Functions of Several Variables: Chain Rules

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Transcription:

Functions of Several Variables: Chain Rules Calculus III Josh Engwer TTU 29 September 2014 Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 1 / 30

PART I PART I: MULTIVARIABLE CHAIN RULES Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 2 / 30

1-1 Chain Rule (from Calculus I) Let y = f (x) C 1 where x = g(t) C 1. Determine the chain rule formula for dy dt. Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 3 / 30

1-1 Chain Rule (from Calculus I) Let y = f (x) C 1 where x = g(t) C 1. Determine the chain rule formula for dy dt. First, sketch the dependency tree for y: Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 4 / 30

1-1 Chain Rule (from Calculus I) Let y = f (x) C 1 where x = g(t) C 1. Determine the chain rule formula for dy dt. Slide down the tree: Then, dy dt = dy dx dx dt Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 5 / 30

1-1 Chain Rule (from Calculus I) Proposition Let y = f (x) C 1 where x = g(t) C 1. Then: dy dt = dy dx dx dt 1-1 means 1 intermediate variable (x) and 1 independent variable (t). Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 6 / 30

1-2 Chain Rule Let z = f (x) C 1 where x = g(s, t) C (1,1). Determine the chain rule formulas for s & t. Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 7 / 30

1-2 Chain Rule Let z = f (x) C 1 where x = g(s, t) C (1,1). Determine the chain rule formulas for s & t. First, sketch the dependency tree for z: Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 8 / 30

1-2 Chain Rule Let z = f (x) C 1 where x = g(s, t) C (1,1). Determine the chain rule formulas for s & t. Slide down the branch of the tree that has s as its bottom node (in red): Then, s = dz x dx s Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 9 / 30

1-2 Chain Rule Let z = f (x) C 1 where x = g(s, t) C (1,1). Determine the chain rule formulas for s & t. Slide down the branch of the tree that has t as its bottom node (in red): Then, t = dz x dx t Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 10 / 30

1-2 Chain Rule Proposition Let z = f (x) C 1 where x = g(s, t) C (1,1). s = dz x dx s Then: t = dz x dx t 1-2 means 1 intermediate variable (x) and 2 independent var s (s, t). Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 11 / 30

2-1 Chain Rule Let z = f (x, y) C (1,1) where x = g(t) C 1 and y = h(t) C 1. Determine the chain rule formula for dz dt. Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 12 / 30

2-1 Chain Rule Let z = f (x, y) C (1,1) where x = g(t) C 1 and y = h(t) C 1. Determine the chain rule formula for dz dt. First, sketch the dependency tree for z: Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 13 / 30

2-1 Chain Rule Let z = f (x, y) C (1,1) where x = g(t) C 1 and y = h(t) C 1. Determine the chain rule formula for dz dt. Slide down each branch of the tree that has t as its bottom node (in red). Then, dz dt = dx x dt + dy y dt Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 14 / 30

2-1 Chain Rule 2-1 means 2 intermediate var s (x, y) and 1 independent variable (t). Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 15 / 30 Proposition Let z = f (x, y) C (1,1) where x = g(t) C 1 and y = h(t) C 1. Then: dz dt = dx x dt + dy y dt

2-2 Chain Rule Let z = f (x, y) C (1,1) where x = g(s, t) C (1,1) and y = h(s, t) C (1,1). Determine the chain rule formula for s & t. Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 16 / 30

2-2 Chain Rule Let z = f (x, y) C (1,1) where x = g(s, t) C (1,1) and y = h(s, t) C (1,1). Determine the chain rule formula for s & t. First, sketch the dependency tree for z: Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 17 / 30

2-2 Chain Rule Let z = f (x, y) C (1,1) where x = g(s, t) C (1,1) and y = h(s, t) C (1,1). Determine the chain rule formula for s & t. Slide down each branch of the tree that has s as its bottom node (in red). s = x x s + y y s Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 18 / 30

2-2 Chain Rule Let z = f (x, y) C (1,1) where x = g(s, t) C (1,1) and y = h(s, t) C (1,1). Determine the chain rule formula for s & t. Slide down each branch of the tree that has t as its bottom node (in red). t = x x t + y y t Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 19 / 30

