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Mathematics

MAT237Y1

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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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