DEPARTMENT OF MATHEMATICS AND APPLIED MATHEMATICS

Mathematics II Advanced Calculus (2AC)

Tutorial Sheet 7: Solutions 17th April 2025

1. Find the cylindrical coordinates (r, θ, z) and the spherical coordinates of the point with cartesian
 √2 √6 √2 
coordinates , , . Explain clearly how you found those coordinates.
8 8 4
 √2 √6 √2 
For the cylindrical coordinates of the point , , , we seek r, θ, z such that
8 8 4
 √ 
2 1

 r cos θ = ,

 cos θ = ,
2
√
 

 √8 


6 2 2 6 8 1 
3
r sin θ = , ⇒ r = + = = and hence sin θ = , .

 √ 8 64 64 64 8 
 √ 2
 z= 2  z= 2

 

 
4 4
Therefore, we find that
√
1 π 2
r= √ , θ= , z= .
2 2 3 4
 √2 √6 √2 
For the spherical coordinates of the point , , , we seek ρ, θ, ϕ such that
8 8 4
 √  √
2 2
ρ cos θ sin ϕ = , cos θ sin ϕ = ,

 

 

 √8 
 √4
6 ⇒ ρ2 = 2 + 6 + 8 = 1 and hence (∗)
 
6
ρ sin θ sin ϕ = , 64 64 64 4 sin θ sin ϕ = ,

 √ 8 
 √ 4
 ρ cos ϕ = 2  cos ϕ = 2

 

 
4 2
So we obtain that
1

sin2 ϕ = , 1
 


 2
√ π  cos θ = ,

π
2 ⇒ ϕ= ⇒ (from (∗) 2
√ ⇒ θ= .

 cos ϕ =
2
, 4  sin θ = 3 ,
 3
2

with 0 ≤ ϕ ≤ π


Therefore, we find that

1 π π
ρ= , θ= , z= .
2 3 4

One could also proceed by inspection, but in that way it is difficult to explain how
the coodinates in the new system are obtained! Notice also that, having found the
coordinates of the point in cylindrical coordinates, one could use the fact that θ in
cylindrical coordinates is the same in spherical and then just find ρ and ϕ.
2. Describe the given sets in the system of cartesian coordinates:
π
(a) A : ϕ = in spherical coordinates. Here we have that 0 ≤ θ ≤ 2π which means complete
3
rotation around the z-axis and also ρ is unbounded from 0 to ∞. Knowing that is a point
(x, y, z) belongs to the set A, then ϕ is the angle between the positive z-axis and the line
joining the point (x, y, z) to the orgin (0, 0, 0). Therefore, the set A is an upward circular
cone. To find the cartesain equation of that cone we just use the fact that
p
x2 + y 2 p π  r1
tan ϕ = ⇒ z = x2 + y 2 / tan = (x2 + y 2 ) .
z 3 3

π
(b) B : ≤ θ ≤ π in cylindrical coordinates.
2
Here, z ∈ R and r ∈ [0, ∞). We find that the set B is the quarter space

x ≤ 0, y ≥ 0, z ∈ (−∞, ∞) .
π
(c) C : θ = in cylindrical coordinates.
4
The set C corresponds to the half-plane x = y with x, y ≥ 0 since x = r cos θ and y = r sin θ.
(d) D = C.

3. A torus T is obtained by rotating a circle C in the xz-plane about the z-axis. If C has radius r
and center (R, 0, 0) with R > r then the parametrization of the torus is given by:

⃗r(s, t) = ((R + r cos s) cos t, (R + r cos s) sin t, r sin s) 0 ≤ s ≤ 2π and 0 ≤ t ≤ 2π .

(i) Show that the torus T is a regular surface.

⃗rs (s, t) = (−r sin s cos t, −r sin s sin t, r cos s), continuous
⃗rt (s, t) = (−(R + r cos s) sin t, (R + r cos s) cos t, 0) continuous .

Now either we evaluate ⃗rs (s, t) × ⃗rt (s, t) and we show that ⃗rs (s, t) × R ⃗ t (s, t) ̸= (0, 0, 0) for
every s, t) ∈ (0, 2π) × (0, 2π) or we notice that ⃗rs (s, t) · ⃗rt (s, t) = 0 for every (s, t) and hence,

⃗rs (s, t) × ⃗rt (s, t) = ⃗rs (s, t) ⃗rt (s, t)

One can easily find that

⃗rs (s, t) = r, ⃗rt (s, t) = R + r cos s

⇒ ⃗rs (s, t) ⃗rt (s, t) = r(R + r cos s) ̸= 0 because r > 0 and R > r while cos s ∈ [−1, 1] .

