Science:Math Exam Resources/Courses/MATH 180/December 2017/Question 08 (b)

MATH 180 December 2017

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

   Other MATH 180 Exams

  •  December 2017  •  

Question 08 (b)

Consider two tanks of water, one above the other, as pictured below. The top tank, which is initially full, is a cylinder of radius ${\textstyle R}$ and height ${\textstyle H}$. The bottom tank, which is initially empty, is a cone of radius ${\textstyle R}$ and identical volume as the top tank. Suppose water drains at a constant rate ${\textstyle C}$ from the top tank to the bottom tank. (b) At the time specified in part (a), is the rate of increase of water depth in the bottom tank equal to, greater than, or less than the rate of decrease of water depth in the top tank? Determine which of these three options is correct, and justify your answer.

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
Find the rate of decrease of water depth in the top tank, using the constant change rate of the volume of the water.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. Let $l$ be the water depth in the top tank at time $t$. Then, the volume of the water in the top tank at $t$ is given by $\pi R^{2}l$ (the volume of the cylinder with radius $R$ and height $l$). According to the question, the volume of water in the top tank is decreasing with the constant rate $C$. In other words, the rate of decrease of the volume of water in the top tank $-{\frac {d(\pi R^{2}l)}{dt}}$ is equal to $C$. Here, we put the minus sign to describe the rate of decrease. Then, we have $$C=-{\frac {d(\pi R^{2}l)}{dt}}=-\pi R^{2}{\frac {dl}{dt}}\implies -{\frac {dl}{dt}}={\frac {C}{\pi R^{2}}}.$$ The second equality follows from that $\pi$ and $R$ are constants and not depending on $t$. Also, since $l$ is the water depth in the top tank at time $t$, $-{\frac {dl}{dt}}$ can be interpreted as the rate of decrease of water depth in the top tank. The minus sign again represents the rate of decrease. Observe that $-{\frac {dl}{dt}}$ is the constant ${\frac {C}{\pi R^{2}}}.$ Compared with the rate of increase of water depth in the bottom tank at the time specified in part (a), which is ${\frac {9C}{\pi R^{2}}}$ (see part (a)), the rate of decrease of water depth in the top tank, ${\frac {C}{\pi R^{2}}}$, is $\color {blue}{\mbox{less than}}$ that. In other words, $${\frac {C}{\pi R^{2}}}\leq {\frac {9C}{\pi R^{2}}}.$$