MATH 105 101
Assignment 1 Solutions
Due date: September 18, 2014
MATH 105 101 Assignment 1 Solutions
1. Find all vectors in R3 of length 3 that are normal to the plane 5x − 12z = 5. (3 marks)
Solution: Let ~v be a vector in R3 of length 3 that is normal to the plane 5x−12z =
5. Then, ~v is parallel to the normal vector of that plane, which is ~n = (5, 0, −12).
So, ~v = c~n for some real number c. Thus, ~v = (5c, 0, −12c). Since ~v has length 3,
we have that:
√
p
3 = |~v | = (5c)2 + 02 + (−12c)2 = 169c2 = 13|c|
3
So, c = ± 13
. Hence, there are exactly two vectors in R3 of length 3 that are normal
to the plane 5x − 12z = 5, which are:
15
36
15
36
, 0, −
,
− , 0,
.
13
13
13
13
2. Find an equation of the plane that is parallel to the xz-plane and passes through the
point (4, 2, −3). (3 marks)
Solution: An equation of the xz-plane is y = 0, and its normal vector is (0, 1, 0).
Since the plane is parallel to the xz-plane, it can be defined using the same normal
vector. So, the equation of the plane that is parallel to the xz-plane and passes
through the point (4, 2, −3) is:
0(x − 4) + 1(y − 2) + 0(z − (−3)) = 0, which can be simplified into y = 2.
3. Determine the domain of the function f (x, y) =
1
√
(ln(y 2 −3))
1−cos (2x)
. (4 marks)
Solution: In order for the function f to be defined, we need to make sure that
p
1 − cos (2x) 6= 0, ln(y 2 − 3) 6= 0 (so that we can divide), y 2 − 3 > 0 (so that
we can take log), and 1 − cos (2x) ≥ 0 (so that we can take the square root). We
proceed to unfold these conditions.
Observe that −1 ≤ cos (2x) ≤ 1 for all values of (x, y), so 1 − cos (2x) ≥ 0 is always
true for any values of (x, y).
Page 1 of 5
MATH 105 101
Assignment 1 Solutions
Due date: September 18, 2014
p
On the other hand, 1 − cos (2x) 6= 0 means that 1 − cos (2x) 6= 0, which is
equivalent to cos (2x) 6= 1. So, 2x 6= 2kπ for any integer k, that is, x 6= kπ for any
integer k.
The condition ln(y 2 − 3) 6= 0 holds for y 2 − 3 6= 1, that is, y 2 6= 4, which means
y 6= ±2.
√
√
Lastly, y 2 − 3 > 0 is equivalent to y 2 > 3. So, y < − 3 or y > 3.
Combining the above conditions, we get that the domain of f is:
√
√
D = {(x, y) ∈ R2 | y 6= ±2, y < − 3 or y > 3, x 6= kπ for any integer k}
4. Determine and sketch the domain of the function f (x, y) = ln (x2 + (y − 1)2 − 9).
Please shade the area of the domain. (3 marks)
Solution: The domain is Df = {(x, y) ∈ R2 : x2 + (y − 1)2 − 9 > 0}, which consists
of those points lying strictly outside the circle centered at (0, 1) with radius 3.
1
5. Find and sketch the level curve of the surface z = f (x, y) = xy − 3 ln (5 − x) for z = 0.
(3 marks)
Solution: The domain of f is:
Df = {(x, y) ∈ R2 : x < 5, y 6= 0}
To find the level curve at z = 0, we solve for (x, y) such that
1
xy − 3 ln (5 − x) = 0.
So, x = 0, or ln (5 − x) = 0. Solving ln (5 − x) = 0, we get 5 − x = 1, so x = 4.
Hence, the level curve at z = 0 consists of two lines x = 0 and x = 4 excluding the
points (0, 0) and (4, 0) (since any point with y = 0 does not lie in the domain of
f ).
2
6. Find the equations of the xy−, yz− and xz−traces of the surface 4x2 − y − (z+1)
= 0.
9
(6 marks)
Page 2 of 5
MATH 105 101
Assignment 1 Solutions
Due date: September 18, 2014
Solution: The xy-trace is the intersection of the surface with the xy-plane z = 0.
Its equation is 4x2 − y − 91 = 0 or y = 4x2 − 91 , which is a parabola being stretched
vertically and shifted downwards by 19 .
The yz-trace is the intersection of the surface with the yz-plane x = 0. Its equation
2
2
= 0 or y = − (z+1)
, which is an inverted parabola being compressed
is −y − (z+1)
9
9
vertically and shifted to the left along the z-axis by 1.
The xz-trace is the intersection of the surface with the xz-plane y = 0. Its equation
2
is 4x2 − (z+1)
= 0, which is a pair of intersecting lines.
9
Page 3 of 5
MATH 105 101
Assignment 1 Solutions
Due date: September 18, 2014
7. Sketch the graph of the function g(x, y) = (x + 1)2 + y 2 − 1, and label at least 3 level
curves. (3 marks)
Solution: To graph the function g(x, y) = (x + 1)2 + y 2 − 1, we will first look at
some of the level curves:
• For z = 0, we get (x + 1)2 + y 2 = 1, which is a circle of radius 1, centered at
(−1, 0).
• For z = 3, we get (x + 1)2 + y 2 = 4, which is a circle of radius 2, centered at
(−1, 0).
Page 4 of 5
MATH 105 101
Assignment 1 Solutions
Due date: September 18, 2014
• For z = 8, we get (x + 1)2 + y 2 = 9, which is a circle of radius 3, centered at
(−1, 0).
So, after sketching those level curves, we get a general idea of the graph of the
surface. In fact, the graph of the function g(x, y) is a paraboloid with vertex at
(−1, 0, −1).
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