Practice Related Rates: Swimming Pool

harpazo

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A swimming pool is 12 meters long, 6 meters wide, and 1 meter deep at the shallow end, and 3 meters deep at the deep end. Water is being pumped into the pool at 1/4 cubic meter per minute, and there is 1 meter of water at the deep end.

(A) What percent of the pool is filled?
(B) At what rate is the water level rising?
 

MarkFL

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All measures of length will be in meters, and measures of time will be in minutes.

Consider the following diagram, a cross-section of the pool:

tml_pool.png


Now, to find the entire volume of the pool, we may consider that the cross section of the pool is a trapezoid and write:

\(\displaystyle V_P=\frac{12}{2}(3+1)6=144\)

To find the volume of water initially present, we may consider the cross section is a triangle:

\(\displaystyle V_W=\frac{1}{2}(1)(6)(6)=18\)

(A) And so the percentage of the pool that is initially filled is:

\(\displaystyle P=100\frac{V_W}{V_P}\%=\frac{25}{2}\%\)

(B) Let \(h\) be in depth of the water at time \(t\), where \(h_0=1\). Now, we need to find the volume of water present, as a function of \(h_0\le h\le2\), which means we need to know the altitude \(a\) of the triangular cross section. By similarity, we may state:

\(\displaystyle \frac{6}{1}=\frac{a}{h}\implies a=6h\)

And so the volume is:

\(\displaystyle V=\frac{1}{2}h(6h)(6)=18h^2\)

Implicitly differentiating w.r.t \(t\), we find:

\(\displaystyle \d{V}{t}=36h\d{h}{t}\)

We are told:

\(\displaystyle \d{V}{t}=\frac{1}{4}\)

And so we have:

\(\displaystyle \frac{1}{4}=36h\d{h}{t}\)

Thus:

\(\displaystyle \d{h}{t}=\frac{1}{144h}\)

Now, we need to consider the simpler case where \(2<h\le3\)

Now, the volume is given by:

\(\displaystyle V=72+12(h-2)(6)=72(h-1)\)

And so:

\(\displaystyle \d{V}{t}=72\d{h}{t}\)

Thus:

\(\displaystyle \d{h}{t}=\frac{1}{288}\)

And so, in conclusion, we have:

\(\displaystyle \d{h}{t}=\begin{cases}\dfrac{1}{36h}, & 1\le h\le2 \\[3pt] \dfrac{1}{288}, & 2<h\le3 \\ \end{cases}\)
 
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harpazo

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This is really hard to grasp.
 

MarkFL

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harpazo

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1. Where in the question do you see (1/2)(6)(6) as the volume of the water?

2. Explain part (B) in an easier way, if possible, especially the inequalities.
 

MarkFL

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1. Where in the question do you see (1/2)(6)(6) as the volume of the water?
The volume of the water is the product of the area of the blue shaded triangle and 6 ft (width of pool).

2. Explain part (B) in an easier way, if possible, especially the inequalities.
Part B is the meat of the problem, to say "explain it in an easier way" basically means to redo the entire problem. Please be more specific about what you don't understand.
 

harpazo

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1. Explain the inequalities.

2. Explain the answer expressed as a piecewise-defined function.
 

MarkFL

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The inequalities, if I am interpreting your query correctly, are necessary because of the geometry of the pool. The bottom is slanted, and so the size of a horizontal slice varies as we move up. If we label a vertical axis as \(h\) and orient this axis such that the origin is at the bottom of the deep end, then as we move up, the bottom of the pool is slanted from \(0\le h\le2\), which means the length of a horizontal slice of the pool is determined by where that slice meets the slanted bottom. And for \(2<h\le3\) the size of a horizontal slice remains fixed.
 
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harpazo

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Thanks.