The rate at which a machine operator's efficiency, $E$ (expressed as a percentage), changes with respect to time $t$ is given by $\frac{d E}{d t}=30-8 t$, where $t$ is the number of hours the operator has been at work.
a) Find $E(t)$, given that the operator's efficiency after working $2 \mathrm{hr}$ is $77 \%$; that is, $E(2)=77$.
b) Use the answer in part (a) to find the operator's efficiency after $3 \mathrm{hr}$, after $5 \mathrm{hr}$.
a) $E(t)=$

Step 1 ：The given equation is a differential equation. To find \(E(t)\), we need to integrate the given equation with respect to \(t\).

Step 2 ：Integrating \(\frac{d E}{d t}=30-8 t\) with respect to \(t\), we get \(E(t) = -4t^2 + 30t + C\), where \(C\) is the constant of integration.

Step 3 ：We can use the initial condition \(E(2) = 77\) to find the constant of integration.

Step 4 ：Substituting \(t = 2\) and \(E(2) = 77\) into the equation, we get \(77 = -4(2)^2 + 30(2) + C\). Solving for \(C\), we find that \(C = 33\).

Step 5 ：Substituting \(C = 33\) back into the equation, we get \(E(t) = -4t^2 + 30t + 33\).

Step 6 ：\(\boxed{E(t) = -4t^2 + 30t + 33}\) is the operator's efficiency as a function of time.