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40. Prove the mean value theorem for functions of two variables.

40. Prove the mean value theorem for functions of two variables.

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
6th Edition
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
Chapter5: A Survey Of Other Common Functions
Section5.4: Combining And Decomposing Functions
Problem 14E: Decay of Litter Litter such as leaves falls to the forest floor, where the action of insects and...
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6.40) my professor says I have to explain the steps in the solved problems in the picture. Not just copy eveything down from the text.

Mean value theorem 6.40. Prove the mean value theorem for functions of two variables. Let f(t) = f(x, + ht, Yo + kt). By the mean value theorem for functions of one variable, F(1) = F(0) = F'(0) 0 < 0 < 1 (1) CHAPTER 6 Partial Derivatives 151 If x = x, + ht, y = y + kt, then F(t) = f(x, y), so that by Problem 6.17, F'(1) =f. (dv/dt) +f,(dy/dt) = hf, + kf, and F(0) = hf,(xo + Oh, yo + Ok) + kf,(xo + Oh, yo + Ok) where 0 <0<1. Thus, (1) becomes f(x, + h, yo + k) – f (Xp» Yo) = hf,(x, + Oh, yo + Ok) + kf,(x, + 0h, yo + Ok) (2) where 0 <0<1 as required. Note that Equation (2), which is analogous to Equation (1) of Problem 6.14, where h= Ax, has the advan- tage of being more symmetric (and also more useful), since only a single number 0 is involved.
Transcribed Image Text:Mean value theorem 6.40. Prove the mean value theorem for functions of two variables. Let f(t) = f(x, + ht, Yo + kt). By the mean value theorem for functions of one variable, F(1) = F(0) = F'(0) 0 < 0 < 1 (1) CHAPTER 6 Partial Derivatives 151 If x = x, + ht, y = y + kt, then F(t) = f(x, y), so that by Problem 6.17, F'(1) =f. (dv/dt) +f,(dy/dt) = hf, + kf, and F(0) = hf,(xo + Oh, yo + Ok) + kf,(xo + Oh, yo + Ok) where 0 <0<1. Thus, (1) becomes f(x, + h, yo + k) – f (Xp» Yo) = hf,(x, + Oh, yo + Ok) + kf,(x, + 0h, yo + Ok) (2) where 0 <0<1 as required. Note that Equation (2), which is analogous to Equation (1) of Problem 6.14, where h= Ax, has the advan- tage of being more symmetric (and also more useful), since only a single number 0 is involved.
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