Question

Given \(x^{2}+y^{2}+z^{2}=6,\) and \(w^{3}+z^{3}=5 x y+12,\) find the following partial derivatives at the point \((x, y, z, w)=(1,-2,1,1)\). $$\left(\frac{\partial z}{\partial x}\right)_{y}, \quad\left(\frac{\partial z}{\partial x}\right)_{w}, \quad\left(\frac{\partial z}{\partial y}\right)_{x}, \quad\left(\frac{\partial z}{\partial y}\right)_{w}, \quad\left(\frac{\partial w}{\partial x}\right)_{z}, \quad\left(\frac{\partial x}{\partial w}\right)_{z}$$.

Short answer

At (1, -2, 1, 1):\(\frac{\partial z}{\partial x} = -1\), \(\frac{\partial z}{\partial y} = 2\), \(\frac{\partial w}{\partial x} = -\frac{10}{3}\), \(\frac{\partial x}{\partial w} = \frac{10}{3}\).

Step-by-step solution

Understand the given functions

We have two functions: 1. \(f(x, y, z) = x^2 + y^2 + z^2 - 6 = 0\) 2. \(g(x, y, z, w) = w^3 + z^3 - 5xy - 12 = 0\).

Calculate the partial derivatives of function \(f\)

Find the partial derivatives of \(f = x^2 + y^2 + z^2 - 6\) with respect to \(x\), \(y\), and \(z\): \(\frac{\partial f}{\partial x} = 2x\)\(\frac{\partial f}{\partial y} = 2y\)\(\frac{\partial f}{\partial z} = 2z\)

Calculate the partial derivatives of function \(g\)

Find the partial derivatives of \(g = w^3 + z^3 - 5xy - 12\) with respect to \(x\), \(y\), \(z\), and \(w\): \(\frac{\partial g}{\partial x} = -5y\)\(\frac{\partial g}{\partial y} = -5x\)\(\frac{\partial g}{\partial z} = 3z^2\)\(\frac{\partial g}{\partial w} = 3w^2\)

Calculate \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\)

Using implicit differentiation on \(f = x^2 + y^2 + z^2 - 6 = 0\):\(2z \frac{\partial z}{\partial x} + 2x = 0 \Rightarrow \frac{\partial z}{\partial x} = -\frac{x}{z}\)\(2z \frac{\partial z}{\partial y} + 2y = 0 \Rightarrow \frac{\partial z}{\partial y} = -\frac{y}{z}\)

Evaluate \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) at the given point

At point \((1, -2, 1, 1)\):\(\frac{\partial z}{\partial x} = -\frac{1}{1} = -1\) and \(\frac{\partial z}{\partial y} = -\frac{-2}{1} = 2\)

Calculate \(\frac{\partial w}{\partial x}\) and \(\frac{\partial x}{\partial w}\)

Using implicit differentiation on \(g = w^3 + z^3 - 5xy - 12 = 0\):\[\frac{\partial g}{\partial x} + \frac{\partial g}{\partial w} \frac{\partial w}{\partial x} = 0 \Rightarrow -5y + 3w^2 \frac{\partial w}{\partial x} = 0 \Rightarrow \frac{\partial w}{\partial x} = \frac{5y}{3w^2}\]Similarly, for \(\frac{\partial x}{\partial w}\):\[\frac{\partial g}{\partial x} + \frac{\partial g}{\partial w} \frac{\partial w}{\partial x} = 0 \Rightarrow \frac{3w^2 \frac{\partial x}{\partial w}}{-5y} = 1 \Rightarrow \frac{\partial x}{\partial w} = -\frac{5y}{3w^2}\]

Evaluate \(\frac{\partial w}{\partial x}\) and \(\frac{\partial x}{\partial w}\) at the given point

At point \((1, -2, 1, 1)\):\(\frac{\partial w}{\partial x} = \frac{5(-2)}{3(1)^2} = -\frac{10}{3}\)\(\frac{\partial x}{\partial w} = -\frac{5(-2)}{3(1)^2} = \frac{10}{3}\)

Key concepts

Multivariable Functions

A multivariable function is a function that depends on more than one variable. In the given exercise, we have two such functions:

1. \( f(x, y, z) = x^2 + y^2 + z^2 - 6 = 0 \)
2. \( g(x, y, z, w) = w^3 + z^3 - 5xy - 12 = 0 \)

Understanding these functions is crucial because they determine the relationships between their variables. For example, in function \( f \), we see how \( x, y, \) and \( z \) are related to one another. Function \( g \) shows a more complex relationship, including an additional variable, \( w \). These relationships are explored through their derivatives.

When dealing with multivariable functions, we often need to understand how a change in one variable affects another. This brings us to the concept of partial derivatives, which capture the rate of change of functions with respect to one variable while keeping others constant.

Implicit Differentiation

Implicit differentiation is a technique to find derivatives of functions that are not given explicitly. In the provided exercise, both functions are implicit because \( z \) and \( w \) are not isolated on one side of the equation. This technique allows us to handle such scenarios effectively.

For example, to find \( \frac{\partial z}{\partial x} \) from the function \( f(x, y, z) = x^2 + y^2 + z^2 - 6 \), we treat \( z \) as a function of \( x \). Then we differentiate both sides with respect to \( x \), while treating \( y \) and \( z \) as dependent variables. This gives us:
\[ 2x + 2z \frac{\partial z}{\partial x} = 0 \]
Solving this for \( \frac{\partial z}{\partial x} \), we get:
\[ \frac{\partial z}{\partial x} = -\frac{x}{z} \]

Implicit differentiation hence helps us find how one variable changes concerning another, even if the function isn't separated explicitly.

Calculus

Calculus is a branch of mathematics focusing on limits, functions, derivatives, integrals, and infinite series. It is the study of change and motion through the use of derivatives and integrals.

In the context of this exercise, calculus helps us understand the relationships between the variables through the concept of partial derivatives. Partial derivatives are a fundamental concept in calculus, particularly in the field of multivariable calculus. They measure how a multivariable function changes as one of its variables changes, while keeping the other variables fixed.
For instance, finding \( \frac{\partial z}{\partial y} \) when given \( f(x, y, z) = x^2 + y^2 + z^2 - 6 \), means differentiating the function \( f \) with respect to \( y \):
\[ 2y + 2z \frac{\partial z}{\partial y} = 0 \]
We solve for \( \frac{\partial z}{\partial y} \) to find:
\[ \frac{\partial z}{\partial y} = -\frac{y}{z} \]

These derivatives are essential for understanding how small changes in one variable affect the overall system, a critical concept in many scientific and engineering disciplines.