2008 Solution - University of Toronto Scarborough...

University of Toronto Scarborough
Department of Computer & Mathematical Sciences
MAT B41H 2008/2009
Term Test Solutions
1. (a) From the lecture notes we have
Let f : U ⊂ R n → R k be a given function. We say that f is
differentiable at a ∈ U if the partial derivatives of f exist at a and if
lim
x
→
a
bardbl f ( x ) − f ( a ) − Df ( a ) ( x − a ) bardbl
bardbl x − a bardbl
= 0 ,
where Df ( a ) is the k × n matrix
parenleftbigg
∂f
i
∂x
j
parenrightbigg
evaluated at a .
Df ( a ) is called the derivative of f at a .
(b) From the lecture notes we have
Extreme Value Theorem . Let D be a compact set in R n and let
f : D ⊂ R n → R be continuous. Then f assumes both a (global)
maximum and a (global) minimum on D .
2. (a) lim
( x,y ) → (1 , - 1)
f ( x, y ) = 3 does not say anything about f (1 , − 1). The limit
says something about what happens as a point is approached, not what
happens at that point.
( x + y ) 2
3

MATB41H Term Test Solutions page 2
3. f ( x, y ) =
x
x
2
+ y
2
.
Domain is { ( x, y ) ∈ R 2 | ( x, y ) negationslash = (0 , 0) } .
Putting f ( x, y ) = c we have
x
x 2 + y 2
= c . For
c = 0, the level curve is x = 0. For c negationslash = 0,
we have x = c ( x
2
+ y
2
) ⇐⇒ x
2
−
x
c
+ y
2
=
0 ⇐⇒
parenleftbigg
x −
1
2 c
parenrightbigg
2
+ y
2
=
parenleftbigg
1
2