Water leaks from a vertical cylindrical tank
through a small hole in its base at a rate
proportional to the square root of the volume of
water remaining. The tank initially contains 125
liters and 17 . liters leak out during the first day.
A. When will the tank be half empty?
B. How much water will remain in the tank
after 5 days? volume =
Liters

Question

Water leaks from a vertical cylindrical tank
through a small hole in its base at a rate
proportional to the square root of the volume of
water remaining. The tank initially contains 125
liters and 17 . liters leak out during the first day.
A. When will the tank be half empty?

B. How much water will remain in the tank
after 5 days? volume =
Liters

UpStudy AI · Accepted Answer

We start with the statement that the rate of leakage is proportional to the square root of the volume. This means that

$$
\frac{dV}{dt} = - k \sqrt{V},
$$

where $$ k > 0 $$ (the negative sign indicates that the volume is decreasing).

### Step 1. Separate Variables and Integrate

Separate variables:

$$
\frac{dV}{\sqrt{V}} = - k\,dt.
$$

Integrate both sides:

$$
\int \frac{dV}{\sqrt{V}} = \int -k\,dt.
$$

We have

$$
\int \frac{dV}{\sqrt{V}} = 2\sqrt{V},
$$

so that

$$
2\sqrt{V} = -k t + C,
$$

where $$ C $$ is the constant of integration.

### Step 2. Use the Initial Condition

The tank initially contains 125 liters. At $$ t=0 $$, $$ V=125 $$. Thus

$$
2\sqrt{125} = C.
$$

Since

$$
\sqrt{125} = 5\sqrt{5},
$$

we have

$$
C = 2\cdot 5\sqrt{5} = 10\sqrt{5}.
$$

So now the equation becomes

$$
2\sqrt{V} = -k t + 10\sqrt{5}.
$$

Solving for $$ \sqrt{V} $$:

$$
\sqrt{V} = -\frac{k}{2}\,t + 5\sqrt{5}.
$$

Thus,

$$
V = \Bigl(-\frac{k}{2}\,t + 5\sqrt{5}\Bigr)^2.
$$

### Step 3. Determine the Constant $$ k $$

We are told that during the first day, 17 liters leak out. Therefore, after 1 day the volume is

$$
125 - 17 = 108 \text{ liters}.
$$

Hence, at $$ t=1 $$:

$$
108 = \Bigl(-\frac{k}{2}(1) + 5\sqrt{5}\Bigr)^2.
$$

Taking square roots (and taking the positive root since volume is positive):

$$
\sqrt{108} = -\frac{k}{2} + 5\sqrt{5}.
$$

Note that

$$
\sqrt{108} = 6\sqrt{3}.
$$

Thus,

$$
6\sqrt{3} = -\frac{k}{2} + 5\sqrt{5}.
$$

Solve for $$ k $$:

$$
-\frac{k}{2} = 6\sqrt{3} - 5\sqrt{5},
$$
$$
\frac{k}{2} = 5\sqrt{5} - 6\sqrt{3},
$$
$$
k = 2\Bigl(5\sqrt{5} - 6\sqrt{3}\Bigr).
$$

### Step 4. Part A. When Will the Tank Be Half Empty?

"Half empty" means the remaining water is half of the initial 125 liters, that is, $$ \frac{125}{2} = 62.5 $$ liters. We need to solve for $$ t $$ when $$ V = 62.5 $$. Using

$$
\sqrt{V} = -\frac{k}{2}\,t + 5\sqrt{5},
$$

we have

$$
\sqrt{62.5} = -\frac{k}{2}\,t + 5\sqrt{5}.
$$

Notice that

$$
\sqrt{62.5} = \sqrt{\frac{125}{2}} = \frac{\sqrt{125}}{\sqrt{2}} = \frac{5\sqrt{5}}{\sqrt{2}}.
$$

Thus,

$$
\frac{5\sqrt{5}}{\sqrt{2}} = -\frac{k}{2}\,t + 5\sqrt{5}.
$$

Solve for $$ t $$:

