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Material Type: Exam; Class: Calculus III--Multivariable; Subject: Mathematics; University: Colgate University; Term: Fall 2005;
Typology: Exams
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Math 113 – Calculus III Exam 1 Practice Problems Fall 2005
Name:
Contour f (x, y) Graph Diagram
x^2 +
3 y 2
x^2 −
y^2
x +
y 2
− x +
y 2
sin(πx)
xy^2 √
x^2 +
3 y 2
x^2
(^2) −y+z
(a) f (x, y) =
sin(x + y) x − y
(b) g(x, y) =
x^2 + y^2 + 1
(c) h(x, y) =
xy x^2 +y^2 (x, y)^6 = (0,^0 0 (x, y) = (0, 0)
z = x^2 − y − xey^ − 1 ,
which we may rewrite as x^2 − y − xey^ − z − 1 = 0, or, by adding 5 to both sides, x^2 − y − xey^ − z + 4 = 5. So we can choose
g(x, y, z) = x^2 − y − xey^ − z + 4.
(b) i. g(x, y, z) = x^2 + x + y^4 + z(z − 1) NO. If we try to solve g(x, y, z) = 0 for z, we first find
z^2 − z + x^2 + x + y^4 = 0,
and to solve this for z, we have to use the quadratic formula:
z =
1 − 4(x^2 + x + y^4 )
which means there will be two z values for each point (x, y) where the expression inside the square root is not negative. So we can not express the surface as the graph of a single function f (x, y). ii. g(x, y, z) = sin(z − y + 2x) NO. The level surface g(x, y, z) = 0 is sin(z − y + 2x) = 0, which implies z−y+2x = nπ, where n is some integer. That is, the level set g(x, y, z) = 0 consists of infinitely many planes of the form
z = − 2 x + y + nπ.
Since there is not a single function of x and y whose graph is this level set, the surface can not be expressed as the graph of a function f (x, y). iii. g(x, y, z) = 1 − ex
(^2) −y+z
YES. Solve for z: g(x, y, z) = 0 =⇒ 1 − ex (^2) −y+z = 0 =⇒ ex (^2) −y+z = 1 =⇒ x^2 − y + z = ln(1) = 0 =⇒ z = y − x^2. So the surface g(x, y, z) = 0 is the graph of f (x, y) = y − x^2.
sin(x + y) x − y The numerator and denominator are both continuous, so the only points where the function would not be continuous are where the denominator is
(b) g(x, y) =
x^2 + y^2 + 1 The numerator and denominator are both continuous, and the denominator is never 0, so there are no points where this function is not continuous.
(c) h(x, y) =
xy x^2 +y^2 (x, y)^6 = (0,^0 0 (x, y) = (0, 0) For (x, y) 6 = (0, 0), the function is a quotient of polynomials, and the de- nominator is not 0 if (x, y) 6 = (0, 0), so the function is continuous for all (x, y) 6 = (0, 0). To determine if the function is continuous at (0, 0), we must first determine if lim(x,y)→(0,0) h(x, y) exists. Let’s check a few paths towards (0, 0). Along the x axis, we have y = 0, and
lim x→ 0 h(x, 0) = lim x→ 0
x^2
Along the y axis, we have x = 0, and we find the same limit, 0. Let’s try along the line y = x. Then
lim x→ 0
h(x, x) = lim x→ 0
x^2 2 x^2
= lim x→ 0
We see that the function approaches the value 0 along the x axis (and along the y axis), but approaches the value 1 / 2 along the line y = x. Since the function approaches different values along two different paths to (0, 0), the limit lim(x,y)→(0,0) h(x, y) does not exist. Therefore the function is not continuous at (0, 0).
