Draining a tank (Torricelli’s law) A cylindrical tank with a cross-sectional area of 101010 m2 is filled to a depth of 252525 m with water. At t=0t=0t=0 s, a drain in the bottom of the tank with an area of 111 m² is opened, allowing water to flow out of the tank. The depth of water in the tank (in meters) at time t≥0t\geq{0}t≥0 is d(t)=(5−0.22t)2d\left(t\right)=\left(5-0.22t\$$right)^2d(t)=(5$$−0.22t)2.b. At what time is the tank empty?

Ch. 1 - Functions

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 1 - Functions

Problem 32b

Chapter 1, Problem 32b

Draining a tank (Torricelli’s law) A cylindrical tank with a cross-sectional area of $10$ m² is filled to a depth of $25$ m with water. At $t=0$ s, a drain in the bottom of the tank with an area of $1$ m² is opened, allowing water to flow out of the tank. The depth of water in the tank (in meters) at time $t$\geq{0}$$ is $d$\left$(t$\right$)=$\left$(5-0.22t$\right$)^2$ .
b. At what time is the tank empty?

Verified step by step guidance

First, understand that the function d(t) = (5 - 0.22t)^2 represents the depth of water in the tank at time t. We need to find the time t when the tank is empty, which means the depth d(t) becomes 0.

Set the equation for the depth of water to zero: (5 - 0.22t)^2 = 0. This equation will help us find the time when the tank is empty.

To solve (5 - 0.22t)^2 = 0, take the square root of both sides to simplify the equation. This gives us 5 - 0.22t = 0.

Solve the linear equation 5 - 0.22t = 0 for t. This involves isolating t by first subtracting 5 from both sides, resulting in -0.22t = -5.

Finally, divide both sides of the equation by -0.22 to solve for t. This will give you the time at which the tank is empty.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Torricelli's Law

Torricelli's Law states that the speed of fluid flowing out of an orifice under the force of gravity is proportional to the square root of the height of the fluid above the opening. This principle is crucial for understanding how the depth of water in the tank decreases over time as it drains. The law can be mathematically expressed as v = √(2gh), where v is the exit speed, g is the acceleration due to gravity, and h is the height of the fluid.

Differential Equations

Differential equations are mathematical equations that relate a function to its derivatives. In the context of this problem, they are used to model the rate of change of the water depth in the tank over time. By setting up a differential equation based on the flow rate out of the tank, one can solve for the depth of water as a function of time, which is essential for determining when the tank will be empty.

Integration

Integration is a fundamental concept in calculus that involves finding the accumulated area under a curve, which can represent quantities like distance, area, or volume. In this problem, integration is used to solve the differential equation derived from Torricelli's Law, allowing us to calculate the total change in water depth over time. This process is key to determining the exact time when the tank will be completely drained.

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Draining a tank (Torricelli’s law) A cylindrical tank with a cross-sectional area of $10$ m² is filled to a depth of $25$ m with water. At $t=0$ s, a drain in the bottom of the tank with an area of $1$ m² is opened, allowing water to flow out of the tank. The depth of water in the tank (in meters) at time $t$\geq{0}$$ is $d$\left$(t$\right$)=$\left$(5-0.22t$\right$)^2$ .

a. Check that $d$\left$(0$\right$)=25$ , as specified.

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Textbook Question

Draining a tank (Torricelli’s law) A cylindrical tank with a cross-sectional area of $10$ m² is filled to a depth of $25$ m with water. At $t=0$ s, a drain in the bottom of the tank with an area of $1$ m² is opened, allowing water to flow out of the tank. The depth of water in the tank (in meters) at time $t$\geq{0}$$ is $d$\left$(t$\right$)=$\left$(5-0.22t$\right$)^2$ .

c. What is an appropriate domain for $d$ ?

319

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