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Q14E

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Essential Calculus: Early Transcendentals
Found in: Page 656
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

Draw a tree diagram of the partial derivatives of the function. The functions are\({\rm{R = f(x, y, z, t)}}\), where\({\rm{x = x(u, v, w), y = y(u, v, w), z = z(u, v, w)}}\), and \({\rm{t = t(u, v, w)}}{\rm{.}}\)

The answer is stated below.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: The chain rule.

With the chain rule, the partial derivative of function \({\rm{f(x, y)}}\) with respect to \({\rm{t}}\) becomes \(\frac{{\partial f}}{{\partial t}} = \frac{{\partial f}}{{\partial x}} \cdot \frac{{\partial x}}{{\partial t}} + \frac{{\partial f}}{{\partial y}} \cdot \frac{{\partial y}}{{\partial t}}.\)

Step 2: Use the chain rule for calculation and obtain the diagram.

The functions are\(R = f(x,y,z,t)\), where \(x = x(u,v,w),y = y(u,v,w){\rm{,}}z = z(u,v,w)\) and \(t = t(u,v,w){\rm{.}}\)

Differentiate \(R\) partially with respect to \(u\) as follows.

\(\begin{aligned}{l}\frac{{\partial R}}{{\partial u}} = \frac{\partial }{{\partial u}}f(x,y,z,t)\\ = \frac{{\partial f}}{{\partial x}} \cdot \frac{{\partial x}}{{\partial u}} + \frac{{\partial f}}{{\partial y}} \cdot \frac{{\partial y}}{{\partial u}} + \frac{{\partial f}}{{\partial z}} \cdot \frac{{\partial z}}{{\partial u}} + \frac{{\partial f}}{{\partial t}} \cdot \frac{{\partial t}}{{\partial u}}\\ = {f_x} \cdot {x_u} + {f_y} \cdot {y_u} + {f_z} \cdot {z_u} + {f_t} \cdot {t_u}\end{aligned}\)

Differentiate \(R\) partially with respect to \(v\) as follows.

\(\begin{aligned}{l}\frac{{\partial R}}{{\partial v}} = \frac{\partial }{{\partial v}}f(x,y,z,t)\\ = \frac{{\partial f}}{{\partial x}} \cdot \frac{{\partial x}}{{\partial v}} + \frac{{\partial f}}{{\partial y}} \cdot \frac{{\partial y}}{{\partial v}} + \frac{{\partial f}}{{\partial z}} \cdot \frac{{\partial z}}{{\partial v}} + \frac{{\partial f}}{{\partial t}} \cdot \frac{{\partial t}}{{\partial v}}\\ = {f_x} \cdot {x_v} + {f_y} \cdot {y_v} + {f_z} \cdot {z_v} + {f_t} \cdot {t_v}\end{aligned}\)

Differentiate \(R\) partially with respect to \(w\) as follows.

\(\begin{aligned}{l}\frac{{\partial R}}{{\partial w}} = \frac{\partial }{{\partial w}}f(x,y,z,t)\\ = \frac{{\partial f}}{{\partial x}} \cdot \frac{{\partial x}}{{\partial w}} + \frac{{\partial f}}{{\partial y}} \cdot \frac{{\partial y}}{{\partial w}} + \frac{{\partial f}}{{\partial z}} \cdot \frac{{\partial z}}{{\partial w}} + \frac{{\partial f}}{{\partial t}} \cdot \frac{{\partial t}}{{\partial w}}\\ = {f_x} \cdot {x_w} + {f_y} \cdot {y_w} + {f_z} \cdot {z_w} + {f_t} \cdot {t_w}\end{aligned}\)

Obtain the tree diagram as shown in Figure 1.

In Figure 1, there are two layers, one with partial derivatives of \(R\) with respect to \(x,y,z,t\) and another with partial derivatives of \(x,y,z,t\)with respect to \(u,v,w\) respectively.

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