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?
N= INT 1/2 (t+1)^(-1/2)
I will be happy to critique your thinking. Solve for the constant of integration, and you have it.
I really don't understand the problem for some reason, but my thinking thus far is INT(t+1)^(-1/2)
1/2 INTu^(-1/2)du
1/2((u^(1/2)/(1/2))du
u^(1/2)du
(t+1)^(1/2)
Anything you can do to help would be great!
That is it. YOu can check it by taking the derivative. DO NOT forget the constant of integration, I think you need it.
Alright, let's break it down step by step to find the solution to the problem.
Given: The rate of performance is given by dN/dt = 1/(2√(t+1)), where N is the number of units completed t hours after beginning a new task.
We are given that 2 units are completed after 3 hours. To find out how many units are completed after 8 hours, we need to integrate the given rate of performance equation with respect to time.
∫dN/dt dt = ∫1/(2√(t+1)) dt
To find the integral of 1/(2√(t+1)), let's use the substitution method.
Let u = t+1. Differentiate both sides with respect to t, we get du/dt = 1.
Rearrange the equation to solve for dt: dt = du.
Substituting √u for √(t+1) and replacing dt with du, the integral becomes:
∫1/(2√(t+1)) dt = ∫1/(2√u) du.
This simplifies to:
1/2 ∫1/√u du.
The integral of 1/√u is 2√u.
Plugging this back in, we have:
1/2 ∫1/√u du = 1/2 (2√u) + C.
Now, substitute t+1 back in for u:
1/2 (2√(t+1)) + C = √(t+1) + C.
Here, C represents the constant of integration. Since we don't have any initial conditions given, we will leave it as C for now.
Therefore, N = √(t+1) + C.
Now, to find the value of C, we will use the condition that 2 units are completed after 3 hours. Substituting t = 3 and N = 2 into the equation, we have:
2 = √(3+1) + C.
Simplifying this equation, we find:
2 = √4 + C = 2 + C.
Subtracting 2 from both sides, we get:
C = 0.
Now, we can substitute C = 0 back into the equation N = √(t+1) + C:
N = √(t+1) + 0.
Therefore, the number of units completed after 8 hours is:
N = √(8+1) = √9 = 3 units.
So, after 8 hours, 3 units will be completed.
I hope this explanation helps you understand the problem and the process to find the solution. If you have any further questions, feel free to ask!