64t] .
y''(1/64)=8064 { -1/2 } [1/e] < 0 So t=1/64 is a Maximum.
Alon.
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Alon Mandes
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Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems. Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .
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