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Mathematics

MAT237Y1

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1. (a) Verify whether or not the following statements are correct. No marks for guessing.
2 2
(i) [3 marks] If S = {( , )∈R :x − ≤1y ≤ 1}\{( ,0): x > 0}, then
int 2 2
S ={(x, y)∈R : x −1< y <1}\{(0,0)}.
int 2 2
NO. S ={(x, y)∈R : x −1< y <1}\{(x,0): x ≥ 0}
(ii) [4 marks] Every continuous function f S ⊂:R → R 2 attains its absolute minimum value
∞
and its absolute maximum value on the set S = U Li, where L ienotes the line segment in
i=1
1 1
R from the origin (0,0) to the point ( , 1 − ) on the circular arc y = 1− x . 2
i i2
NO. By the Extreme Value Theorem, continuous function is sure to attain its absolute
minimum value and its absolute maximum value on the set S if S is compact.
1 1
Consider the sequence of points x = k ( , 1− 2 ) ⊂ S , k→∞x =k(0,1)∉S so by the
k k
Bolzano-Weierstrass S is not compact. Hence the statement is false.
2 1.(c) [5 marks] Consider the area A of the parallelogram generated by the vectors
u = (2,3,0), = ( 1,5,0) Use differentials to answer the following question:
“To which non-zero component of the vectors, v is the value of the area A
most sensitive?”
(That is a small change of that component causes the biggest change of the value of A)
2 3
HINT: Show first that A =
−1 5
2
2 2 2 3 2 3 2 3 2 3
A = u× v =| 0 +0 + − 1 5 = | − 1 5 | = −1 5 since − 1 5 =13 > 0
x y
Consider the functionA( , . , , ) z w = xw− yz Then
dA = wdx − zdy − ydz + xdw. At the point (2,3,−1,5) we get dA = 5dx + dy −3dz + 2dw.
Hence a small change of dx causes the largest change in A, and consequently A is most
sensitive to the entry 2 of the vectoru.
3 2. (a) [4 marks] The direction of the greatest increase in height at some po0nt0P 0x 0 y ,z )
on a hill, 20 per 100m, is towards the southwest. At this point, in the direction toward
the west, how steep is the hill?
Remark: Give the answer in meters per 100m, positive direction of y-axis points north.
Suppose that the hill is in the shape of the graph of the differentiable fuz = f x y)
The greatest increase in height is in the direction of a gradient and this direction is given by
the vector ∇ f = 1 (−1,−1) which is in southwest direction, and the value of the change is
|| f || 2
also given to be20 = || || Hence∇f = 20 ( 1, 1) (−10 2 ,−10 2 ).
100 100 2 100 100
The west is in the direction of= (−1,0), so
10 2 10 2 10 2
∂u= (∇f )⋅u = −( ,− )⋅ −1,0)= . Hence the rate of change in the west
100 100 100
direction is 102 m per 100m (approximately 14m per 100m)
π 20 2 10 2
Remark: The shortest way would be ∂ fW= ∇f (u =)∇f || || cos = =
4 100 2 100
(b) [4 marks] Suppose that the hill of part (a) in some neighborhood of the point
P0(400,200,1000) has the shape of a graph of the differentiable function z = f (x, y).
Can the curve C : ( ) (400 2cos ,200 sin2 ) t ,0 ≤ t ≤ 2π, be a level curve of f ?
Justify your answer.
This can happen only when the tangent to the curve at the point (400,200) is orthogonal
1
to the direction of the gradient u = (−1,−1). The point (400,200) corresponds to
2
π 3π
the values t = and t = on the curve. The tangent vector to the curve at the point
2 2
π ′
t = 2 is v = g (π /2) = (2sint,2cos2tt=π / 2(2,− 2). Sinceu⋅ v = 0, the curve C can
be the level curve of f locally near t =. The tangent vector to the curve at the point
2
3π 3π
t = is v = (−2,−2) so C cannot be a level curve neart = . So the curve C would
2 2
not be a level curve of f globally (it intersects itself).
4 2 2
3. [4 marks each part] Let f (x, y) = x(3x − 2y ) if (x, y) ≠ (0.0) and f (0,0) = 0.
x + y 2
(a) Prove, using the “ ε −δ ” definition of continuity, that f is continuous at t

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