1. Subtract $$ 5\sqrt{5} $$ from both sides:

$$
   \frac{5\sqrt{5}}{\sqrt{2}} - 5\sqrt{5} = -\frac{k}{2}\,t.
   $$

2. Multiply both sides by $$-\frac{2}{k}$$:

$$
   t = \frac{2}{k}\Bigl(5\sqrt{5} - \frac{5\sqrt{5}}{\sqrt{2}}\Bigr).
   $$

Factor out $$ 5\sqrt{5} $$:

$$
t = \frac{10\sqrt{5}}{k}\Bigl(1 - \frac{1}{\sqrt{2}}\Bigr).
$$

Recall that

$$
k=2\Bigl(5\sqrt{5} - 6\sqrt{3}\Bigr).
$$

Thus, the time when the tank is half empty is

$$
t = \frac{10\sqrt{5}(1-\frac{1}{\sqrt{2}})}{2(5\sqrt{5} - 6\sqrt{3})} = \frac{5\sqrt{5}(1-\frac{1}{\sqrt{2}})}{5\sqrt{5} - 6\sqrt{3}}.
$$

This is the exact answer for part A.

### Step 5. Part B. How Much Water Remains After 5 Days?

We use the expression

$$
V(t) = \Bigl(-\frac{k}{2}\,t + 5\sqrt{5}\Bigr)^2.
$$

At $$ t = 5 $$ days:

$$
V(5) = \Bigl(-\frac{k}{2}\cdot 5 + 5\sqrt{5}\Bigr)^2.
$$

First, compute

$$
-\frac{k}{2}\cdot 5 = -\frac{5k}{2}.
$$

Substitute $$ k = 2\Bigl(5\sqrt{5}-6\sqrt{3}\Bigr) $$:

$$
\frac{5k}{2} = \frac{5\cdot 2\Bigl(5\sqrt{5}-6\sqrt{3}\Bigr)}{2} = 5\Bigl(5\sqrt{5}-6\sqrt{3}\Bigr).
$$

Thus,

$$
V(5) = \Bigl(5\sqrt{5} - 5\Bigl(5\sqrt{5}-6\sqrt{3}\Bigr)\Bigr)^2.
$$

Simplify the inside:

$$
5\sqrt{5} - 5\Bigl(5\sqrt{5}-6\sqrt{3}\Bigr) = 5\sqrt{5} - 25\sqrt{5} + 30\sqrt{3} = -20\sqrt{5} + 30\sqrt{3}.
$$

Thus, the volume after 5 days is

$$
V(5) = \Bigl(30\sqrt{3} - 20\sqrt{5}\Bigr)^2.
$$

This is the exact answer for part B.

### Final Answers

A. Time when the tank is half empty:

$$
t = \frac{5\sqrt{5}\Bigl(1-\frac{1}{\sqrt{2}}\Bigr)}{5\sqrt{5} - 6\sqrt{3}} \quad \text{days}.
$$

B. Volume of water remaining after 5 days:

$$
V(5) = \Bigl(30\sqrt{3} - 20\sqrt{5}\Bigr)^2 \quad \text{liters}.
$$

For a numerical approximation:

- Compute $$5\sqrt{5} \approx 11.18$$ and $$6\sqrt{3} \approx 10.39$$. Then
  $$
  t \approx \frac{11.18(1-0.7071)}{11.18-10.39} = \frac{11.18 \cdot 0.2929}{0.79} \approx \frac{3.27}{0.79} \approx 4.14 \text{ days}.
  $$

- For $$V(5)$$, $$20\sqrt{5}\approx 44.72$$ and $$30\sqrt{3}\approx 51.96$$, so
  $$
  30\sqrt{3} - 20\sqrt{5}\approx 51.96 - 44.72 \approx 7.24,
  $$
  and
  $$
  V(5) \approx (7.24)^2 \approx 52.42 \text{ liters}.
  $$

Thus, the tank is half empty after about 4.14 days, and about 52.4 liters of water remain after 5 days.
A. The tank will be half empty after approximately 4.14 days. B. After 5 days, about 52.4 liters of water will remain in the tank.

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Solución

Step 1. Separate Variables and Integrate

Step 2. Use the Initial Condition

Step 3. Determine the Constant

Step 4. Part A. When Will the Tank Be Half Empty?

Step 5. Part B. How Much Water Remains After 5 Days?

Final Answers

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