Problem 49 - Calculus Single Variable

Question

According to Torricelli's Law, if there is a hole of area \(a\) at the bottom of a tank, then the volume \(V(t)\) and height \(y(t)\) of water in the tank at time \(t\) are related by $$ \frac{d V}{d t}=-a \cdot \sqrt{2 g y} $$ where \(g\) is equal to gravitational acceleration at Earth's surface. (Notice that the differential equation is dimensionally correct - both sides bear units of \(L^{3} / T\) where \(L\) is length and \(t\) is time.) Use the Fundamental Theorem of Calculus together with the Chain Rule to show that $$ A(y) \frac{d y}{d t}=-a \cdot \sqrt{2 g y} $$

Step-by-Step Solution

Answer

Rates of change of volume and height are linked by the hole's area: \(A(y) \frac{d y}{d t} = -a \sqrt{2 g y}\).

Step 1: Understand the Given Information

We are given the differential equation \(\frac{d V}{d t}=-a \cdot \sqrt{2 g y}\) which describes the rate of change of volume \(V(t)\) in a tank with respect to time \(t\). Here, \(a\) is the area of the hole at the bottom of the tank, and \(g\) is the gravitational acceleration.

Step 2: Relate Volume and Height of Water

Assume the volume \(V\) of water in the tank is expressed as a function of the height \(y\): \(V = A(y)\cdot y\), where \(A(y)\) is the cross-sectional area of the water at height \(y\). This relationship indicates that \(V\) changes with \(y\).

Step 3: Apply the Chain Rule

We differentiate \(V = A(y)\cdot y\) with respect to time \(t\) using the chain rule: \(\frac{dV}{dt} = A(y)\cdot \frac{dy}{dt} + y\cdot \frac{dA}{dt}\). However, if \(A(y)\) is a function that does not explicitly depend on time \(t\), the derivative \(\frac{dA}{dt} = 0\). Thus, \(\frac{dV}{dt} = A(y)\cdot \frac{dy}{dt}\).

Step 4: Substitute and Simplify

Substitute \(\frac{dV}{dt} = A(y)\cdot \frac{dy}{dt}\) from Step 3 into the original differential equation \(\frac{d V}{d t}=-a \cdot \sqrt{2 g y}\). This gives \(A(y)\cdot \frac{dy}{dt} = -a \cdot \sqrt{2 g y}\).

Step 5: Conclusion

The equation \(A(y) \frac{d y}{d t}=-a \cdot \sqrt{2 g y}\) is obtained, showing the desired relationship between the changing height \(y\) of the water, the area of the cross-section at height \(y\), and the area of the hole at the bottom.

Key Concepts

Differential Equations in Torricelli's Law

Differential equations are vital in describing how quantities change over time. In our Torricelli's Law scenario, the differential equation \( \frac{d V}{d t}=-a \cdot \sqrt{2 g y} \) reflects how the volume \( V(t) \) of water in a tank decreases as water flows out through a hole. The left side of the equation, \( \frac{dV}{dt} \), represents the rate at which the volume changes over time. This rate of change depends on the size of the hole \( a \) and the current height of water \( y(t) \), with \( g \), the gravitational acceleration, acting as a constant force. Differential equations allow us to solve complex problems by predicting behaviors of dynamic systems, such as how fast liquid drains from a tank. When we integrate this equation, we can predict the volume of water left at any given time. This breathes life into mathematics, giving us real-world applications.

Understanding the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the world of differentiation and integration, serving as a central piece of calculus. It tells us that differentiation and integration are inverse processes. If you differentiate a function first and then integrate it, you return to the original function. In our problem, we want to link volume and height, which are related by integration and differentiation. By recognizing the change of volume \( V(t) \) over time as a derivative, we can apply integration to progress backward to the total volume or determine \( y(t) \), given changes in \( V(t) \). This concept allows us to transform the differential form into a solvable integral one, stepping through calculations with confidence. Understanding this theorem is key to unraveling the dynamic relationships in changing systems.

Application of the Chain Rule

The Chain Rule is an essential tool in calculus for finding derivatives of composite functions. It's what allows us to tackle the derivative of a function that is itself dependent on another variable. Imagine a variable that changes based on not one, but several layers of relationships—like in our exercise, where volume is indirectly affected by time through height.Using the chain rule, we rewrite the derivative of the volume \( V = A(y)\cdot y \) with respect to time \( t \). Here, both \( A(y) \) and \( y \) are functions of \( y \), which itself is a function of \( t \) (time). The chain rule helps earn \( \frac{dV}{dt} = A(y)\cdot \frac{dy}{dt} \), ensuring we account for all factors affecting the change. This method simplifies complex layered derivatives into understandable parts, essential for correctly solving our water flow problem.

Role of Gravitational Acceleration

Gravitational acceleration, represented as \( g \), is a factor inherent to the physical laws governing motion and fluid dynamics on Earth. It is approximately \( 9.8 \mathrm{m/s^2} \) and radically impacts how quickly water exits through the hole in the tank.In the equation \( \frac{dV}{dt} = -a \cdot \sqrt{2 g y} \), its presence under the square root signifies gravity's role in determining the flow rate. When water discharges, gravity accelerates each particle, accelerating the leakage as a function of height—greater height implies stronger gravitational potential, hence faster flows. Understanding gravitational acceleration's part in this equation enhances our grasp of real-world applications and forces acting within and upon our system, culminating in the accurate reflection of water flow behaviors due to gravity.