(A) What percent of the pool is filled?

(B) At what rate is the water level rising?

- Thread starter harpazo
- Start date

(A) What percent of the pool is filled?

(B) At what rate is the water level rising?

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:

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

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:

Let me know where I first lost you, and I will try to explain further.This is really hard to grasp.

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

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