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Chapter 12: Problem 29

Use a tree diagram to write the required Chain Rule formula. $$\begin{aligned} &u=f(v), \text { where } v=g(w, x, y), w=h(z), x=p(t, z), \text { and }\\\ &y=q(t, z) . \text { Find } \partial u / \partial z \end{aligned}$$

Short Answer

Expert verified
Question: Determine the expression for the derivative of $$u$$ with respect to $$z$$ using the Chain Rule, given functions: 1. $$u = f(v)$$ 2. $$v = g(w, x, y)$$ 3. $$w = h(z)$$ 4. $$x = p(t, z)$$ 5. $$y = q(t, z)$$ Answer: $$\frac{\partial u}{\partial z} = \frac{\partial f(v)}{\partial v} \left(\frac{\partial g(w, x, y)}{\partial w}\frac{\partial h(z)}{\partial z} + \frac{\partial g(w, x, y)}{\partial x}\frac{\partial p(t, z)}{\partial z} + \frac{\partial g(w, x, y)}{\partial y}\frac{\partial q(t, z)}{\partial z}\right)$$

Step by step solution

01

Draw Tree Diagram

Draw the tree diagram based on the given equations like this: ``` u | v / | \ w x y | / \ / \ z t z t z ```
02

Apply Chain Rule

The multivariable Chain Rule states that: $$\frac{\partial u}{\partial z} = \frac{\partial u}{\partial v} \cdot \frac{\partial v}{\partial z}$$ Since $$v = g(w, x, y)$$, we have $$\frac{\partial v}{\partial z} = \frac{\partial v}{\partial w}\frac{\partial w}{\partial z} + \frac{\partial v}{\partial x}\frac{\partial x}{\partial z} + \frac{\partial v}{\partial y}\frac{\partial y}{\partial z}$$. Putting it all together, we get: $$\frac{\partial u}{\partial z} = \frac{\partial u}{\partial v} \left(\frac{\partial v}{\partial w}\frac{\partial w}{\partial z} + \frac{\partial v}{\partial x}\frac{\partial x}{\partial z} + \frac{\partial v}{\partial y}\frac{\partial y}{\partial z}\right)$$
03

Differentiation

Now, differentiate each fraction with respect to its corresponding variable: 1. $$\frac{\partial u}{\partial v} = \frac{\partial f(v)}{\partial v}$$ 2. $$\frac{\partial v}{\partial w} = \frac{\partial g(w, x, y)}{\partial w}$$ 3. $$\frac{\partial w}{\partial z} = \frac{\partial h(z)}{\partial z}$$ 4. $$\frac{\partial v}{\partial x} = \frac{\partial g(w, x, y)}{\partial x}$$ 5. $$\frac{\partial x}{\partial z} = \frac{\partial p(t, z)}{\partial z}$$ 6. $$\frac{\partial v}{\partial y} = \frac{\partial g(w, x, y)}{\partial y}$$ 7. $$\frac{\partial y}{\partial z} = \frac{\partial q(t, z)}{\partial z}$$
04

Substitute The Differentiation

Substitute back these derivatives into the Chain Rule formula: $$\frac{\partial u}{\partial z} = \frac{\partial f(v)}{\partial v} \left(\frac{\partial g(w, x, y)}{\partial w}\frac{\partial h(z)}{\partial z} + \frac{\partial g(w, x, y)}{\partial x}\frac{\partial p(t, z)}{\partial z} + \frac{\partial g(w, x, y)}{\partial y}\frac{\partial q(t, z)}{\partial z}\right)$$ That is the required Chain Rule formula for $$\frac{\partial u}{\partial z}$$.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
Partial derivatives play a crucial role in understanding how functions change with respect to one variable while keeping others constant. They are particularly essential in multivariable calculus where functions depend on several variables. When you take the partial derivative of a function concerning one of its variables, you treat the other variables as constants.
For example, consider a function \( f(x, y) \). The partial derivative with respect to \( x \) is denoted as \( \frac{\partial f}{\partial x} \) and to find this, you assume \( y \) is constant while differentiating. Similarly, \( \frac{\partial f}{\partial y} \) assumes \( x \) is constant. This concept helps us analyze how changes in individual variables influence the overall function.