(iii) Write down the cartesian equation of the tangent plane to the torus T at the point (0, R, r).
π
The given point (0, R, r) corresponds to s = t = . So, the normal vector to the tangent
2
plane is
π π  π π  ⃗i ⃗j ⃗k
⃗rs , × ⃗rt , = 0 −r 0 = (0, 0, −rR) .
2 2 2 2
−R 0 0
 Hence, the tangent plane has cartesian equation z = r.

4. Let S be the surface described by the vector function ⃗r : [0, 3π 2

] × [ π4 , 2π
3
] defined by
h 3π i h π 2π i
⃗r(s, t) := (2 sin s sin t, 3 cos t, cos s sin t) ∀(s, t) ∈ 0, × , .
2 4 3
(a) Show that the surface S is regular. Notice that
⃗rs (s, t) = (2 cos s sin t, 0, − sin s sin t) and ⃗rt (s, t) = (2 sin s cos t, −3 sin t, cos s cos t)
which are continuous and
⃗i ⃗j ⃗k
⃗rs (s, t) × ⃗rt (s, t) = 2 cos s sin t 0 − sin s sin t
2 sin s cos t −3 sin t cos s cos t
= (−3 sin s sin2 t, −2 cos2 s sin t cos t − 2 sin2 s sin t cos t, 6 cos s sin2 t)
= (−3 sin s sin2 t, − sin(2t), −6 cos s sin2 t) .
Now,

 −3 sin s sin2 t = 0,
⃗rs (s, t) × ⃗rt (s, t) = (0, 0, 0) ⇔ − sin(2t) = 0,
6 cos s sin2 t = 0 .

( π
t = (from the second equation) and hence sin t = 1
⇒ 2
sin s = cos s = 0 (from first and second equations)
and this is impossible. Therefore,
h 3π i h π 2π i
⃗rs (s, t) × ⃗rt (s, t) ̸= (0, 0, 0) for every (s, t) ∈ 0, × ,
2 4 3
and hence, being ⃗rs and ⃗rt continuous, we get that the surface S is regular.
(b) Write down a cartesian equation of the surface S.
We set
x = 2 sin s sin t, y = 3 cos t, z = cos s sin t .
Looking at these equations, one could notice that the trig. identity cos2 (α) + sin2 (α) = 1 is
likely to be the most appropriate. The first attempt is to consider evaluating x2 + y 2 + z 2
which gives
x2 + y 2 + z 2 = 4 sin2 (s) sin2 (t) + 9 cos2 (t) + cos2 (s) sin2 (t)
Looking at the terms in the left hand side in the order: first, third and second term, we
rather consider
x2 y 2
+ + z 2 = sin2 (s) sin2 (t) + cos2 (t) + cos2 (s) sin2 (t)
4 9
= (sin2 (s) + cos2 (s)) sin2 (t) + cos2 (t) = sin2 (t) + cos2 (t) = 1 .
Hence, the surface S is a piece of the ellipsoid
x2 y 2
+ + z2 = 1 .
4 9
 (c) Write down
√ the cartesian equation of the tangent plane to the surface S at the point
(0, 3/2, 3/2).
h i h i √
Let us first seek (s, t) ∈ 0, 3π
2
× π 2π
,
4 3
such that ⃗
r (s, t) = (0, 3/2, 3/2). That is
 
 2 sin s sin t = 0,  t = π3 (second equation)
(
s = 0,
3 cos t = 23 , √ ⇒ sin s = 0, ⇒ π
3 t=
cos s sin t = 2 cos s = 1 3
 

√
Now, the normal vector of the tangent plane to the surface S at the point (0, 3/2, 3/2) is
√
⃗rs (0, π/3) × ⃗rt (0, π/3) = (0, − 3/2, −9/2)
and hence the tagent plane has equation
√  √ 
3 3 9 3 √
0(x − 0) − y− − z− =0 ⇔ y + 3z 3 = 6 .
2 2 2 2
5. Given a differentiable function f : R3 → R. The normal to the tangent plane of any level surface
f (x, y, z) = c at the point (x0 , y0 , z0 ) is given by ∇f (x0 , y0 , z0 ).
(i) We have a level surface of the functions f (x, y, z) = x2 + 2xy − y 2 + z 2 . So, the normal to
the tangent plane at the point (−1, 2, −3) is ∇f (−1, 2, −3).
∇f (x, y, z) = (2x + 2y, 2x − 2y, 2z); so, ∇f (−1, 2, −3) = (2, −6, −6). Hence, the tangent
plane at the point (−1, 2, −3) is
2(x + 1) − 6(y − 2) − 6(z + 3) = 0 ⇔ x − 3y − 3z = 2 .

