Question

In the implicit relationship \(F(x, y, z)=0,\) any two of the variables may be considered independent, which then determines the third variable. To avoid confusion, we use a subscript to indicate which variable is held fixed in a derivative calculation; for example \(\left(\frac{\partial z}{\partial x}\right)_{y}\) means that \(y\) is held fixed in taking the partial derivative of \(z\) with respect to \(x\). (In this context, the subscript does not mean a derivative.) a. Differentiate \(F(x, y, z)=0\) with respect to \(x\) holding \(y\) fixed to show that \(\left(\frac{\partial z}{\partial x}\right)_{y}=-\frac{F_{x}}{F_{z}}\). b. As in part (a), find \(\left(\frac{\partial y}{\partial z}\right)_{x}\) and \(\left(\frac{\partial x}{\partial y}\right)_{z}\). c. Show that \(\left(\frac{\partial z}{\partial x}\right)_{y}\left(\frac{\partial y}{\partial z}\right)_{x}\left(\frac{\partial x}{\partial y}\right)_{z}=-1\). d. Find the relationship analogous to part (c) for the case \(F(w, x, y, z)=0\).

Short answer

Question: Prove that \(\left(\frac{\partial z}{\partial x}\right)_{y}\left(\frac{\partial y}{\partial z}\right)_{x}\left(\frac{\partial x}{\partial y}\right)_{z} = -1\) if \(F(x, y, z) = 0\), and find the analogous relationship for the case \(F(w, x, y, z) = 0\). Solution: Using the partial derivatives obtained in the step-by-step solution, we showed that \(\left(\frac{\partial z}{\partial x}\right)_{y}\left(\frac{\partial y}{\partial z}\right)_{x}\left(\frac{\partial x}{\partial y}\right)_{z} = -1\). Additionally, for the case \(F(w, x, y, z) = 0\), we found the analogous relationship to be \(\left(\frac{\partial w}{\partial x}\right)_{y, z}\left(\frac{\partial x}{\partial y}\right)_{w, z}\left(\frac{\partial y}{\partial z}\right)_{w, x}\left(\frac{\partial z}{\partial w}\right)_{x, y} = 1\).

Step-by-step solution

Differentiate F(x, y, z) with respect to x

Since y is held fixed, we treat it as a constant while differentiating. Apply the chain rule: \[\frac{\partial F}{\partial x} + \frac{\partial F}{\partial z} \frac{\partial z}{\partial x} = 0\]

Solve for \(\left(\frac{\partial z}{\partial x}\right)_{y}\)

Rearrange the equation to isolate \(\frac{\partial z}{\partial x}\): \[\left(\frac{\partial z}{\partial x}\right)_{y} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}}=-\frac{F_x}{F_z}\] b. Find \(\left(\frac{\partial y}{\partial z}\right)_{x}\) and \(\left(\frac{\partial x}{\partial y}\right)_{z}\) using the same method.

Differentiate F(x, y, z) with respect to z

Holding x fixed and differentiating, we get: \[\frac{\partial F}{\partial z} + \frac{\partial F}{\partial y} \frac{\partial y}{\partial z} = 0\] \[\left(\frac{\partial y}{\partial z}\right)_{x}=-\frac{\frac{\partial F}{\partial z}}{\frac{\partial F}{\partial y}}=-\frac{F_z}{F_y}\]

Differentiate F(x, y, z) with respect to y

Holding z fixed and differentiating, we get: \[\frac{\partial F}{\partial y} + \frac{\partial F}{\partial x} \frac{\partial x}{\partial y} = 0\] \[\left(\frac{\partial x}{\partial y}\right)_{z}=-\frac{\frac{\partial F}{\partial y}}{\frac{\partial F}{\partial x}}=-\frac{F_y}{F_x}\] c. Show that \(\left(\frac{\partial z}{\partial x}\right)_{y}\left(\frac{\partial y}{\partial z}\right)_{x}\left(\frac{\partial x}{\partial y}\right)_{z} = -1\).

Substitute the expressions for the partial derivatives

Using the expressions derived in parts (a) and (b), substitute into the given equation and simplify: \[-\frac{F_x}{F_z}\cdot-\frac{F_z}{F_y}\cdot-\frac{F_y}{F_x} = (-1)^3\]

Simplify the expression

After cancellation of the terms, we get: \[-1^{3} = -1\] d. Find the relationship analogous to part (c) for the case \(F(w, x, y, z) = 0\).

