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Draw a branch diagram and write a Chain Rule formula for each derivative. \(\frac{\partial w}{\partial s} \text { for } w=g(x, y), \quad x=h(r, s, t), \quad y=k(r, s, t)\)

Short Answer

Expert verified
\( \frac{\partial w}{\partial s} = \frac{\partial w}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial w}{\partial y} \cdot \frac{\partial y}{\partial s} \).

Step by step solution

01

Identify Components

First, we need to identify the functions involved: the primary function is \( w = g(x, y) \), where \( x = h(r, s, t) \) and \( y = k(r, s, t) \). The aim is to determine \( \frac{\partial w}{\partial s} \).
02

Draw Branch Diagram

Draw a branch diagram starting from \( w \). Branch \( w \) into \( x \) and \( y \) (since \( w \) depends on them), and further branch each \( x \) and \( y \) into \( r, s, t \), because they are functions of \( r, s, t \). The path relevant to \( \frac{\partial w}{\partial s} \) passes through both \( x \) and \( y \) with respect to \( s \).
03

Define Chain Rule Formula

The chain rule for partial derivatives here is given by:\[\frac{\partial w}{\partial s} = \frac{\partial w}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial w}{\partial y} \cdot \frac{\partial y}{\partial s}\]
04

Apply the Derivatives

Calculate each component:- \( \frac{\partial w}{\partial x} \) and \( \frac{\partial w}{\partial y} \) are partial derivatives of \( g \) with respect to \( x \) and \( y \), respectively.- \( \frac{\partial x}{\partial s} \) is derived from \( x = h(r, s, t) \), and \( \frac{\partial y}{\partial s} \) from \( y = k(r, s, t) \).Combine the results using the formula from the previous step.

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Key Concepts

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

Partial Derivatives
When dealing with multivariable functions, partial derivatives help us understand how a function changes with respect to one variable while keeping the others constant. This is important in contexts like the given exercise. Suppose we have a function, say, \( w = g(x, y) \), where \( x \) and \( y \) are functions of other variables such as \( r, s, \text{ and } t \). We can find how \( w \) changes with respect to \( s \) using partial derivatives.
In our exercise, to find \( \frac{\partial w}{\partial s} \), we first compute the partial derivatives with respect to the intermediate variables \( x \) and \( y \). These are \( \frac{\partial w}{\partial x} \) and \( \frac{\partial w}{\partial y} \), respectively. Then, we also need \( \frac{\partial x}{\partial s} \) and \( \frac{\partial y}{\partial s} \), which tell us how \( x \) and \( y \) themselves change with \( s \). Combining these using the chain rule allows for the complete computation of \( \frac{\partial w}{\partial s} \). By understanding partial derivatives, we gain insight into complex, dynamic systems in multiple dimensions.
Branch Diagram
A branch diagram is a visual tool that aids in organizing and understanding the relationships between functions and variables in complex systems, commonly used in chain rule applications. Think of it like a family tree, where one function branches into others, showing dependencies and pathways to follow.
In the given exercise, to determine \( \frac{\partial w}{\partial s} \), a branch diagram helps us trace the path of \( w \) as it depends on \( x \) and \( y \), with both of these further depending on \( r, s, \text{ and } t \). The diagram starts at \( w \) and branches into \( x \) and \( y \). Then each branches further into \( r, s, \text{ and } t \). For our exercise, we focus on the branches related to \( s \). This helps visualize which derivatives need to be computed first and follow the correct order of operations with the chain rule.
By using a branch diagram, we can simplify complex ideas and make the calculation process much more manageable.
Multivariable Calculus
Multivariable calculus is an extension of calculus in one dimension to higher dimensions, considering functions with multiple variables. This is where concepts like partial derivatives and the chain rule become vital. When dealing with functions that depend on several variables, we must consider how each variable influences the function independently and together.
In our exercise, the function \( w = g(x, y) \) depends on two variables, \( x \) and \( y \), and then each of those depends on \( r, s, \text{ and } t \). Multivariable calculus allows for the examination of rates of change and interactions between variables through tools like partial derivatives.
By applying the chain rule to compute \( \frac{\partial w}{\partial s} \), we navigate this complexity and find the rate of change with respect to \( s \) while accounting for the interactions of \( r, s, \text{ and } t \) in the transformations of \( x \) and \( y \). The power of multivariable calculus lies in its ability to handle these intricate systems, giving us deeper insights and more accurate predictions in fields ranging from physics to economics.

