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Making trees Use a tree diagram to write the required Chain Rule formula. \(u=f(v),\) where \(v=g(w, x, y), w=h(z), x=p(t, z),\) and \(y=q(t, z) .\) Find \(\partial u / \partial z\)

Short Answer

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Question: Use a tree diagram to write the Chain Rule formula for the partial derivative of \(u\) with respect to \(z\), considering the following relations: \(u = f(v)\), \(v = g(w, x, y)\), \(w = h(z)\), \(x = l(t, z)\), and \(y = m(t, z)\). Answer: \(\frac{\partial{u}}{\partial{z}} = \frac{\partial{u}}{\partial{v}}\frac{\partial{v}}{\partial{w}}\frac{\partial{w}}{\partial{z}} + \frac{\partial{u}}{\partial{v}}\frac{\partial{v}}{\partial{x}}\frac{\partial{x}}{\partial{t}}\frac{\partial{t}}{\partial{z}} + \frac{\partial{u}}{\partial{v}}\frac{\partial{v}}{\partial{x}}\frac{\partial{x}}{\partial{z}} + \frac{\partial{u}}{\partial{v}}\frac{\partial{v}}{\partial{y}}\frac{\partial{y}}{\partial{t}}\frac{\partial{t}}{\partial{z}} + \frac{\partial{u}}{\partial{v}}\frac{\partial{v}}{\partial{y}}\frac{\partial{y}}{\partial{z}}.\)

Step by step solution

01

Draw a tree diagram with derivatives

First, we need to draw a tree diagram to represent the dependencies between \(u\), \(v\), \(w\), \(x\), \(y\), and \(z\). u | v /|\ w x y | | | zmmmmm / \ t z The arrows represent the partial derivatives. For example, \(\frac{\partial{v}}{\partial{w}}\), \(\frac{\partial{v}}{\partial{x}}\) and \(\frac{\partial{v}}{\partial{y}}\).
02

Write down the Chain Rule formula using the tree diagram

We will follow the tree diagram to write the Chain Rule formula for \(\frac{\partial{u}}{\partial{z}}\). From the tree diagram we can find three different paths between \(u\) and \(z\): 1. \(u \rightarrow v \rightarrow w \rightarrow z\) 2. \(u \rightarrow v \rightarrow x \rightarrow t \rightarrow z\) 3. \(u \rightarrow v \rightarrow x \rightarrow z\) 4. \(u \rightarrow v \rightarrow y \rightarrow t \rightarrow z\) 5. \(u \rightarrow v \rightarrow y \rightarrow z\) Applying the Chain Rule, the partial derivative \(\frac{\partial{u}}{\partial{z}}\) is the sum of multiplying the derivatives along each path: \(\frac{\partial{u}}{\partial{z}} = \frac{\partial{u}}{\partial{v}}\frac{\partial{v}}{\partial{w}}\frac{\partial{w}}{\partial{z}} + \frac{\partial{u}}{\partial{v}}\frac{\partial{v}}{\partial{x}}\frac{\partial{x}}{\partial{t}}\frac{\partial{t}}{\partial{z}} + \frac{\partial{u}}{\partial{v}}\frac{\partial{v}}{\partial{x}}\frac{\partial{x}}{\partial{z}} + \frac{\partial{u}}{\partial{v}}\frac{\partial{v}}{\partial{y}}\frac{\partial{y}}{\partial{t}}\frac{\partial{t}}{\partial{z}} + \frac{\partial{u}}{\partial{v}}\frac{\partial{v}}{\partial{y}}\frac{\partial{y}}{\partial{z}}.\) Since we are asked to find the Chain Rule formula only, we can stop here. In case the given functions were provided, we would have to compute the relevant partial derivatives and substitute them into the formula to obtain the final result.

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Key Concepts

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

Tree Diagram
A tree diagram in calculus is a visual representation of the functional relationships between variables. It helps to determine how to apply the Chain Rule to the functions involved, especially in complex scenarios where several functions are dependent on a mixture of variables.

Consider the exercise dealing with the function u as a function of variables w, x, and y, which in turn are functions of z and possibly t. The tree diagram starts with u at the top, branching down to v, then further branching to w, x, and y, with subsequent branches to z and t as required. Each branch is associated with a derivative, creating a pathway that assists in the application of the Chain Rule to find the derivative of u with respect to z.
Partial Derivatives
Partial derivatives are an integral component of multivariable calculus. They represent the rate of change of a function with respect to one variable, while keeping the other variables constant. In our exercise, the partial derivatives are the 'links' that connect the nodes on the tree diagram.

