# PracticeRelated Rates: Swimming Pool

#### harpazo

##### Pure Mathematics
Banned
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

##### La Villa Strangiato
Staff member
Moderator
Math Helper
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: 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:

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

Last edited:
• anemone and harpazo

#### harpazo

##### Pure Mathematics
Banned
This is really hard to grasp.

#### MarkFL

##### La Villa Strangiato
Staff member
Moderator
Math Helper
This is really hard to grasp.
Let me know where I first lost you, and I will try to explain further.

• anemone and harpazo

#### harpazo

##### Pure Mathematics
Banned
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

##### La Villa Strangiato
Staff member
Moderator
Math Helper
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

##### Pure Mathematics
Banned
1. Explain the inequalities.

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

#### MarkFL

##### La Villa Strangiato
Staff member
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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