Respuesta :
Answer:
a) [tex]P(t)=0.1\,\sqrt{t^2+4t} +13.65[/tex]
b) P(2) = 14 units
c) from hour 4 to hour 5, an average worker can complete about 0.1 of a unit.
Step-by-step explanation:
Notice that the problem gives the derivative of the production function, and also an extra piece of information (P(2) = 14) - also called initial condition - that allows us to find the actual production with any additional constant.
Let's work on the "antiderivative" of the function:
[tex]P'(t)= \frac{0.1\,(t+2)}{\sqrt{t^2+4t} }[/tex]
Using the change of variables [tex]u=\sqrt{t^2+4t}[/tex] , [tex]du[/tex] becomes [tex]\frac{1}{2}\frac{\,2\,(t+2)}{\sqrt{t^2+4t} }[/tex]
Then, the family of antiderivatives becomes:
[tex]P(t)=0.1\,\sqrt{t^2+4t} +C[/tex]
We should be able to determine the constant C using the initial condition for the problem: [tex]P(2)=14[/tex]
Then we evaluate:
[tex]P(2)=0.1\,\sqrt{2^2+4*2} +C\\14=0.1\,\sqrt{12} +C\\C=14-0.1\,\sqrt{12} \\C=14-0.35\\C=13.65[/tex]
Then the function requested in point a) is:
[tex]P(t)=0.1\,\sqrt{t^2+4t} +13.65[/tex]
Point b) "Find the number of units an average worker can complete 2 hours after beginning work" was already given as the initial condition, so P(2) = 14 units.
Point c) asks for the number of units an average worker can complete from hour 4 to hour 5, so we calculate the difference:
[tex]P(5)-P(4)=14.32 - 14.22= 0.10\,\, units[/tex]
So the average worker can complete one tenths of a unit between hour 4 and hour 5.