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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