Problem 56 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

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

Step-by-Step Solution

Answer

We find the first partial derivative of the kinetic energy equation, K, with respect to mass (m) as \(\frac{\partial K}{\partial m} = \frac{1}{2} v^2\), and the second partial derivative of K with respect to velocity (v) as \(\frac{\partial^2 K}{\partial v^2} = m\). Then, by multiplying these two derivatives, we get \(\frac{\partial K}{\partial m} \cdot \frac{\partial^2 K}{\partial v^2} = \left(\frac{1}{2} v^2\right) \cdot (m) = \frac{1}{2}mv^2 = K\), which verifies that the given condition holds true.

Step 1: Find the first partial derivative with respect to mass

Differentiate the kinetic energy equation \[K = \frac{1}{2}mv^2\] with respect to m to find the first partial derivative. Using the constant rule (as v is not dependent on m), this gives: \[\frac{\partial K}{\partial m} = \frac{1}{2} v^2\]

Step 2: Find the second partial derivative with respect to velocity

Differentiate the kinetic energy equation \[K = \frac{1}{2}mv^2\] with respect to v to find the first partial derivative. Using the constant rule (as m is not dependent on v), this gives: \[\frac{\partial K}{\partial v} = mv \] Now, differentiate \[\frac{\partial K}{\partial v} = mv\] with respect to v once more to find the second partial derivative. Using the constant rule again, this gives: \[\frac{\partial^2 K}{\partial v^2} = m\]

Step 3: Compute the product of the two derivatives

Calculate the product of the two derivatives found in Step 1 and Step 2. \[\frac{\partial K}{\partial m} \cdot \frac{\partial^2 K}{\partial v^2} = \left(\frac{1}{2} v^2\right) \cdot (m)\]

Step 4: Verification

Check if the result obtained in Step 3 is equal to the original kinetic energy equation, \[K = \frac{1}{2}mv^2\]. From Step 3, we have: \[\frac{\partial K}{\partial m} \cdot \frac{\partial^2 K}{\partial v^2} = \left(\frac{1}{2} v^2\right) \cdot (m) = \frac{1}{2}mv^2\] Thus, we have shown that \[\frac{\partial K}{\partial m} \cdot \frac{\partial^2 K}{\partial v^2} = K\].

Key Concepts

Partial Derivative

A \textbf{partial derivative} represents how a function changes as one of its variables moves infinitesimally, while holding other variables constant. It is a fundamental tool in multivariable calculus, which is essential when dealing with functions of several variables.

For example, in the kinetic energy equation, when we find the partial derivative of kinetic energy with respect to mass \(m\), we treat the velocity \(v\) as if it were a constant. This is indicated by \(\frac{\partial K}{\partial m}\) and it captures the rate at which kinetic energy changes with small changes in mass alone.

Kinetic Energy Formula

The kinetic energy formula \(K = \frac{1}{2}mv^2\) quantifies the energy that a body possesses due to its motion. It denotes that kinetic energy is directly proportional to the mass \(m\) of the object and the square of its velocity \(v\).

Kinetic energy is a vital concept in physics, providing insights into how much work an object can do by virtue of its motion. Understanding this relationship is crucial because it underpins physical principles from the conservation of energy to the analysis of dynamic systems.

Mathematical Differentiation

Mathematical differentiation is the process of finding the rate at which a function changes at any given point. It's a core operation in calculus, used for finding minimum and maximum values of functions, solving rate problems, and much more.

In the context of the kinetic energy equation, differentiation allows us to see how kinetic energy changes in relation to the object's mass and velocity. As demonstrated in the original exercise, differentiation is used twice to explore these dynamic relationships.

Multivariable Calculus

Multivariable calculus extends the concepts of single variable calculus to functions of several variables. It includes partial derivatives, gradient vectors, and multiple integrals among other things.

Problems that involve rates of change in multiple dimensions, like the changes in kinetic energy with respect to mass and velocity, are precisely where multivariable calculus shines. The exercise provided makes use of this by calculating partial derivatives with respect to each variable independently and showing how these different rates of change interact to describe the behavior of kinetic energy.