For instance, the tree diagram shows variables w, x, and y depending on z and t. This means that the function u indirectly depends on z through multiple pathways. We calculate partial derivatives like ∂v/∂w, ∂w/âˆÁ©, and ∂x/∂t for each link in these paths to eventually find ∂u/âˆÁ© by applying the Chain Rule.
Multivariable Calculus
Multivariable calculus extends the concepts of single-variable calculus to functions of several variables. This field addresses functions like u=f(v) and v=g(w, x, y), which in our exercise depend on multiple variables. In this context, understanding how variations in one variable affect another becomes complex and requires the use of partial derivatives and the Chain Rule to elucidate these relationships.

The tree diagram simplifies this complexity by visualizing all the variables and their interdependent functions. The ability to identify and differentiate along each path in the diagram allows us to apply the Chain Rule effectively, paving the way for accurate calculation of derivatives in higher dimensions.
Chain Rule Formula
The Chain Rule is a fundamental tool in calculus for finding the derivative of composite functions. It states that the derivative of a composite function is the product of the derivative of the outer function and the derivative of the inner function. In multivariable calculus, this rule extends to the situation where multiple functions are nested within each other.

As illustrated in the solution for our exercise, the Chain Rule formula for ∂u/âˆÁ© is the sum of the derivatives along each pathway from u to z. Each term in the sum is a product of partial derivatives that corresponds to a path in the tree diagram. These paths represent the sequential application of the Chain Rule across multiple variables. The Chain Rule enables us to systematically approach the differentiation of complex, inter-related functions in a clear and logical manner.

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Most popular questions from this chapter

a. Determine the domain and range of the following functions. b. Graph each function using a graphing utility. Be sure to experiment with the window and orientation to give the best perspective on the surface. $$p(x, y)=1-|x-1|+|y+1|$$

Use what you learned about surfaces in Sections 13.5 and 13.6 to sketch a graph of the following functions. In each case, identify the surface and state the domain and range of the function. $$H(x, y)=\sqrt{x^{2}+y^{2}}$$

Find the values of \(K\) and \(L\) that maximize the following production functions subject to the given constraint, assuming \(K \geq 0\) and \(L \geq 0\) Given the production function \(P=f(K, L)=K^{a} L^{1-a}\) and the budget constraint \(p K+q L=B,\) where \(a, p, q,\) and \(B\) are given, show that \(P\) is maximized when \(K=\frac{a B}{p}\) and \(L=\frac{(1-a) B}{q}\)

Second Derivative Test Suppose the conditions of the Second Derivative Test are satisfied on an open disk containing the point \((a, b) .\) Use the test to prove that if \((a, b)\) is a critical point of \(f\) at which \(f_{x}(a, b)=f_{y}(a, b)=0\) and \(f_{x x}(a, b) < 0 < f_{\text {vy }}(a, b)\) or \(f_{y y}(a, b) < 0 < f_{x x}(a, b),\) then \(f\) has a saddle point at \((a, b)\)

Potential functions Potential functions arise frequently in physics and engineering. A potential function has the property that a field of interest (for example, an electric field, a gravitational field, or a velocity field) is the gradient of the potential (or sometimes the negative of the gradient of the potential). (Potential functions are considered in depth in Chapter \(17 .)\) The electric field due to a point charge of strength \(Q\) at the origin has a potential function \(\varphi=k Q / r,\) where \(r^{2}=x^{2}+y^{2}+z^{2}\) is the square of the distance between a variable point \(P(x, y, z)\) and the charge, and \(k>0\) is a physical constant. The electric field is given by \(\mathbf{E}=-\nabla \varphi,\) where \(\nabla \varphi\) is the gradient in three dimensions. a. Show that the three-dimensional electric field due to a point charge is given by $$\mathbf{E}(x, y, z)=k Q\left\langle\frac{x}{r^{3}}, \frac{y}{r^{3}}, \frac{z}{r^{3}}\right\rangle$$ b. Show that the electric field at a point has a magnitude \(|\mathbf{E}|=\frac{k Q}{r^{2}} .\) Explain why this relationship is called an inverse square law.

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