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)  •  

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[hide] 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.

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

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[show] Solution

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, $${\frac {C}{\pi R^{2}}}\leq {\frac {9C}{\pi R^{2}}}.$$