Problem 22 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the first partial derivatives of the function. \(g(u, v, w)=\frac{2 u w w}{u^{2}+v^{2}+w^{2}}\)

Step-by-Step Solution

Answer

The first partial derivatives of the function \(g(u, v, w) = \frac{2uw^2}{u^2 + v^2 + w^2}\) are: \(\frac{\partial g}{\partial u} = \frac{(2w^2)(u^2 + v^2 + w^2) - (2uw^2)(2u)}{(u^2 + v^2 + w^2)^2}\) \(\frac{\partial g}{\partial v} = \frac{(0)(u^2 + v^2 + w^2) - (2uw^2)(2v)}{(u^2 + v^2 + w^2)^2}\) \(\frac{\partial g}{\partial w} = \frac{(4uw)(u^2 + v^2 + w^2) - (2uw^2)(2w)}{(u^2 + v^2 + w^2)^2}\)

Step 1: Write down the given function

First, let's write down the given function: \(g(u, v, w) = \frac{2uw^2}{u^2 + v^2 + w^2}\)

Step 2: Find the partial derivative with respect to u (∂g/∂u)

To find the partial derivative with respect to u, differentiate g(u, v, w) with respect to u, treating v and w as constants: \(\frac{\partial g}{\partial u} = \frac{\partial}{\partial u} \left(\frac{2uw^2}{u^2 + v^2 + w^2}\right)\) We can use the quotient rule for differentiation, which is given by: \(\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}\) In our case, f(u) = 2uw^2, g(u) = u^2 + v^2 + w^2, f'(u) = 2w^2 (derivative of 2uw^2 with respect to u, treating w as a constant), and g'(u) = 2u (derivative of u^2 + v^2 + w^2 with respect to u, treating v and w as constants). Using the quotient rule, we get: \(\frac{\partial g}{\partial u} = \frac{(2w^2)(u^2 + v^2 + w^2) - (2uw^2)(2u)}{(u^2 + v^2 + w^2)^2}\)

Step 3: Find the partial derivative with respect to v (∂g/∂v)

To find the partial derivative with respect to v, differentiate g(u, v, w) with respect to v, treating u and w as constants: \(\frac{\partial g}{\partial v} = \frac{\partial}{\partial v}\left(\frac{2uw^2}{u^2 + v^2 + w^2}\right)\) In this case, f(v) = 2uw^2, g(v) = u^2 + v^2 + w^2, f'(v) = 0 (derivative of 2uw^2 with respect to v, treating u and w as constants) and g'(v) = 2v (derivative of u^2 + v^2 + w^2 with respect to v, treating u and w as constants). Using the quotient rule, we get: \(\frac{\partial g}{\partial v} = \frac{(0)(u^2 + v^2 + w^2) - (2uw^2)(2v)}{(u^2 + v^2 + w^2)^2}\)

Step 4: Find the partial derivative with respect to w (∂g/∂w)

To find the partial derivative with respect to w, differentiate g(u, v, w) with respect to w, treating u and v as constants: \(\frac{\partial g}{\partial w} = \frac{\partial}{\partial w}\left(\frac{2uw^2}{u^2 + v^2 + w^2}\right)\) In this case, f(w) = 2uw^2, g(w) = u^2 + v^2 + w^2, f'(w) = 4uw (derivative of 2uw^2 with respect to w, treating u as a constant) and g'(w) = 2w (derivative of u^2 + v^2 + w^2 with respect to w, treating u and v as constants). Using the quotient rule, we get: \(\frac{\partial g}{\partial w} = \frac{(4uw)(u^2 + v^2 + w^2) - (2uw^2)(2w)}{(u^2 + v^2 + w^2)^2}\)

Step 5: Conclusion

We have found the first partial derivatives of the function \(g(u, v, w) = \frac{2uw^2}{u^2 + v^2 + w^2}\) with respect to each of the variables u, v, and w: \(\frac{\partial g}{\partial u} = \frac{(2w^2)(u^2 + v^2 + w^2) - (2uw^2)(2u)}{(u^2 + v^2 + w^2)^2}\) \(\frac{\partial g}{\partial v} = \frac{(0)(u^2 + v^2 + w^2) - (2uw^2)(2v)}{(u^2 + v^2 + w^2)^2}\) \(\frac{\partial g}{\partial w} = \frac{(4uw)(u^2 + v^2 + w^2) - (2uw^2)(2w)}{(u^2 + v^2 + w^2)^2}\)

Key Concepts

Multivariable Calculus

Multivariable calculus is the extension of single-variable calculus to functions of several variables. It involves concepts like partial derivatives and multiple integrals. In the context of multivariable calculus, a function can rely on two or more variables – such as the function in our exercise.

Understanding the behavior of these multivariable functions is crucial in many fields like physics, engineering, and economics. Analyzing how a function changes with respect to each variable independently while keeping the other variables constant is where the concept of partial derivatives comes into play. With our function g(u, v, w), each partial derivative represents the rate of change along an individual axis (u, v, or w), potentially impacting the overall behavior of the function.

Quotient Rule for Differentiation

Differentiation is a fundamental tool in calculus that gives us the rate at which a quantity changes. When we have a function that is a ratio of two differentiable functions, the quotient rule for differentiation becomes particularly useful. The quotient rule is reflected in the expression:
\[\begin{equation}\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}.\end{equation}\]
This rule is critical for finding the partial derivatives of functions like g(u, v, w), especially when they are expressed as one variable-dependent term divided by another. As we must treat all other variables as constants while differentiating with respect to one variable, applying the quotient rule simplifies what could otherwise be a complex and time-consuming process. This meticulous approach ensures accurate computation and deepens our understanding of how the function varies as each variable changes.

First Partial Derivative

The first partial derivative of a function with respect to a given variable represents the sensitivity of the function's output in relation to small changes in that variable. In simple terms, it answers the question: 'How does the function's value change as I tweak this one variable, keeping everything else fixed?'

Take for example our function g(u, v, w) from the exercise. To determine how g varies with u while v and w are held constant, we calculate the first partial derivative with respect to u. Symbolically, this is written as \(\frac{\partial g}{\partial u}\). Similarly, to understand how g changes with v (holding u and w constant), or with w (holding u and v constant), we calculate the first partial derivatives with respect to v and w, respectively. Having found the first partial derivatives for all variables, we gain a complete picture of how the function responds to changes in each individual variable within its domain.