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.
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.
Thank you!!
To find the maximum charge q, we need to determine the value of t that maximizes q.
Given: q = t / (t^2 + 1)
To find the maximum of q, we can take the derivative of q with respect to t and set it equal to zero, then solve for t.
Let's find the derivative of q with respect to t:
q' = d/dt (t / (t^2 + 1))
Using the quotient rule:
q' = [(t^2 + 1)(1) - (t)(2t)] / (t^2 + 1)^2
= (t^2 + 1 - 2t^2) / (t^2 + 1)^2
= (1 - t^2) / (t^2 + 1)^2
Setting q' equal to zero:
(1 - t^2) / (t^2 + 1)^2 = 0
Since the numerator cannot be zero (1 - t^2 ≠ 0 for all t), we can conclude that the only possible solution is that the denominator equals zero:
(t^2 + 1)^2 = 0
Taking the square root of both sides:
t^2 + 1 = 0
Subtracting 1 from both sides:
t^2 = -1
Since there is no real number t that satisfies t^2 = -1, there is no critical point where the derivative is equal to zero. This means that q does not have a maximum or minimum within the given function.
Therefore, there is no maximum charge q for the given function q(t) = t / (t^2 + 1).
To find the maximum charge q in the given electrical circuit as a function of time, we need to analyze the behavior of the function q(t) = t/(t^2 + 1) and determine the value of t at which q is maximized.
First, note that the function q(t) is defined for all real values of t, except when the denominator t^2 + 1 equals zero (which is never the case since t^2 + 1 is always positive). So, we can proceed to find the maximum charge without any restrictions on the domain of t.
To find the maximum value of q(t), we can use calculus and find the critical points of the function. The critical points occur when the derivative of q with respect to t, q'(t), is equal to zero or is not defined.
Let's find the derivative of q(t) with respect to t:
q(t) = t/(t^2 + 1)
Taking the derivative using the quotient rule:
q'(t) = (1*(t^2 + 1) - t*(2t))/(t^2 + 1)^2
= (t^2 + 1 - 2t^2)/(t^2 + 1)^2
= (1 - t^2)/(t^2 + 1)^2
Now, we need to solve the equation q'(t) = 0 to find the critical points:
(1 - t^2)/(t^2 + 1)^2 = 0
To solve this equation, we set the numerator equal to zero:
1 - t^2 = 0
Simplifying gives:
t^2 = 1
Taking the square root of both sides yields two possible solutions:
t = 1 or t = -1
Now we need to check whether these values are the local maximum or minimum points. To do this, we can analyze the second derivative q''(t) of the function q(t). If q''(t) is negative at t = 1, then there is a maximum at t = 1. Similarly, if q''(t) is positive at t = -1, then there is a minimum at t = -1. If neither condition is satisfied, then there is no maximum or minimum.
Let's find the second derivative of q(t):
q''(t) = [(1 - t^2)'*(t^2 + 1)^2 - (1 - t^2)*(t^2 + 1)^2'] / (t^2 + 1)^4
= [(-2t)*(t^2 + 1)^2 - (1 - t^2)*(4t*(t^2 + 1))] / (t^2 + 1)^4
= [-2t(t^2 + 1)^2 - 4t(t^2 + 1) + 4t^3 + 4t] / (t^2 + 1)^4
= [-2t(t^2 + 1)^2 - 4t(t^2 + 1) + 4t(t^2 + 1)] / (t^2 + 1)^4
= [-2t(t^2 + 1)^2 - 4t(t^2 + 1) + 4t(t^2 + 1)] / (t^2 + 1)^4
= [-2t(t^2 + 1)^2 - 4t(t^2 + 1) + 4t(t^2 + 1)] / (t^2 + 1)^4
Substituting t = 1 into q''(t):
q''(1) = [-2(1)(1^2 + 1)^2 - 4(1)(1^2 + 1) + 4(1)(1^2 + 1)] / (1^2 + 1)^4
= [-2(1)(2^2) - 4(1)(2) + 4(1)(2)] / (2^4)
= [-8 - 8 + 8] / 16
= -8/16
= -1/2
Since the second derivative q''(t) evaluated at t = 1 (-1/2) is negative, this means q(t) has a local maximum at t = 1.
Similarly, substituting t = -1 into q''(t):
q''(-1) = [-2(-1)((-1)^2 + 1)^2 - 4(-1)((-1)^2 + 1) + 4(-1)((-1)^2 + 1)] / ((-1)^2 + 1)^4
= [-2(-1)(2^2) - 4(-1)(2) + 4(-1)(2)] / (2^4)
= [8 - 8 - 8] / 16
= -8/16
= -1/2
Since the second derivative q''(t) evaluated at t = -1 (-1/2) is also negative, this means q(t) also has a local maximum at t = -1.
Now that we have identified the critical points t = 1 and t = -1 as the local maximums, we can analyze the behavior of q(t) as t approaches positive and negative infinity to determine if there is a global maximum.
As t approaches positive infinity, the denominator t^2 + 1 becomes dominant, and q(t) approaches zero. Therefore, q(t) does not have a global maximum as t goes to positive infinity.
As t approaches negative infinity, the denominator t^2 + 1 becomes dominant again, and q(t) approaches zero. Thus, q(t) does not have a global maximum as t goes to negative infinity.
In conclusion, the maximum charge q in the electrical circuit is achieved at the local maximum points t = 1 and t = -1. Substituting these values into the expression for q(t):
q(1) = 1/(1^2 + 1) = 1/2 coulombs
q(-1) = -1/((-1)^2 + 1) = -1/2 coulombs
Therefore, the maximum charge q in the electrical circuit is 1/2 coulombs.