Science:Math Exam Resources/Courses/MATH102/December 2015/Question 19 (c)

MATH102 December 2015

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Question 19 (c)

In deciding how long a resident’s shift in the emergency room should be, the Chief of Staff at Vancouver General Hospital would like to minimize the average rate at which errors are made. Let $E(t)$ be the number of errors made by a resident from the start of a shift until t hours into the shift. The instantaneous rate of change of errors made is $E'(t)=4-t+{\frac {1}{16}}t^{2}$ (c) What is the average rate of change of $E(t)$ from the start of a shift ($t=0$) up until time $t$?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Recall the definition of average rate of change. Then use this definition with part b) to calculate the average rate of change for this problem.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The average rate of change from $x_{1}$ to $x_{2}$, where $x_{1}<x_{2}$, is ${\frac {E(x_{2})-E(x_{1})}{x_{2}-x_{1}}}$ Setting $x_{1}=0$ and $x_{2}=t$, we obtain ${\frac {E(t)-E(0)}{t-0}}$ Recall from part b) that $E(t)=4t-{\frac {1}{2}}t^{2}+{\frac {1}{48}}t^{3}$ and that $E(0)=0$ Using this formula, we find that the average rate of change is ${\frac {E(t)-E(0)}{t-0}}={\frac {4t-{\frac {1}{2}}t^{2}+{\frac {1}{48}}t^{3}}{t}}=\color {blue}4-{\frac {1}{2}}t+{\frac {1}{48}}t^{2}$