Answer:
[tex]\frac{dP}{dt} (1) = -4.02\,\frac{\%}{day}[/tex], [tex]\frac{dP}{dt} (10) = -0.25\,\frac{\%}{day}[/tex]
Step-by-step explanation:
The complete statement is:
The percent [tex]p[/tex] of defective parts produced by a new employee [tex]t[/tex] days after the employee starts work can be modeled by:
[tex]p(t) = \frac{t+1810}{50\cdot (t+2)}[/tex]
Find the rates of change of [tex]p[/tex] at [tex]t = 1[/tex] and [tex]t = 10[/tex] (Round your answer to two decimal places.)
The rate of change of the function is the first derivative of the function in time. By rules of differentiation, we obtain the expression of the rate of change of the percent of defective parts:
[tex]\frac{dP}{dt} = \frac{50\cdot (t+2)-50\cdot (t+1810)}{50^{2}\cdot (t+2)^{2}}[/tex]
[tex]\frac{dP}{dt} = \frac{1}{50\cdot (t+2)}-\frac{t+1810}{50\cdot (t+2)^{2}}[/tex] (1)
The rates of change of P at [tex]t = 1[/tex] and [tex]t = 10[/tex] are, respectively:
[tex]\frac{dP}{dt} (1) = -4.02\,\frac{\%}{day}[/tex]
[tex]\frac{dP}{dt} (10) = -0.25\,\frac{\%}{day}[/tex]