(ii) The tangent line to the curve C at the point (2, 1, 6) is the line of intersection of the tangent
planes to f and g at the point (2, 1, 6).
The tangent planes to f and g at the point (2, 1, 6) have normal vectors (−fx (2, 1), −fy (2, 1), 1)
and (−gx (2, 1), −gy (2, 1), 1). Therefore the vector direction of the tangent line to the curve
C at the point (2, 1, 6) is given by (−fx (2, 1), −fy (2, 1), 1) × (−gx (2, 1), −gy (2, 1), 1) .

fx (x, y) = 2x, fy (x, y) = 4y ⇒ fx (2, 1) = 4, fy (2, 1) = 4,
gx (x, y) = 4x, gy (x, y) = −6y ⇒ gx (2, 1) = 8, gy (2, 1) = −6 .
⃗i ⃗j ⃗k
(−fx (2, 1), −fy (2, 1), 1) × (−gx (2, 1), −gy (2, 1), 1) = −4 −4 1 = (−10, −4, −56) .
−8 6 1
Therefore, we find that the vector equation to the tangent line is
 
⃗r(λ) = (2, 1, 6) + λ (−fx (2, 1), −fy (2, 1), 1) × (−gx (2, 1), −gy (2, 1), 1) , λ∈R
= (2, 1, 6) + λ(−10, −4, −56), λ ∈ R

6. Consider the function f : R2 → R defined by

x(y − 1)3

if (x, y) ̸= (kπ, 1),


sin2 (x) + |y − 1|5/2

f (x, y) := k ∈ Z.


0 if (x, y) = (kπ, 1) .

(a) Show that the function f is continuous at the point (0, 1).
We want to check whether lim f (x, y) = f (0, 1) = 0.
(x,y)→(0,1)
Looking at the structure of the function f , we notice that
|x(y − 1)3 | |x| |y − 1|3 p
0 ≤ |f (x, y)| = ≤ ≤ |x| |y − 1| → 0 as (x, y) → (0, 1) .
sin2 (x) + |y − 1|5/2 |y − 1|5/2
Hence, by the squeeze theorem, we find that lim f (x, y) = 0 = f (0, 1). Thus, the
(x,y)→(0,1)
function f is continuous at the pint (0, 1).
(b) Say whether the partial derivatives fx (01, ) and fy (0, 1) exist. If they exist, then find them.
For the partial derivatives fx (0, 1) and fy (0, 1), we want to find the limits
f (x, 1) − f (0, 1) 0
lim = lim = 0 := fx (0, 1)
x→0 x x→0 x
f (0, 1 + y) − f (0, 1) 0
lim = lim = 0 := fy (0, 1) .
y→0 y y→0 y

(c) Explain clearly why the function f is differentiable at the point (0, 1).
We want to show that
f (x, y) − f (0, 1) − fx (0, 1)(x − 0) − fy (0, 1)(y − 1)
lim p = 0.
(x,y)→(0,1) (x − 0)2 + (y − 1)2
Notice that
f (x, y) − f (0, 1) − fx (0, 1)(x − 0) − fy (0, 1)(y − 1)
lim p
(x,y)→(0,1) (x − 0)2 + (y − 1)2
x(y − 1)3
= lim p
(x,y)→(0,1) (sin2 (x) + |y − 1|5/2 ) × (x − 0)2 + (y − 1)2
(y − 1)3 x
= lim 2 5/2
×p
(x,y)→(0,1) sin (x) + |y − 1| (x − 0)2 + (y − 1)2
Notice that
" #
|(y − 1)3 | |x| |y − 1|3 p
0≤ × ≤ ≤ |y − 1| → 0 as (x, y) → (0, 1) .
sin2 (x) + |y − 1|5/2
p
x2 + (y − 1)2 |y − 1|5/2
| {z }
≤1

Hence, by the squeeze theorem,

f (x, y) − f (01, ) − fx (0, 1)(x − 0) − fy (0, 1)(y − 1)
lim p = 0.
(x,y)→(0,1) (x − 0)2 + (y − 1)2
Therefore, the function f is differentiable at the point (0, 1).
(d) Write down the cratesian equation of the tangent plane to the graph of f at the point (0, 1, 0).
The tangent plane to the graph of f at the point (0, 1, 0) has cartesian equation
z = f (0, 1) + fx (0, 1)(x − 0) + fy (0, 1)(y − 1) = 0 ⇔ z = 0.
| {z } | {z } | {z }
=0 =0 =0

Therefore, the tangent plane is the xy-plane.