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 of q.
To find the maximum charge (q) in the electrical circuit, we need to find the maximum value of the given function q(t) = t/(t^2 + 1).
To do this, we can find the derivative of q(t) with respect to t and set it equal to zero to find the critical points. Then, we can check these critical points to determine which one corresponds to the maximum value of q.
Step 1: Find the derivative of q(t):
q'(t) = [d/dt(t)] / (t^2 + 1) - t[d/dt(t^2 + 1)] / (t^2 + 1)^2
= 1 / (t^2 + 1) - (t)(2t) / (t^2 + 1)^2
= 1 / (t^2 + 1) - 2t^2 / (t^2 + 1)^2
= (1 - 2t^2) / (t^2 + 1)^2
Step 2: Set q'(t) equal to zero and solve for t:
(1 - 2t^2) / (t^2 + 1)^2 = 0
The numerator can only be zero when 1 - 2t^2 = 0.
Solving for t^2, we get:
t^2 = 1/2
t = ±√(1/2)
t = ±(1/√2)
Step 3: Check the critical points to determine the maximum charge:
To determine if these critical points correspond to a maximum, we can use the second derivative test.
q''(t) = [d^2/dt^2(t - 2t^3)] / (t^2 + 1)^2
= (1 - 6t^2) / (t^2 + 1)^2
For t = 1/√2:
q''(1/√2) = (1 - 6(1/2)) / ((1/2) + 1)^2
= -2 / (3/2)^2
= -2 / (9/4)
= -8/9
For t = -1/√2:
q''(-1/√2) = (1 - 6(1/2)) / ((1/2) + 1)^2
= -8/9
Since the second derivative is negative for both critical points, this means that q(t) has a relative maximum at t = 1/√2 and t = -1/√2.
Step 4: Calculate the maximum charge:
We can substitute the values of t back into the original function to find the maximum charge.
For t = 1/√2:
q(1/√2) = (1/√2) / ((1/√2)^2 + 1)
= (1/√2) / (1/2 + 1)
= (1/√2) / (3/2)
= (1/√2) * (2/3)
= √2 / 3 ≈ 0.471 coulombs
For t = -1/√2:
q(-1/√2) = (-1/√2) / ((-1/√2)^2 + 1)
= (-1/√2) / (1/2 + 1)
= (-1/√2) / (3/2)
= (-1/√2) * (2/3)
= -√2 / 3 ≈ -0.471 coulombs
Therefore, the maximum charge in the electrical circuit is approximately 0.471 coulombs.
To find the maximum charge of q, we need to find the maximum value of the given function q(t) = t / (t^2 + 1).
1. First, set the derivative of q(t) equal to zero to find the critical points:
q'(t) = 0
2. Differentiate q(t) with respect to t:
q'(t) = [d/dt (t)] / (t^2 + 1) - [t * d/dt(t^2 + 1)] / (t^2 + 1)^2
= 1 / (t^2 + 1) - 2t^2 / (t^2 + 1)^2
3. Set q'(t) equal to zero and solve for t:
1 / (t^2 + 1) - 2t^2 / (t^2 + 1)^2 = 0
Multiply both sides by (t^2 + 1)^2 to eliminate the fractions:
(t^2 + 1) - 2t^2 = 0
t^2 + 1 - 2t^2 = 0
-t^2 + 1 = 0
t^2 = 1
t = ±√1
t = ±1
4. Now we have the critical points t = -1 and t = 1. To determine which is a maximum or minimum, we can use the second derivative test.
5. Differentiate q'(t) with respect to t to find the second derivative:
q''(t) = d^2/dt^2 [1 / (t^2 + 1) - 2t^2 / (t^2 + 1)^2]
= [d/dt (1)] / (t^2 + 1)^2 - [d/dt (2t^2)] / (t^2 + 1)^2
= 0 - 4t / (t^2 + 1)^2
= -4t / (t^2 + 1)^2
6. Test the second derivative at the critical points:
q''(-1) = -4(-1) / ((-1)^2 + 1)^2 = 4 / 4 = 1 (positive)
q''(1) = -4(1) / ((1)^2 + 1)^2 = -4 / 4 = -1 (negative)
7. Since q''(-1) > 0 and q''(1) < 0, we can conclude that t = -1 corresponds to a local maximum, and t = 1 corresponds to a local minimum.
8. To find the global maximum, we need to evaluate q(t) at the critical point and at the endpoints of the given interval (if any).
q(-1) = (-1) / ((-1)^2 + 1) = -1 / 2
q(1) = (1) / ((1)^2 + 1) = 1 / 2
9. Comparing q(-1), q(1), and the endpoints (if any), we can determine the maximum charge of q:
The maximum charge is q(-1) = -1 / 2 coulombs.