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A 13,824 -liter cistern is empty when water begins flowing into it (at \( t=0 \) ) at a rate (in \( \mathrm{L} / \mathrm{min} \) ) given by \( Q^{\prime}(t)=12 \sqrt{t} \), where \( t \) is measured in minutes. a. How much water flows into the cistern in 1.75 hours? 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?

Ask by Davies Carter. in the United States
Mar 19,2025

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**(a) Water Flow in 1.75 Hours** Convert 1.75 hours to minutes: \[ 1.75 \times 60 = 105 \text{ minutes} \] Calculate the total water flow: \[ V = 8 \times 105^{3/2} \approx 8610 \text{ liters} \] **(b) Water in the Tank at Any Time \( t \geq 0 \)** The amount of water \( W(t) \) is: \[ W(t) = 8t^{3/2} \quad \text{liters} \] **(c) Time to Fill the Tank** The tank will be full when: \[ 8t^{3/2} = 13,\!824 \implies t = 144 \text{ minutes} \text{ or } 2.4 \text{ hours} \]

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To solve your questions regarding the cistern and the water flow, let's break it down step by step: ### a. How much water flows into the cistern in 1.75 hours? To find out how much water flows into the cistern over 1.75 hours, we first need to convert hours to minutes: \[ t = 1.75 \text{ hours} = 1.75 \times 60 \text{ minutes} = 105 \text{ minutes} \] Next, we need to compute the total amount of water entering the tank from \( t = 0 \) to \( t = 105 \) minutes. This involves integrating the flow rate \( Q'(t)=12\sqrt{t} \): \[ Q(t) = \int_0^{105} 12\sqrt{t} \, dt \] Calculating the integral: \[ Q(t) = 12 \cdot \frac{2}{3} t^{3/2} \Big|_0^{105} = 8 t^{3/2} \Big|_0^{105} = 8 \left(105^{3/2} - 0\right) \] \[ = 8 \cdot 105 \cdot \sqrt{105} \approx 8 \cdot 105 \cdot 10.24695 \approx 8618.4 \text{ liters} \] So, approximately **8618.4 liters** of water flow into the cistern in 1.75 hours. ### b. Find the function that gives the amount of water in the tank at any time \( t \geq 0 \). To develop the function that represents the amount of water in the tank as a function of time \( t \), we continue from the integral we performed: \[ Q(t) = \int_0^t 12\sqrt{u} \, du \] Calculating \( Q(t) \): \[ Q(t) = 12 \cdot \frac{2}{3}u^{3/2} \Big|_0^t = 8u^{3/2} \Big|_0^t = 8t^{3/2} \] Thus, the function giving the amount of water in the tank at any time \( t \) is: \[ W(t) = 8t^{3/2} \] ### c. When will the tank be full? To find out when the tank will be full, set \( W(t) = 13824 \): \[ 8t^{3/2} = 13824 \] Dividing both sides by 8: \[ t^{3/2} = 1728 \] Taking both sides to the power of \( \frac{2}{3} \): \[ t = 1728^{\frac{2}{3}} = 144 \] Thus, the tank will be full at **144 minutes**, which is **2 hours and 24 minutes** after the water begins to flow. Feel free to ask any clarifying questions or further details!

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