University of Cape Town

Department of Mathematics and Applied Mathematics

Mathematics II Advanced Calculus (2AC)
Class Test 1 15th March 2012

Solutions

Section A

1. Given the curve C described by the vector function

r(t) := (t sin t + cos t, t cos t − sin t, 2t), 0 ≤ t ≤ 2.

z2
(a) Show that the curve C lies on the surface with equation x2 + y 2 − = 1.
4
We need to show that (t sin t + cos t)2 + (t cos t − sin t)2 + 41 (2t)2 = 1 for every t ∈ [0, 2π].XX
[2]
(b) The point (−1, −π, 2π) corresponds to t = π.X
r 0 (t) = (sin t + t cos t − sin t, cos 0
√ t − t sin t − cos t, 2) = (t cos t, −t sin t, 2). Hence, r (π) =
0
(−π, 0, 2) and hence |r (π)| = π 2 + 4. Therefore
1
T (π) = √ (−π, 0, 2)X
π2 + 4
[2]
√
4π 2 + 4 + π 4
(c) Show that the curvature of the curve C at the point in (a) is equal to .
(π 2 + 4)3/2
r 00 (t) = (cos t − t sin t, − sin t − t cos t, 0). So, r 00 (π) = (−1, π, 0)X. Thus,
¯ ¯
¯ i j k ¯
¯ ¯
¯ ¯
¯ ¯
r (π) × r (π) = ¯¯ −π 0 2 ¯¯ = (−2π, −2, −π 2 ) .X
0 00

¯ ¯
¯ ¯
¯ −1 π 0 ¯

So, √
|r 0 (π) × r 00 (π)| 4π 2 + 4 + π 4
κ(π) = = .X
|r 0 (π)|3 (π 2 + 4)3/2
[3]
(d) Write down the cartesian equation of the normal plane to the curve C at the point in (a).
The normal vector to the normal to the curve C at the point (−1, −π, 2π) is the vector r 0 (π).
So we get the cartesian equation

−π(x + 1) + 2(z − 2π) = 0 ⇔ −xπ + 2z = 5π .XX

[2]
2. Find the value(s) of the real number λ such that the function
 x

 (e − 1) sin y
 if x 6= 0 and y 6= 0,
f (x, y) := xy


 λ if x = 0 or y = 0

is continuous at (0, 0).

· x ¸
(ex − 1) sin y e −1 sin y
lim f (x, y) = lim = lim × lim = 1X .
(x,y)→(0,0) (x,y)→(0,0) xy (x,y)→(0,0) x (x,y)→(0,0) y

So in order for the function f to be continuous we seek λ such that lim f (x, y) = f (0, 0)X.
(x,y)→(0,0)
Thus we find that λ = 1X.
[3]
 3 y

 x ye
 6 if (x, y) 6= (0, 0),
3. Consider the function f (x, y) := x + y2



0 if (x, y) = (0, 0) .

(a) The domain of f is R2 .X [1]

3
(b) The limit of f (x, y) when (x, y) approaches (0, 0) along the curve y = x is given by
2
x6 e x 1 3 1
lim 6
= lim ex = .X
x→0 2x x→0 2 2
[1]
(c) The function f is not continuous at (0, 0) because the limit along the curve y = x3 is different
from f (0, 0).XX [2]
(d) Find fx (0, 0) and fy (0, 0) if they exist.
f (x, 0) − f (0, 0) 0
limx→0 = lim = 0. So, fx (0, 0) = 0X. Similarly, fy (0, 0) = 0.X
x x→0 x
[2]
(c) Is the function f differentiable at (0, 0)? Explain clearly your answer.
The function f is not differentiable at (0, 0) because f is not continuous at (0, 0).XX
[2]
 Section B

In this section, indicate only the correct answer by filling in a, b, c, d, or e in the relevant box
provided on the attached sheet. Working will not be marked.

³ 1 ´
2 −πt
1. The domain of the vector function r(t) := ln(t − 2), ,e √ is
arcsin( t2 − 4)
√ √
(a) (−∞, − 2) ∪ ( 2, 2) ∪ (2 + ∞); (b) (−∞, −2) ∪ (2 + ∞); (c) (2, +∞);
√ √ √
(d) All t 6= ± 2, ±2; (e) [− 5, −2) ∪ (2, 5].
Solution: e
[2]
Z y
2. The domain of the function f (x, y) := sin(t2 )dt is the set
x
(a) {(x, y) ∈ R2 : x ≤ y}; (b) {(x, y) ∈ R2 : y 6= 0}; (c) {(x, y) ∈ R2 : x ≥ 0 and y ≥ 0};

(d) R2 (e) {(x, y) ∈ R2 : y > 0}.

