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

MATH102 December 2015

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Question 19 (b)

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 $E'(t)=4-t+{\frac {1}{16}}t^{2}$ (b) What is the total number of errors, $E(t)$, made $t$ hours into a shift?

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
To find $E(t)$ , first determine $E(0)$. Then, use the Fundamental Theorem of Calculus to find $E(t)$ from $E'(t)$.

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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. It is impossible for the resident to make any errors at time $t=0$, so $E(0)=0$. Next, recall that the fundamental theorem of calculus states that $f(v)-f(u)=\int _{u}^{v}f'(x)dx$ where $f'(x)$ is the derivative of $f(x)$. Now, calculate $E(s)$ for any $s>0$ as follows. Let $s>0$. Then in the above equation, set $f=E$, $u=0$, $v=s$, and $x=t$. By the fundamental theorem of calculus $E(s)-E(0)=\int _{0}^{s}E'(t)dt$ As $E(0)=0$, it follows that $E(s)=\int _{0}^{s}E'(t)dt=\int _{0}^{s}4-t+{\frac {1}{16}}t^{2}$ To calculate the above definite integral, we calculate the following indefinite integral (below, recall that 48 is 16 times 3). $\int 4-t+{\frac {1}{16}}t^{2}dt=\int 4dt-\int tdt+\int {\frac {1}{16}}t^{2}dt=4t-{\frac {1}{2}}t^{2}+{\frac {1}{48}}t^{3}$ Now compute $\int _{0}^{s}4-t+{\frac {1}{16}}t^{2}=(4s-{\frac {1}{2}}s^{2}+{\frac {1}{48}}s^{3})-(4*0-{\frac {1}{2}}0^{2}+{\frac {1}{48}}0^{3})$ Hence $E(s)=4s-{\frac {1}{2}}s^{2}+{\frac {1}{48}}s^{3}$. By simply changing notation (write t instead of s) we can re-write our solution as $\color {blue}E(t)=4t-{\frac {1}{2}}t^{2}+{\frac {1}{48}}t^{3}$ which is the answer to the question.