Asked by Liz
The total charge in an electrical circuit as a function of time is given by q=t/t^2+1 coulombs. Find the maximum charge q.
Please provide your reasoning.
Please provide your reasoning.
Answers
Answered by
Elena
q = t/(t^2 +1) = t•(t^2 +1)^-1.
The derivative of this expression is
dq/dt = 1•(t^2+1)^-1 - t•(t^2 +1)^-2•2•t = 1/(t^2+1) -2•t^2/(t^2 + !)^2 =
=(t+1-2•t^2)/(t^2+1)^2 .
The maximum of this function at
(t+1-2•t^2)/(t^2+1)^2 = 0,
t+1-2•t^2= 0
2•t^2 – t -1 = 0.
Two roots: t1 =-0.5 (impossible for time), t2 = 1 s.
q = t/(t^2 +1) = 1/(1+1) = 0.5 Coulomb.
The derivative of this expression is
dq/dt = 1•(t^2+1)^-1 - t•(t^2 +1)^-2•2•t = 1/(t^2+1) -2•t^2/(t^2 + !)^2 =
=(t+1-2•t^2)/(t^2+1)^2 .
The maximum of this function at
(t+1-2•t^2)/(t^2+1)^2 = 0,
t+1-2•t^2= 0
2•t^2 – t -1 = 0.
Two roots: t1 =-0.5 (impossible for time), t2 = 1 s.
q = t/(t^2 +1) = 1/(1+1) = 0.5 Coulomb.
Answered by
Liz
Thank you!!
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.