Problem 62 - Calculus with Concepts in Calculus

Question

The kinetic energy \(K\) of a body with mass \(m\) and velocity \(v\) is given by \(K=\frac{1}{2} m v^{2}\). Show that $$ \frac{\partial K}{\partial m} \frac{\partial^{2} K}{\partial v^{2}}=K $$

Step-by-Step Solution

Answer

The relation \(\frac{\partial K}{\partial m} \frac{\partial^{2} K}{\partial v^{2}} = K\) is confirmed.

Step 1: Understand the Expression for Kinetic Energy

The kinetic energy of a body is given by the formula \(K = \frac{1}{2} m v^2\). This relates kinetic energy \(K\), mass \(m\), and velocity \(v\). We are tasked with confirming that \(\frac{\partial K}{\partial m} \frac{\partial^{2} K}{\partial v^{2}} = K\).

Step 2: Differentiate K with Respect to m

To find \(\frac{\partial K}{\partial m}\), we differentiate \(K\) with respect to \(m\):\[\frac{\partial K}{\partial m} = \frac{\partial}{\partial m} \left( \frac{1}{2} m v^2 \right) = \frac{1}{2} v^2.\]

Step 3: Differentiate K Respect to v, Twice

First, find \(\frac{\partial K}{\partial v}\):\[\frac{\partial K}{\partial v} = \frac{\partial}{\partial v} \left( \frac{1}{2} m v^2 \right) = m v.\]Now, differentiate \(m v\) again with respect to \(v\):\[\frac{\partial^2 K}{\partial v^2} = \frac{\partial}{\partial v}(m v) = m.\]

Step 4: Multiply Results of Derivatives

We multiply the results obtained:\[\frac{\partial K}{\partial m} \frac{\partial^2 K}{\partial v^2} = \left( \frac{1}{2} v^2 \right) m = \frac{1}{2} m v^2.\]

Step 5: Compare Result with Original Expression for Kinetic Energy

The expression \(\frac{1}{2} m v^2\) from Step 4 is the original expression for kinetic energy \(K\). Thus, it confirms that the given relationship holds:\[\frac{\partial K}{\partial m} \frac{\partial^{2} K}{\partial v^{2}} = K.\]

Key Concepts

Partial Derivatives

Partial derivatives are an essential concept in calculus, especially when dealing with functions of multiple variables. The idea is to find the rate of change of a function with respect to one variable while keeping all other variables constant. This is particularly useful in physics, where quantities often depend on more than one variable.

• A partial derivative with respect to a variable measures how the function changes as that particular variable changes, assuming all other variables remain constant.
• For example, in the expression for kinetic energy \( K = \frac{1}{2} m v^2 \), \( \frac{\partial K}{\partial m} \) represents the change in kinetic energy with respect to mass \( m \).

Partial derivatives allow us to explore how different factors contribute to the overall behavior of a physical system. They are denoted by the \( \partial \) symbol, distinguishing them from ordinary derivatives which involve only a single variable.

Differentiation

Differentiation is the mathematical process used to find the derivative, or the rate at which a quantity changes. It is foundational in calculus, allowing us to understand how one variable affects another. This process helps uncover the relationships between variables in a function.

• Differentiation can be performed with respect to one variable in a function of multiple variables, leading to concepts such as partial derivatives.
• In our exercise, differentiating the kinetic energy \( K \) with respect to \( m \) or twice with respect to \( v \) helps uncover fundamental relationships in physics.

Understanding differentiation enables us to predict and calculate changes in physical systems. It is crucial in many scientific fields, making it a vital tool for scientists and engineers alike.

Physics Concepts

Physics is a vast field that seeks to understand the natural laws governing the universe. Within physics, several core concepts help explain how matter and energy interact. Kinetic energy is one of these key concepts, representing the energy of motion.

• In this exercise, the formula for kinetic energy \( K = \frac{1}{2} m v^2 \) shows how mass and velocity are interrelated in generating kinetic energy.
• Physics uses mathematical expressions and transformations, such as using partial derivatives, to probe and confirm these relationships.

These mathematical tools let us not only describe but also manipulate and predict phenomena, demonstrating the utility of physics in real-world applications. As you study physics, developing a strong grasp of these concepts aids your understanding of both simple and complex systems.