Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 12 (c)

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MATH100 B December 2024

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q5  •  Q6  •  Q7  •  Q8  •  Q9 (a)  •  Q9 (b)  •  Q10  •  Q11 (a)  •  Q11 (b)  •  Q11 (c)  •  Q11 (d)  •  Q12 (a)  •  Q12 (b)  •  Q12 (c)  •  Q13  •  Q14  •  

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Question 12 (c)

Suppose a chemical is being converted to a product in tank. The amount of the chemical is $S$, which is nonnegative and changes with respect to time. If the chemical is being added to the tank at a constant nonnegative rate $R$, then the amount of the chemical in the tank satisfies the differential equation $$\frac{dS}{dt}=R-\frac{S}{2+S}.$$ Describe in words what happens to the amount of the chemical in the tank over a long period of time if $0<R<1$. Support your argument with a phase diagram.

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
First, solve for the steady state. Then determine the sign of $\frac{dS}{dt}$ for $S$ on either side of the steady state by considering very small and very large values of $S$.

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Solution

Let $0<R<1$. Solving for the steady state $0=R-\frac{S}{2+S}$, we obtain $S=\frac{2R}{1-R}$. For $S$ very close to $0$, we would have $R-\frac{S}{2+S}$ close to $R$, which is positive, so $\frac{dS}{dt}>0$ for $S<\frac{2R}{1-R}$. On the other hand, for $S$ very large, we would have $\frac{S}{2+S}$ close to $1$, so $R-\frac{S}{2+S}$ is negative, and therefore $\frac{dS}{dt}<0$. The situation is described by the following phase diagram:

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