# s6 - University of Toronto Scarborough Department of...

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University of Toronto Scarborough Department of Computer & Mathematical Sciences MAT B41H 2010/2011 Solutions #6 1. f ( x, y, z, w ) = e xyz sin( xw ) ∂f ∂x = yz e xyz sin( xw ) + w e xyz cos( xw ), 2 f ∂z ∂x = ( y + xy 2 z ) e xyz sin( xw ) + xyw e xyz cos( xw ) and 3 f ∂w ∂z ∂x = ( xy + x 2 y 2 z ) e xyz cos( xw ) + xy e xyz cos( xw ) - x 2 yw e xyz sin( xw ) = (2 xy + x 2 y 2 z ) e xyz cos( xw ) - x 2 yw e xyz sin( xw ). ∂f ∂z = xy e xyz sin( xw ), 2 f ∂w ∂z = x 2 y e xyz cos( xw ), and 3 f ∂x ∂w ∂z = 2 xy e xyz cos( xw )+ x 2 y 2 z e xyz cos( xw ) - x 2 yw e xyz sin( xw ) = (2 xy + x 2 y 2 z ) e xyz cos( xw ) - x 2 yw e xyz sin( xw ). ∂f ∂w = x e xyz cos( xw ), 2 f ∂x ∂w = (1+ xyz ) e xyz cos( xw ) - wx e xyz sin( xw ) and 3 f ∂z ∂x ∂w = ( xy + x 2 y 2 z ) e xyz cos( xw ) + xy e xyz cos( xw ) - x 2 yw e xyz sin( xw ) = (2 xy + x 2 y 2 z ) e xyz cos( xw ) - x 2 yw e xyz sin( xw ). We now observe that 3 f ∂w ∂z ∂x = 3 f ∂x ∂w ∂z = 3 f ∂z ∂x ∂w . 2. (a) We regard z as a function of x and y and differentiate F ( x, y, z ) = 0 w.r.t. x giving ∂F ∂x ∂x ∂x + ∂F ∂y ∂y ∂x + ∂F ∂z ∂z ∂x = 0. Now ∂x ∂x = 1 and ∂y ∂x = 0, since y is independent of x . Hence we have ∂F ∂x + 0 + ∂F ∂z ∂z ∂x = 0 = ∂F ∂z ∂z ∂x = - ∂F ∂x = ∂z ∂x = - ∂F/∂x ∂F/∂z , if ∂F ∂z negationslash = 0, as required. (b) If we differentiate F ( x, y, z ) = 0 w.r.t. y in place of x we will get 0 + ∂F ∂y + ∂F ∂z ∂z ∂y = 0 or ∂z ∂y = - ∂F/∂y ∂F/∂z , if ∂F ∂z negationslash = 0, as required. (c) Using the formulas from parts (a) and (b) we have ∂z ∂x = - y e x 2 e y - e z and ∂z ∂y = - e x + 2 z e y 2 e y - e z .
MATB41H Solutions # 6 page 2 3. f ( x, y ) - tan parenleftbigg x y parenrightbigg . f ( π, 4) = 1. f x = parenleftbigg 1 y parenrightbigg sec 2 parenleftbigg x y parenrightbigg , f x ( π, 4) = 1 2 ; f y = parenleftbigg - x y 2 parenrightbigg sec 2 parenleftbigg x y parenrightbigg , f y ( π, 4) = - π 8 ; f xx = parenleftbigg 2 y 2 parenrightbigg sec 3 parenleftbigg x y parenrightbigg sin parenleftbigg x y parenrightbigg , f xx ( π, 4) = 1 4 ; f xy = - parenleftbigg 2 x y 3 parenrightbigg sin parenleftbigg x y parenrightbigg sec 3 parenleftbigg x y parenrightbigg - parenleftbigg 1 y 2 parenrightbigg sec 2 parenleftbigg x y parenrightbigg , f xy ( π, 4) = - π 16 - 1 8 ; f yy = parenleftbigg 2 x 2 y 4 parenrightbigg sin parenleftbigg x y parenrightbigg sec 3 parenleftbigg x y parenrightbigg + parenleftbigg 2 x y 3 parenrightbigg sec 2 parenleftbigg x y parenrightbigg , f yy ( π, 4) = π 2 64 + π 16 . Now T 2 = f ( π, 4) + f x ( π, 4)( x - π ) + f y ( π, 4)( y - 4) + 1 2 !