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7. Suppose that f (x, y) = e" cos TY and u? – v², y = u? + v?. x = Using the chain rule, compute the values of af af ди at the point (u, v) = (1, 1). Solution. • According to the chain rule af Əx af Əy af dy dy du' af ôf Əx fe ди Əx du dv Əx dv dy dv • We have af e cos TY, af = -Te" sin TY dy and ду dy 2и, -2v, 2и, = 2v. %3D ди du • At (u, v) = (1,1) we have (x, Y) (0, 2) and fe e° cos 27 = 1, af — те' sin 2т — ду ду 0, dy 2, du 2, -2, = 2. • It follows that af af = 1. 2+0· 2 = 2, = 1.(-2) +0 ·2 = -2. dv

7. Suppose that f (x, y) = e" cos TY and u? – v², y = u? + v?. x = Using the chain rule, compute the values of af af ди at the point (u, v) = (1, 1). Solution. • According to the chain rule af Əx af Əy af dy dy du' af ôf Əx fe ди Əx du dv Əx dv dy dv • We have af e cos TY, af = -Te" sin TY dy and ду dy 2и, -2v, 2и, = 2v. %3D ди du • At (u, v) = (1,1) we have (x, Y) (0, 2) and fe e° cos 27 = 1, af — те' sin 2т — ду ду 0, dy 2, du 2, -2, = 2. • It follows that af af = 1. 2+0· 2 = 2, = 1.(-2) +0 ·2 = -2. dv

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Can you explain the steps in computing the circled part. Baby steps pls

7. Suppose that f (x, y) = e® cos TY and x = u? – v?, y = u? + v². Using the chain rule, compute the values of af af ди at the point (u, v) = (1, 1). Solution. • According to the chain rule af əx dv ôf ðy + af dy af af əx Əx du af Əx dv • We have af af e cos TY, -Te" sin TY ду and ду ду 2u, 2и, -2v, 2v. du ди dv • At (u, v) = (1,1) we have (x, y) (0, 2) and af e° cos 27 af —те' sin 2т 3D0, 1, ду dy ду 2, ди = 2. 2, • It follows that af = 1. 2+0. 2 = 2, ди af = 1. (-2) +0 · 2 = -2. dv ||
Transcribed Image Text:7. Suppose that f (x, y) = e® cos TY and x = u? – v?, y = u? + v². Using the chain rule, compute the values of af af ди at the point (u, v) = (1, 1). Solution. • According to the chain rule af əx dv ôf ðy + af dy af af əx Əx du af Əx dv • We have af af e cos TY, -Te" sin TY ду and ду ду 2u, 2и, -2v, 2v. du ди dv • At (u, v) = (1,1) we have (x, y) (0, 2) and af e° cos 27 af —те' sin 2т 3D0, 1, ду dy ду 2, ди = 2. 2, • It follows that af = 1. 2+0. 2 = 2, ди af = 1. (-2) +0 · 2 = -2. dv ||
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