Solution: d
[2]

3. Which of the following statements is(are) true?

x2
(i) lim = 0.
(x,y)→(0,0) y

(ii) If √C is the curve

√ described by the vector function r(t) := (2t2 − 3, t2 + 2, 4t − 3) with
− 2 ≤ t ≤ 3, then the direction to the tangent line to the curve C at the point (−1, 3, 1)
is given by the vector (2, 1, 2).
(iii) If the function f : R2 → R is such that fx (0, 0) = fy (0, 0) = 0, then f is continuous at (0, 0).
(iv) The curve in (ii) is regular.
√ π
(v) If the curve C is described by r(t) = (cos(2t), sin(2t), t 12) with 0 ≤ t ≤ , then length of
3
C is equal to 4π.

(a) (i), (ii), (iv); (b) (i), (ii), (iii); (c) (ii), (iv); (d) (i), (ii), (iii), (iv); (e) All.

Solution: c
[4]

4. The curve intersection of the cylinder x2 + y 2 = 1 with the plane x + 2y − z = 1 is described by

the vector function

(a) r(t) := (sin t, cos t, cos t + 2 sin t − 1) with 0 ≤ t ≤ 2π;

(b) r(t) := (cos t, sin t, cos t + 2 sin t − 1) with 0 ≤ t ≤ 2π;

(c) r(t) := (sin t, cos t, cos t + 2 sin t) with 0 ≤ t ≤ π;

(e) r(t) := (sin t, cos t, 2 sin t − 1) with 0 ≤ t ≤ 2π.

Solution: b
[3]
5. Consider the function f (x, y, z) = arccos(xy + yz + xz). At every point (x, y, z) in R3 such that
(xy + yz + xz)2 6= 1, we find that the gradient ∇f (x, y, z) of at (x, y, z) is equal to
à !
y+z x+z y+x
(a) p ,p ,p ;
1 − (xy + yz + xz)2 1 − (xy + yz + xz)2 1 − (xy + yz + xz)2
(b) ((y + z) arcsin(xy + yz + xz), (x + z) arcsin(xy + yz + xz), (y + x) arcsin(xy + yz + xz));
à !
−y − z −x − z −y − x
(c) p ,p ,p ;
1 − (xy + yz + xz)2 1 − (xy + yz + xz)2 1 − (xy + yz + xz)2
µ ¶
y+z x+z y+x
(d) , , ;
1 + (xy + yz + xz)2 1 + (xy + yz + xz)2 1 + (xy + yz + xz)2
µ ¶
−y − z −x − z −y − x
(e) , , .
1 + (xy + yz + xz)2 1 + (xy + yz + xz)2 1 + (xy + yz + xz)2
Solution: c
[3]

6. Write the equation x2 + y 2 + (z − 3)2 = 9 in spherical coordinates.

(a) ρ = 6 cos φ; (b) ρ2 = 9 + 6ρ cos φ; (c) ρ2 = 9; (d) ρ2 = 3; (e) ρ2 = sin φ.

Solution: a
[2]

7. The tangent line to the curve described by r(t) := (2t + π, et sin t, et cos t) at the point (π, 0, 1),
has parametric equations:

(a) x = π + 2λ, y = λ, z = 1 + λ, λ ∈ Q; (b) x = π + 2λ, y = λ, z = 1 + λ, λ ∈ R;

(c) x = πλ, y = 0, z = λ, λ ∈ R; (d) x = π + 2λ, y = 0, z = λ, λ ∈ R;

(e) x = π + 2λ, y = λ, z = 1 + λ, λ ∈ N.
Solution: b [3]

8. The curve described by the vector function r(t) = (et sin t, 5, et cos t) with −π ≤ t ≤ 0, has length
equal to
√ √ √ √
(a) π 2; (b) 2; (c) 2(e−π − 1); (d) 2(1 − e−π ); (e) 2(1 − e−π ).

Solution: e
[3]