4. A tank contains 125 gallons of heating oil at $$ t=0 $$. During the interval $$ 0 \leq t \leq 12 $$ hours, heating oil is pumped into the tank at the rate $$ H(t)=2+\frac{10}{(1+\ln (t+1))} $$ gallons per hour. During the same time interval, heating oil is removed from the tank at the rate of $$ R(t)=12 \sin \left(\frac{t^{2}}{47}\right) $$ gallons per hour. Use your calculator to evaluate when necessary. a. How many gallons of heating oil are pumped into the tank during the time interval $$ 0 \leq t \leq 12 $$ hours? Show work. b. Is the level of heating oil in the tank rising or falling at $$ t=6 $$ hours? Show work. c. How many gallons of heating oil are in the tank at time $$ t=12 $$ hours? Show work. d. What is the average rate that oil is pumped out of the tank during the time interval $$ 0 \leq t \leq 12 $$. Show work.

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UpStudy AI · Accepted Answer

To determine the **average rate** at which oil is pumped **out** of the tank during the time interval $$ 0 \leq t \leq 12 $$ hours, we will analyze the outflow rate function $$ R(t) $$.

### Given:
$$
R(t) = 12 \sin\left(\frac{t^2}{47}\right) \quad \text{gallons per hour}
$$

### Objective:
Find the **average rate** of $$ R(t) $$ over the interval $$ [0, 12] $$.

### Approach:
The **average value** of a function $$ f(t) $$ over the interval $$ [a, b] $$ is given by:
$$
\text{Average value} = \frac{1}{b - a} \int_{a}^{b} f(t) \, dt
$$

Applying this to $$ R(t) $$:
$$
\text{Average rate} = \frac{1}{12 - 0} \int_{0}^{12} 12 \sin\left(\frac{t^2}{47}\right) \, dt = \int_{0}^{12} \sin\left(\frac{t^2}{47}\right) \, dt
$$

### Numerical Integration:
The integral $$ \int_{0}^{12} \sin\left(\frac{t^2}{47}\right) \, dt $$ does not have an elementary antiderivative, so we'll approximate it using **Simpson's Rule** with $$ n = 8 $$ subintervals for better accuracy.

1. **Divide the interval** $$ [0, 12] $$ into $$ n = 8 $$ subintervals, each of width $$ h = 1.5 $$ hours.

2. **Compute sample points** and evaluate $$ f(t) = \sin\left(\frac{t^2}{47}\right) $$ at these points:
    $$
    \begin{array}{c|c}
    t & f(t) = \sin\left(\frac{t^2}{47}\right) \
    \hline
    0 & 0 \
    1.5 & \sin\left(\frac{2.25}{47}\right) \approx 0.0479 \
    3 & \sin\left(\frac{9}{47}\right) \approx 0.1910 \
    4.5 & \sin\left(\frac{20.25}{47}\right) \approx 0.4179 \
    6 & \sin\left(\frac{36}{47}\right) \approx 0.6952 \
    7.5 & \sin\left(\frac{56.25}{47}\right) \approx 0.9283 \
    9 & \sin\left(\frac{81}{47}\right) \approx 0.9882 \
    10.5 & \sin\left(\frac{110.25}{47}\right) \approx 0.7081 \
    12 & \sin\left(\frac{144}{47}\right) \approx 0.0773 \
    \end{array}
    $$

3. **Apply Simpson's Rule**:
    $$
    \int_{0}^{12} \sin\left(\frac{t^2}{47}\right) dt \approx \frac{h}{3} \left[ f_0 + 4(f_1 + f_3 + f_5 + f_7) + 2(f_2 + f_4 + f_6) + f_8 \right]
    $$
    Plugging in the values:
    $$
    \approx \frac{1.5}{3} \left[ 0 + 4(0.0479 + 0.4179 + 0.9283 + 0.7081) + 2(0.1910 + 0.6952 + 0.9882) + 0.0773 \right]
    $$
    $$
    \approx 0.5 \times [0 + 4(2.1012) + 2(1.8744) + 0.0773] = 0.5 \times [0 + 8.4048 + 3.7488 + 0.0773] = 0.5 \times 12.2309 = 6.1154
    $$

### Conclusion:
The **average rate** at which oil is pumped out of the tank over the 12-hour period is approximately **6.12 gallons per hour**.

**Final Answer:**

Problem d Answer:

The average rate oil is pumped out from time 0 to 12 hours is approximately 6.12 gallons per hour.
The average rate oil is pumped out from time 0 to 12 hours is approximately 6.12 gallons per hour.

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