2-2 Chain Rule Proposition Let z = f (x, y) C (1,1) where x = g(s, t) C (1,1) and y = h(s, t) C (1,1). Then: s = x x s + y y s t = x x t + y y t 2-2 means 2 intermediate var s (x, y) and 2 independent var s (s, t). Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 20 / 30

1-3 Chain Rule Proposition Let z = f (x) C 1 where x = g(r, s, t) C (1,1,1). Then: r = dz x dx r s = dz x dx s t = dz x dx t 1-3 means 1 intermediate variable (x) and 3 independent var s (r, s, t). Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 21 / 30

3-1 Chain Rule Proposition Let w = f (x, y, z) C (1,1,1) s.t. x = g(t) C 1, y = h(t) C 1, z = p(t) C 1. Then: dw dt = w dx x dt + w y dy dt + w dz dt 3-1 means 3 intermediate var s (x, y, z) and 1 independent variable (t). Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 22 / 30

2-3 Chain Rule Proposition Let z = f (x, y) C (1,1) s.t. x = g(r, s, t) C (1,1,1), y = h(r, s, t) C (1,1,1). Then: r = x x r + y y r s = x x s + y y s t = x x t + y y t 2-3 means 2 intermediate var s (x, y) and 3 independent var s (r, s, t). Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 23 / 30

3-2 Chain Rule Proposition Let w = f (x, y, z) C (1,1,1) s.t. x = g(s, t), y = h(s, t), z = p(s, t). Then: w s = w x x s + w y y s + w s w t = w x x t + w y y t + w t 3-2 means 3 intermediate var s (x, y, z) and 2 independent var s (s, t). Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 24 / 30

3-3 Chain Rule Proposition Let w = f (x, y, z) C (1,1,1) s.t. x = g(r, s, t), y = h(r, s, t), z = p(r, s, t). Then: w r = w x x r + w y y r + w r w s = w x x s + w y y s + w s w t = w x x t + w y y t + w t 3-3 means 3 intermediate var s (x, y, z) and 3 independent var s (r, s, t). Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 25 / 30

PART II PART II: IMPLICIT DIFFERENTIATION Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 26 / 30

Implicit Differentiation (2-Variable Function) Proposition Let F(x, y) = 0 s.t. F C (1,1) and y is implicitly a function of x. dy dx = F x F y, provided F y 0 Then: df/dx = 0 = F x + F y = F y = dy dy dx = 0 dy dx = F x F dx = x F y = F x F y Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 27 / 30

Implicit Differentiation (3-Variable Function) Proposition Let F(x, y, z) = 0 s.t. F C (1,1,1) and z is implicitly a function of (x, y). Then: x = F x F z, provided F z 0 y = F y F z, provided F z 0 F/ x = 0 = F x + F = F = x = 0 x = F x F x = x F ( Similarly for / y) = F x F z Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 28 / 30

Related Rates...Revisited WEX 11-5-1: A rectangle is changing in such a way that its length l is decreasing at a rate of 4 mm/min and its width w is increasing at a rate of 10 mm/min. At what rates are its area & perimeter changing when the length is 5 mm and the width is 12 mm? 1 st, realize that the independent variable is time (t). 2 nd, recall the formulas for area & perimeter: A = lw P = 2l + 2w. 3 rd, use 2-1 Chain Rule: da = A dl dt l dt + A dw dp = P w dt dt l 4 th, extract info: l = 5, w = 12, dl dt 5 th, compute the partials: A da dt dp dt = A dl l = P l dt + A dw = w dt dl dt + P dw = w dt dw = 4, = 10. dt l = w = 12, A w = l = 5 )( ) ( )( ) 4 + 5 10 = 2 mm 2 /min ( 12 ( )( ) ( )( ) 2 4 + 2 10 = 12 mm/min P l = 2, dl dt + P w dw dt P w = 2 Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 29 / 30

Fin Fin. Josh Engwer (TTU) Functions of Several Variables: Chain Rules 29 September 2014 30 / 30

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