2xy
.
x2 + y 2
Solution. As f is given by quotient of two polynomials functions, we can
evaluate the polynomials for any (x, y) ∈ R2 . However, if the denominator x2 + y 2 ̸= 0, one can evaluate f (x, y), and hence the domain of f =
{ (x, y) | x2 + y 2 ̸= 0} = R2 \ {(0, 0)}, i.e. the set of all points in R2 except (0, 0).
(ii) For any point (x, y) ∈ D, if either one of x and y is zero, then f (x, y) = 0.
Suppose that m = xy where x and y are not zero, then we have the following
2mx2
2m
f (x, y) = f (x, mx) = 2
=
. In order to determine the range R,
2
2
x +m x
1 + m2
we apply the discriminant method as follows: for any value u in the range, one
2m
has u = 1+m
2 for some real number m ∈ R, then one knows that the quadratic
equation um2 − 2m + u = 0 in m has at least one real root, so its discriminant
∆ = (−2)2 − 4(u)(u) = 4(1 − u2 ) ≥ 0. Hence one has −1 ≤ u ≤ 1. So the range
of f is the closed interval [−1, 1],
Remark. It would be easier to use the polar coordinates (x, y) = (r cos θ, r sin θ).
1. Determine (i) the domain and (ii) the range of the function f (x, y) =
2. Sketch the k-level curve the function of g(x, y) = 9x2 − 4y 2 if k = −9, −4, 0, 4, 9.
Solution. Recall that the level curve Lc of g(x, y) at c is given by
the set of points (x, y) which takes g-value equal to c, i.e. Lc =
{ (x, y) | g(x, y) = c }.
(x, y) lying on or outside the ellipse 4x2 + 9y 2 = 36.
(ii) The domain of g(x, y, z) consists of all points in 3-dimensional space R3 except
the origin (0, 0, 0), i.e. Dom(f ) = R2 \ {(0, 0, 0)}.
6. Draw the graph of (i) f (x, y) = x and (ii) g(x, y) = y 2 .
Solution. Recall that the graph of z = f (x, y) is the set of points in R3 given
by (x, y, f (x, y)), where (x, y) is the input and the third coordinate f (x, y) is the
corresponding output.
(i)
(ii)
7. Consider the line ℓ : x = 3 − t, y = −1 + 2t, z = 5 + 8t. Find the equations of
the projection of the line ℓ onto (i) the xy-plane.
Solution. (i) Let P (x, y, z) be any point in the space, after projecting it onto
the xy-plane, its image Q(x, y, 0). So the image ℓ0 of the line ℓ is given by
ℓ0 : x = 3 − t, y = −1 + 2t, z = 0.
3. √
Describe (i) the domain and (ii) the graph of the function f (x, y) = −1 +
16 − x2 − y 2 .
Solution. (i) The domain D of f (x, y) is given by 16 − x2 − y 2 ≥ 0, i.e. D is the
closed circular disc with center at (0, 0) with radius 4.
(ii) The graph of f is a set { (x, y,√
f (x, y) ) | (x, y) ∈ D }, so if (x, y, z) in the graph
of f, one has z = f (x, y) = −1 + 16 − x2 − y 2 , i.e. (z + 1)2 = 16 − (x2 + y 2 ), i.e.
the graph of f consists of the points in the upper hemisphere x2 +y 2 +(z−1)2 = 42
with center at (0, 0, 1) and radius 4.
8. Consider the line ℓ : x = 3 − t, y = −1 + 2t, z = 5 + 10t. (i) Determine the point
P where the line ℓ meets the plane S : x − y + 3z + 8 = 0.
(ii) Find the equation(s) of the projection of the line ℓ onto the plane
S : x − y + 3z + 8 = 0.
Solution. (i) First we find the point P0 of intersection of the line ℓ and the plane
S as follows: 0 = (3 − t) − (−1 + 2t) + 3(5 + 10t) + 8 = 27t + 27 i.e. 27t = −27
and t = −1. The intersection point is P (x, y, z) = (2, −3, −5). (ii) The image of
the projection of the line ℓ onto S means that any point Q on ℓ will projected to
S along the normal direction n of S, hence the image R lies on S, and the vector
−−→
QR is parallel to n.
4. Let f (x, y) = (x2 + y 2 )1/3 be a function. Find out the domain and the range of
the function f (x, y).
Solution.
For any point (x, y) in xy-plane,
i.e. (x, y) ∈ R2 , we have x2 + y 2 ≥ 0,
√
√
3
3
2
2
2
so x + y is a real number and x + y 2 ≥ 0, and hence f (x, y) ≥ 0. From
this, we know domain of f is the entire xy-plane R2 . And we expect the range
of f is equal to the set of all non-negative real numbers, i.e. any non-negative
number ℓ (ℓ ≥ 0) can be expressed
as the value√of f evaluated at some point (x, y)
√
3
in R2 . In fact, f (ℓ3/2 , 0) = 3 (ℓ3/2 )2 + 02 = ℓ3 = ℓ.
√
5. Determine the domain of (i) f (x, y) =
4x2 + 9y 2 − 36, and g(x, y, z) =
√ 2xyz2 2 .
x +y +z
√
Solution. (i) The domain of f (x, y) = 4x2 + 9y 2 − 36 consists of all the points
When t = 0, we have a point Q(3, −1, 5) on ℓ, but Q is not on the plane S.
It follows from the equation of S that n = (1, −1, 3) is normal to S. Let ℓ′ be
a line through Q with direction vector n, then the equation of ℓ′ is given by
−−→
OQ + tn = (3 + t, −1 − t, 5 + 3t). Suppose ℓ0 meets the plane S at a point R,
then 0 = (3 + t) − (−1 − t) + 3(5 + 3t) + 8 = 11t + 27, i.e. t = 27/11, and so
R(3+27/11, −1−27/11, 5+3×27/11) = (60/11, −38/11, 136/11). As both points
P (2, −3, −5) and R(60/11, −38/11, 136/11) are on S, then the ℓ′ is the line P R,
y+3
x−2
z+5
Then the equation of the line ℓ0 : is given by 60/11−2
= −38/11+3
= 136/11+5
.
The following are taken from an old test paper.
1
x2
9. Let u(x, t) = √ exp(− ). The domain of u is
4t
t
A. the entire xy-plane R2 B. { (x, t) | x ∈ R, and t > 0 }
C. { (x, t) ∈ R2 | t ̸= 0 } D. { (x, t) | x > 0, and t > 0 } E. None of above
Solution.
(x + y)2
is
x2 + y 2
B. R2 \ {(0, 0)} C. x2 + y 2 ≤ 1
10. The domain of the function f (x, y) =
A. the entire xy-plane R2
E. None of the above
4xy
is
x2 + y 2
A. [−1, 1] B. [−2, 2] C. all positive real numbers
E. None of the above.
D. x2 + y 2 ≥ 1
11. The range of f (x, y) =
D. all negative real numbers