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Department
Mathematics
Course
MAT237Y1
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All Professors
Semester
Fall

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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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