A 4,096 -liter cistern is empty when water begins flowing into it (at $$ t=0 $$ ) at a rate (in $$ \mathrm{L} / \mathrm{min} $$ ) given by $$ \mathrm{Q}^{\prime}(\mathrm{t})=12 \sqrt{t} $$, where t is measured in minutes. a. How much water flows into the cistern in 0.75 hour? b. Find the function that gives the amount of water in the tank at any time $$ t \geq 0 $$. c. When will the tank be full? a. To find the amount of water that has flowed into the cistern within the given hours, what process is needed? Choose the correct choice below. A. Find the area under the $$ Q^{\prime}(t) $$ curve between 0 and the given value. Bind the area under the $$ Q^{( }(t) $$ curve between the given value and $$ t $$. Cind the area under the $$ Q^{\prime}(t) $$ curve between 0 and the given value. D. Find the area under the $$ \mathrm{Q}^{\prime}(t) $$ curve between the given value and $$ t $$. In 0.75 hour, $$ \square $$ liters of water flows into the cistern. (Type an exact answer, using radicals as needed.)

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To find the amount of water that has flowed into the cistern within the given hours, we need to find the area under the $$ Q^{\prime}(t) $$ curve between 0 and the given value.

Given:
- The cistern is empty when water begins flowing into it at $$ t=0 $$.
- The rate at which water flows into the cistern is given by $$ Q^{\prime}(t)=12 \sqrt{t} $$, where t is measured in minutes.
- We need to find the amount of water that has flowed into the cistern within 0.75 hours.

First, we need to convert 0.75 hours to minutes:
$$ 0.75 \text{ hours} = 0.75 \times 60 \text{ minutes} = 45 \text{ minutes} $$

Now, we need to find the area under the $$ Q^{\prime}(t) $$ curve between 0 and 45 minutes to determine the amount of water that has flowed into the cistern within 0.75 hours.

The area under the curve can be calculated using the integral of $$ Q^{\prime}(t) $$ with respect to time $$ t $$ from 0 to 45 minutes.

Let's calculate the integral:
$$ \int_{0}^{45} 12 \sqrt{t} \, dt $$

Now, we can calculate the integral to find the amount of water that has flowed into the cistern within 0.75 hours.
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
  $$\int_{0}^{45} 12\sqrt{t} dt$$
- step1: Simplify:
  $$\int_{0}^{45} 12t^{\frac{1}{2}} dt$$
- step2: Evaluate the integral:
  $$\int 12t^{\frac{1}{2}} dt$$
- step3: Use properties of integrals:
  $$12\times \int t^{\frac{1}{2}} dt$$
- step4: Evaluate the integral:
  $$12\times \frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}$$
- step5: Simplify:
  $$12\times \frac{2}{3}t^{\frac{3}{2}}$$
- step6: Multiply the numbers:
  $$8t^{\frac{3}{2}}$$
- step7: Return the limits:
  $$\left(8t^{\frac{3}{2}}\right)\bigg |_{0}^{45}$$
- step8: Calculate the value:
  $$8\times 45^{\frac{3}{2}}$$
- step9: Transform the expression:
  $$8\times 135\sqrt{5}$$
- step10: Multiply:
  $$1080\sqrt{5}$$
The amount of water that has flowed into the cistern within 0.75 hours is $$1080\sqrt{5}$$ liters.

To find the function that gives the amount of water in the tank at any time $$ t \geq 0 $$, we need to integrate $$ Q^{\prime}(t) $$ with respect to time $$ t $$ from 0 to $$ t $$.

Let's calculate the integral:
$$ \int_{0}^{t} 12 \sqrt{t} \, dt $$

Now, we can calculate the integral to find the function that gives the amount of water in the tank at any time $$ t \geq 0 $$.
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
  $$\int_{0}^{t} 12\sqrt{t} dt$$
- step1: Simplify:
  $$\int_{0}^{t} 12t^{\frac{1}{2}} dt$$
- step2: Evaluate the integral:
  $$\int 12t^{\frac{1}{2}} dt$$
- step3: Use properties of integrals:
  $$12\times \int t^{\frac{1}{2}} dt$$
- step4: Evaluate the integral:
  $$12\times \frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}$$
- step5: Simplify:
  $$12\times \frac{2}{3}t^{\frac{3}{2}}$$
- step6: Multiply the numbers:
  $$8t^{\frac{3}{2}}$$
- step7: Transform the expression:
  $$8\sqrt{t^{3}}$$
- step8: Simplify the expression:
  $$8t\sqrt{t}$$
The function that gives the amount of water in the tank at any time $$ t \geq 0 $$ is $$ 8t\sqrt{t} $$ liters.

To find when the tank will be full, we need to find the time $$ t $$ at which the amount of water in the tank is equal to the capacity of the tank, which is 4096 liters.

Let's set up the equation:
$$ 8t\sqrt{t} = 4096 $$

Now, we can solve the equation to find the time when the tank will be full.
Solve the equation by following steps:
- step0: Solve for $$t$$:
  $$8t\sqrt{t}=4096$$
- step1: Find the domain:
  $$8t\sqrt{t}=4096,t\geq 0$$
- step2: Divide both sides:
  $$t\sqrt{t}=512$$
- step3: Raise both sides to the $$2$$-th power$$:$$
  $$\left(t\sqrt{t}\right)^{2}=512^{2}$$
- step4: Evaluate the power:
  $$t^{3}=512^{2}$$
- step5: Simplify the equation:
  $$\sqrt[3]{t^{3}}=\sqrt[3]{512^{2}}$$
- step6: Calculate:
  $$t=\sqrt[3]{512^{2}}$$
- step7: Calculate:
  $$t=64$$
- step8: Check if the solution is in the defined range:
  $$t=64,t\geq 0$$
- step9: Find the intersection:
  $$t=64$$
- step10: Check the solution:
  $$t=64$$
The tank will be full when $$ t = 64 $$ minutes.

Therefore, the amount of water that has flowed into the cistern within 0.75 hours is $$1080\sqrt{5}$$ liters, the function that gives the amount of water in the tank at any time $$ t \geq 0 $$ is $$ 8t\sqrt{t} $$ liters, and the tank will be full after 64 minutes.
a. To find the amount of water that has flowed into the cistern within 0.75 hours, calculate the area under the $$ Q^{\prime}(t) $$ curve between 0 and 45 minutes. b. The function that gives the amount of water in the tank at any time $$ t \geq 0 $$ is $$ 8t\sqrt{t} $$ liters. c. The tank will be full after 64 minutes. In 0.75 hours, $$1080\sqrt{5}$$ liters of water flows into the cistern.

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