Differentiate and express the partial derivatives

Keep in mind the process from parts a and b. Differentiate F(w, x, y, z) with respect to one variable while holding another fixed and express the partial derivatives similarly. We obtain: \[\left(\frac{\partial w}{\partial x}\right)_{y, z} = -\frac{F_x}{F_w}\] \[\left(\frac{\partial x}{\partial y}\right)_{w, z} = -\frac{F_y}{F_x}\] \[\left(\frac{\partial y}{\partial z}\right)_{w, x} = -\frac{F_z}{F_y}\] \[\left(\frac{\partial z}{\partial w}\right)_{x, y} = -\frac{F_w}{F_z}\]

Find the analogous relationship

Using the same logic as in part (c), multiply the relations obtained above and simplify: \[\left(\frac{\partial w}{\partial x}\right)_{y, z} \left(\frac{\partial x}{\partial y}\right)_{w, z} \left(\frac{\partial y}{\partial z}\right)_{w, x} \left(\frac{\partial z}{\partial w}\right)_{x, y} = (-1)^4\] The analogous relationship for the case \(F(w, x, y, z) = 0\) is: \[\left(\frac{\partial w}{\partial x}\right)_{y, z}\left(\frac{\partial x}{\partial y}\right)_{w, z}\left(\frac{\partial y}{\partial z}\right)_{w, x}\left(\frac{\partial z}{\partial w}\right)_{x, y} = 1\]

Key concepts

Partial Differentiation

Partial differentiation focuses on the way multi-variable functions change. Rather than looking at the change concerning all variables, we pick one variable to vary while the others stay constant. Imagine you're exploring a graph of a surface. Partial differentiation lets you look at how the surface changes in the direction of one axis, while sitting still along the other axes.

When you see a notation like \( \frac{\partial z}{\partial x} \), it indicates the partial derivative of \( z \) with respect to \( x \). The simplest way to understand this is by thinking of slicing the surface parallel to the \( y\)- and \( z\)- axes, then checking how the height changes along \( x \) as you move. Another important aspect is using subscripts, such as \( \left( \frac{\partial z}{\partial x} \right)_{y} \), to show which variable remains constant. In this exercise, it tells us that \( y \) stays unchanged while differentiating with respect to \( x \).

Partial differentiation is essential because it helps us study functions with more than one variable, which shows up a lot in physics and engineering. It’s used in optimizing functions and finding equations governing physical systems.

Implicit Differentiation

Implicit differentiation is a technique used when dealing with equations where the relationship between the variables isn't explicitly solved for one variable in terms of the others. For example, the exercise gives us an implicit relationship \( F(x, y, z) = 0 \). Instead of solving for \( z \) as a function of \( x \) and \( y \), implicit differentiation allows you to find the derivative directly from this implicit form.

Here's how it works: differentiate both sides of the equation with respect to a specific variable, treating other variables as constants when required. This might look confusing initially because you have to think about how each part of the equation changes. You apply the rules of differentiation along with the chain rule, which we'll talk about more in another section.

Using implicit differentiation, followed by algebraic manipulation, can reveal relationships between partial derivatives. For example, from the equation \( \frac{\partial F}{\partial x} + \frac{\partial F}{\partial z} \frac{\partial z}{\partial x} = 0 \), you can solve for this partial derivative \( \left( \frac{\partial z}{\partial x} \right)_{y} \). It emphasizes understanding derivatives not just for their numerical values, but for what they tell us about the interrelation of variables.

Chain Rule

The chain rule is a powerful tool for differentiation, particularly useful when dealing with composite functions. In this context, it's applied when you're differentiating a variable that depends on another variable, which, in turn, depends on yet another variable.

For instance, when you differentiate \( F(x, y, z) \) with respect to \( x \) while \( y \) is fixed, you're implicitly differentiating through the chain rule. You apply it because \( z \) might depend on \( x \) and other variables. Thus, you need to account for how \( \frac{\partial z}{\partial x} \) changes in response to \( x \). The chain rule allows you to break down these derivative paths, and it's expressed as:

• Differentiate the outer function first, which gives you \( \frac{\partial F}{\partial z} \).
• Then multiply by the derivative of the inner function, \( \frac{\partial z}{\partial x} \).

This cascade of derivatives forms the "chain" seen in the rule's name.

Understanding the chain rule is essential for tackling more complex multivariable problems. It helps untangle dependencies in implicit functions like those you've dealt with here, providing clearer insights and solutions.