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Chapter 4: Problem 68

The rate at which a machine operator's efficiency, \(E\) (expressed as a percentage), changes with respect to time \(t\) is given by \(\frac{d E}{d t}=40-10 t\) 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 hr is \(72 \% ;\) that is, \(E(2)=72\) b) Use the answer to part (a) to find the operator's efficiency after \(4 \mathrm{hr}\); after \(8 \mathrm{hr}\).

Short Answer

Expert verified
The operator's efficiency after 4 hours is 92%, and after 8 hours is 12%.

Step by step solution

01

Integrate the differential equation

Given the differential equation \frac{dE}{dt} = 40 - 10t\, integrate both sides with respect to \t\: \[ \int \frac{dE}{dt} dt = \int (40 - 10t) dt \] This simplifies to \[ E(t) = 40t - 5t^2 + C \]
02

Find the constant of integration, C

Use the initial condition given, \E(2) = 72\, to find the constant \C\. Substitute \t = 2\ and \E(2) = 72\ into the integrated equation: \[ 72 = 40(2) - 5(2)^2 + C \] Simplify to find \C\: \[ 72 = 80 - 20 + C \] \[ 72 = 60 + C \] \[ C = 12 \]
03

Write the particular solution for E(t)

Substitute \C\ back into the integrated equation: \[ E(t) = 40t - 5t^2 + 12 \]
04

Calculate the efficiency after 4 hours

Substitute \t = 4\ into the equation \E(t)\: \[ E(4) = 40(4) - 5(4)^2 + 12 \] Simplify: \[ E(4) = 160 - 80 + 12 \] \[ E(4) = 92 \]
05

Calculate the efficiency after 8 hours

Substitute \t = 8\ into the equation \E(t)\: \[ E(8) = 40(8) - 5(8)^2 + 12 \] Simplify: \[ E(8) = 320 - 320 + 12 \] \[ E(8) = 12 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate of Change
The rate of change is a fundamental concept in calculus. It measures how a quantity changes over time.
In this exercise, we are given the rate of change of efficiency with respect to time, represented as \(\frac{dE}{dt} = 40 - 10t\).
This tells us that as time progresses, the efficiency changes. Here,

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