The rate of performance is given by dN/dt=1/(2sqroot(t+1)) where N is the number of unity completed t hours after beginning a new tak.
If 2 units are completed afer 3 hours, how many units are completed after 8 hours?
see your other post under another name above.
To find the number of units completed after 8 hours, we need to solve the given differential equation dN/dt = 1/(2√(t+1)).
First, let's integrate both sides of the equation:
∫dN = ∫1/(2√(t+1)) dt
Integrating the left side of the equation is straightforward:
N = ∫1/(2√(t+1)) dt
To integrate the right side, we can make a substitution by letting u = t + 1. Then, du/dt = 1 ⇒ dt = du.
So, the equation becomes:
N = ∫1/(2√u) du
Now, we can apply the power rule of integration:
N = (1/2) * ∫u^(-1/2) du
Using the power rule, we have:
N = (1/2) * (2u^(1/2)) + C
Simplifying further:
N = u^(1/2) + C
Since we know that 2 units are completed after 3 hours, we can solve for C.
N(3) = 2
Substituting t = 3 into the equation:
2 = (√(3+1)) + C
2 = 2 + C
C = 0
Now, we have the particular solution:
N = (√(t+1))
To find the number of units completed after 8 hours, substitute t = 8 into the equation:
N(8) = (√(8+1))
N(8) = (√9)
N(8) = 3
Therefore, after 8 hours, a total of 3 units will be completed.