Understanding partial derivatives is key because they allow us to extend the powerful concept of derivatives to functions of multiple variables. In the context of the Chain Rule, we differentiate each factor separately and observe how they compose together to impact the original function.
Multivariable Calculus
Multivariable calculus extends the ideas of calculus to functions of several variables. This is a major departure from the simpler one-variable calculus, as it involves vectors, matrices, and more complex forms of differentiation such as partial derivatives.
The Chain Rule in multivariable calculus is an extension of the single-variable Chain Rule and allows us to take derivatives of composite functions that involve multiple variables. When a function depends on a composite of other functions, you need to account for how each function impacts the whole.
This was seen in the exercise where the Chain Rule was used to find \( \frac{\partial u}{\partial z} \). By breaking down \( u \) into its dependencies on \( v \), \( w \), \( x \), and \( y \) using separate functions, we applied the Chain Rule to consider each chain's effect.

This makes multivariable calculus an indispensable tool in fields like physics, engineering, and economics where systems rely on multiple variables.
Tree Diagram
A tree diagram serves as a visual aid in organizing and understanding the relationships between different variables in multivariable functions. It allows you to clearly map how variables are related and depict the flow of partial derivatives.
In the given exercise, the tree diagram started with \( u \) and branched out to show its dependency on \( v \), which is influenced by \( w \), \( x \), and \( y \). Each of these subsequently depends on other variables like \( z \), making the diagram complex but insightful.

A tree diagram is helpful for organizing your understanding of how each component affects the other, allowing you to systematically apply the Chain Rule. It prevents confusion which is common in multivariable calculus problems and ensures that all paths of derivation are considered.
The visual representation makes it easier to substitute and multiply the required derivatives, ensuring no part of the function's dependency is overlooked.

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Most popular questions from this chapter

Let \(f\) be a differentiable function of one or more variables that is positive on its domain. a. Show that \(d(\ln f)=\frac{d f}{f}.\) b. Use part (a) to explain the statement that the absolute change in \(\ln f\) is approximately equal to the relative change in \(f.\) c. Let \(f(x, y)=x y,\) note that \(\ln f=\ln x+\ln y,\) and show that relative changes add; that is, \(d f / f=d x / x+d y / y.\) d. Let \(f(x, y)=x / y,\) note that \(\ln f=\ln x-\ln y,\) and show that relative changes subtract; that is \(d f / f=d x / x-d y / y.\) e. Show that in a product of \(n\) numbers, \(f=x_{1} x_{2} \cdots x_{n},\) the relative change in \(f\) is approximately equal to the sum of the relative changes in the variables.

Let \(x, y,\) and \(z\) be non-negative numbers with \(x+y+z=200\) a. Find the values of \(x, y,\) and \(z\) that minimize \(x^{2}+y^{2}+z^{2}\) b. Find the values of \(x, y,\) and \(z\) that minimize \(\sqrt{x^{2}+y^{2}+z^{2}}\). c. Find the values of \(x, y,\) and \(z\) that maximize \(x y z\) d. Find the values of \(x, y,\) and \(z\) that maximize \(x^{2} y^{2} z^{2}\).

Find the dimensions of the rectangular box with maximum volume in the first octant with one vertex at the origin and the opposite vertex on the ellipsoid \(36 x^{2}+4 y^{2}+9 z^{2}=36\).

Find the points at which the plane \(a x+b y+c z=d\) intersects the \(x-y-\), and \(z\) -axes.

Given a differentiable function \(w=f(x, y, z),\) the goal is to find its maximum and minimum values subject to the constraints \(g(x, y, z)=0\) and \(h(x, y, z)=0\) where \(g\) and \(h\) are also differentiable. a. Imagine a level surface of the function \(f\) and the constraint surfaces \(g(x, y, z)=0\) and \(h(x, y, z)=0 .\) Note that \(g\) and \(h\) intersect (in general) in a curve \(C\) on which maximum and minimum values of \(f\) must be found. Explain why \(\nabla g\) and \(\nabla h\) are orthogonal to their respective surfaces. b. Explain why \(\nabla f\) lies in the plane formed by \(\nabla g\) and \(\nabla h\) at a point of \(C\) where \(f\) has a maximum or minimum value. c. Explain why part (b) implies that \(\nabla f=\lambda \nabla g+\mu \nabla h\) at a point of \(C\) where \(f\) has a maximum or minimum value, where \(\lambda\) and \(\mu\) (the Lagrange multipliers) are real numbers. d. Conclude from part (c) that the equations that must be solved for maximum or minimum values of \(f\) subject to two constraints are \(\nabla f=\lambda \nabla g+\mu \nabla h, g(x, y, z)=0\) and \(h(x, y, z)=0\).
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