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Department
Mathematics
Course
MAT237Y1
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Fall

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2 3 2 u 2 2 1.(a) [5 marks] Define f : R → R by f(u,v) = (u + v, e ,uv + v). Suppose g : R → R 1 1 −1  is of class C , g(1,1) = (0,1), anDg(1,1) =  . Compute D (f g)(1,1). 2 1  2u 1  1 − 1 0 1  1 − 1 2 1  D(f g)(1,1) = Df(0,1)Dg(1,1) = e u 0    = 1 0   = 1 −1    2 1    2 1     v u +1 (0,1) 1 1 3 0  2 2 ∂ u (b) [5 marks] Suppose u = ( , ),x = 2 ,y = st. Assuming f is of class C , find ∂s∂t in terms of x , y and the partial derivatives of f over x and y . ∂u ∂f ∂x ∂f ∂y ∂f ∂f ∂f = + = 0+ s =s ∂t ∂x ∂t ∂y ∂t ∂x ∂y ∂y ∂ ∂ u ∂ ∂f ∂f ∂ 2f ∂ x ∂ f ∂y ∂f ∂ f ∂ f ( )= (s ) = + s( + 2 )= +2s + st 2 = ∂s ∂t ∂s ∂y ∂y ∂x y s ∂y ∂s ∂y ∂x∂y ∂y 2 2 = ∂f + x ∂ f + y ∂ f ∂y ∂x y ∂y 2 2 1 2. Assume that the equation F(x, y.z) = 0, where F is of class C , defines z implicitly as a function of x and y , that is z = g(x, y). Suppose F(1,−2,2) = 0 and∇F(1,−2,2) = (1,3,2). (a) [4 marks] Evaluate ∂ g(1,−2) in the direction of u= 1 ( 1,1). u 2 ∇F(1,−2,2) = (1,3,2) ⇒ ∂F (1,−2,2) =1, ∂F (1,−2,2) = 3 , ∂F (1,−2,2) = 2 ∂x ∂y ∂z Hence ∂g (1,−2) = −1 , ∂g (1,−2) = − 3 and ∂x 2 ∂y 2 ∂ g(1,−2) = (− 1 ,− 3)⋅ 1 ( 1,1) = − 1 u 2 2 2 2 (b) [6 marks] Suppose the surface S is parametrized by f(u,v) = (eu+v,u −v,u + v). Is the surfaceS 1 z = g(x, y) tangent toS at the point (1,−2,2) ? Justify. Remark: Two surfaces S and S are tangent at the point ∈S ∩ S if the tangent planes 1 1 at a to both surfaces coincide. By the assumption (1,−2,2)∈S and (1,−2,2)∈S since it corresponds to (u,v) = (−1,1). 1 The normal to the surface S1 at the point (1,−2,2) isn1 =∇F(1,−2,2) = (1,3,2). The normal to S is i j k i j k ∂f ∂f u+v n = ( ∂u × ∂v)(−1,1) = e 1 2u = 1 1 − 2 =(−1,−3,−2) eu+v −1 1 1 −1 1 −1,1) and it is parallel n1 . Hence the tangent planes coincide and the surfaces are tangent to each other at the point (1,−2,2). 3 3. (a)[6 marks] Suppose that u and v are functions of x and y (i.e. u = u(x, y), v = v(x, y)) defined implicitly by the system of equations xy +uv − = 0 , xu + yv − 2 = 0. Find ∂u (0,1) if it is defined. If it is not defined, explain why. ∂y First note that substitutix = 0 and y =1 to the system of equations we get v = 2 and u =1 . Hence considering the first equation as F(x, y,u,v) = 0 and the second as ∂( F G v u G(x, y,u,v) = 0 we get (0,1,1,2) = = 2⋅1−0⋅1= 2 ≠ 0, so the system ∂(u v) x y (0,1,1,2) can be solved for u and v in terms of x and y in the neighborhood of (0,1,1,2) by Implicit
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