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.