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Most popular questions from this chapter

In Exercises \(61-64,\) find an equation for the level surface of the function through the given point. $$ f(x, y, z)=\ln \left(x^{2}+y+z^{2}\right), \quad(-1,2,1) $$

Parametrized Surfaces Just as you describe curves in the plane parametrically with a pair of equations \(x=f(t), y=g(t)\) defined on some parameter interval \(I,\) you can sometimes describe surfaces in space with a triple of equations \(x=f(u, v), y=g(u, v), z=h(u, v)\) defined on some parameter rectangle \(a \leq u \leq b, c \leq v \leq d .\) Many computer algebra systems permit you to plot such surfaces in parametric mode. (Parametrized surfaces are discussed in detail in Section 16.5.) Use a CAS to plot the surfaces in Exercises \(77-80 .\) Also plot several level curves in the \(x y\) -plane. $$ \begin{array}{l}{x=2 \cos u \cos v, \quad y=2 \cos u \sin v, \quad z=2 \sin u} \\\ {0 \leq u \leq 2 \pi, \quad 0 \leq v \leq \pi}\end{array} $$

Display the values of the functions in Exercises \(37-48\) in two ways: (a) by sketching the surface \(z=f(x, y)\) and \((b)\) by drawing an assortment of level curves in the function's domain. Label each level curve with its function value. $$ f(x, y)=1-|y| $$

Let $$f(x, y)=\left\\{\begin{array}{ll}{x y \frac{x^{2}-y^{2}}{x^{2}+y^{2}},} & {\text { if }(x, y) \neq 0} \\ {0,} & {\text { if }(x, y)=0}\end{array}\right.$$ a. Show that \(\frac{\partial f}{\partial y}(x, 0)=x\) for all \(x,\) and \(\frac{\partial f}{\partial x}(0, y)=-y\) for all \(y\) b. Show that \(\frac{\partial^{2} f}{\partial y \partial x}(0,0) \neq \frac{\partial^{2} f}{\partial x \partial y}(0,0)\) The graph of \(f\) is shown on page 788 . The three-dimensional Laplace equation $$\frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}+\frac{\partial^{2} f}{\partial z^{2}}=0$$ is satisfied by steady-state temperature distributions \(T=f(x, y, z)\) in space, by gravitational potentials, and by electrostatic potentials. The two-dimensional Laplace equation $$\frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}=0$$ obtained by dropping the \(\partial^{2} f / \partial z^{2}\) term from the previous equation, describes potentials and steady-state temperature distributions in a plane (see the accompanying figure). The plane (a) may be treated as a thin slice of the solid (b) perpendicular to the \(z\) -axis.

Use a CAS to perform the following steps implementing the method of Lagrange multipliers for finding constrained extrema: a. Form the function \(h=f-\lambda_{1} g_{1}-\lambda_{2} g_{2},\) where \(f\) is the function to optimize subject to the constraints \(g_{1}=0\) and \(g_{2}=0\) . b. Determine all the first partial derivatives of \(h\) , including the partials with respect to \(\lambda_{1}\) and \(\lambda_{2},\) and set them equal to \(0 .\) c. Solve the system of equations found in part (b) for all the unknowns, including \(\lambda_{1}\) and \(\lambda_{2}\) . d. Evaluate \(f\) at each of the solution points found in part (c) and select the extreme value subject to the constraints asked for in the exercise. Maximize \(f(x, y, z)=x^{2}+y^{2}+z^{2}\) subject to the constraints \(2 y+4 z-5=0\) and \(4 x^{2}+4 y^{2}-z^{2}=0.\)

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