Tutorial Solutions: 4MTH271 Coordinate
Systems and Space Curves
Felix Silwimba
University of Zululand
March 25, 2025
Coordinate Transformations: Rectangular, Cylindrical, and Spherical
p
√
2
1. (−2, 2, 4) to cylindrical: r = (−2)2 + 22 = 2 2, tan θ = −2
= −1,
√
3π
3π
θ = 4 (2nd quadrant), z = 4. Answer: (2 2, 4 , 4).
√
3 3
5π
3
2. (3, 5π
x = 3 cos 5π
6 , −1) to rectangular: 6 = − 2 , y = 3 sin 6 = 2 , z =
√
−1. Answer:
− 3 2 3 , 32 , −1 .
√
3. (5, π3 , π6 ) to rectangular: x = 5 sin π6 cos π3 = 54 , y = 5 sin π6 sin π3 = 5 4 3 ,
√
√
√
z = 5 cos π6 = 5 2 3 . Answer: 54 , 5 4 3 , 5 2 3 .
√
√
√
√
2
4. (1, − 3, 2) to spherical: ρ = 1 + 3 + 4 = 8 = 2 2, cos ϕ = 2√
= √12 ,
2
√
√
√ 5π π ϕ = π4 , tan θ = −1 3 = − 3, θ = 5π
3 . Answer: 2 2, 3 , 4 .
p
√
√
2
2
5. (6, 3π
45 = 3 5, θ = 3π
, cos ϕ =
4 , −3) to spherical:ρ = 6 + (−3) =
4
√
−3
√ = − √1 , ϕ = cos−1 − √1 . Answer: 3 5, 3π , cos−1 − √1
.
4
3 5
5
5
5
Transformation of Equations
1. z = 2x2 + 2y 2 : z = 2(r cos θ)2 + 2(r sin θ)2 = 2r2 (cos2 θ + sin2 θ) = 2r2 .
p
2. ρ = 4 sin ϕ: ρ2 = 4ρ sin ϕ, x2 + y 2 + z 2 = 4 x2 + y 2 , square both sides:
(x2 + y 2 + z 2 )2 = 16(x2 + y 2 ).
2
2
2
2
2
2
3. z 2 = 3x2 + 3y 2 : ρ2 cos2 ϕ =
3ρ sin ϕ(cos θ + sin θ), cos ϕ = 3 sin ϕ,
tan2 ϕ = 13 , ϕ = tan−1
√1
3
.
4. r = 2 cos θ: r2 = 2r cos θ, x2 + y 2 = 2x, x2 − 2x + y 2 = 0.
5. x2 + y 2 − 4x = 0: r2 = 4r cos θ.
1
Domains, Limits, and Smoothness of Vector Functions
√
√
1. ⃗r(t) = ⟨ t − 1, ln(t + 2), t2 ⟩: t − 1 needs t ≥ 1, ln(t + 2) needs t > −2,
t2 all real. Domain: [1, ∞).
1
2. limt→1− ⟨ t−1
, t2 , et ⟩ = ⟨−∞, 1, e⟩ (first component diverges to −∞).
√
3. ⃗r(t) = ⟨t2 − 1, t + 1⟩, ⃗r′ (t) = ⟨2t, 2√1t+1 ⟩, defined for t > −1, never zero,
smooth on (−1, ∞).
4. ⃗r(t) = ⟨cos t, √1t , sin t⟩: cos t, sin t all real, √1t needs t > 0. Domain: (0, ∞).
5. limt→0 ⟨et − 1, sint t , t cos t⟩ = ⟨0, 1, 0⟩ ( sint t → 1 by L’Hôpital’s rule).
Derivatives of Space Curves and Basic Sketching
1. ⃗r′ (t) = ⟨sin t + t cos t, cos t − t sin t, 2t⟩.
′
2. ⃗r(t) = ⟨t2 , et , ln t⟩,D ⃗r′ (t) = ⟨2t, et , 1t ⟩, ⃗r′ (1)
E = ⟨2, e, 1⟩, |⃗r (1)| =
√
5 + e2 , T⃗ (1) = √ 2 , √ e , √ 1
.
5+e2
5+e2
√
4 + e2 + 1 =
5+e2
3. ⃗r(t) = ⟨t3 , 1 − t⟩, ⃗r′ (t) = ⟨3t2 , −1⟩, ⃗r(2) = ⟨8, −1⟩, ⃗r′ (2) = ⟨12, −1⟩.
Sketch: ⃗r(2) from (0, 0) to (8, −1), ⃗r′ (2) from (8, −1) in direction ⟨12, −1⟩.
4. ⃗r(t) = ⟨2 − t, 3t, t2 ⟩, ⃗r′ (t) = ⟨−1, 3, 2t⟩, ⃗r′ (0) = ⟨−1, 3, 0⟩.
5. ⃗r(0) = ⟨0, 1, 0⟩, ⃗r′ (t) = ⟨2 cos 2t, −2 sin 2t, 1⟩, ⃗r′ (0) = ⟨2, 0, 1⟩. Sketch:
⃗r(0) to (0, 1, 0), ⃗r′ (0) from (0, 1, 0) in direction ⟨2, 0, 1⟩.
Integration, Arc Length, and Parameterization
1.
R1
⟨t2 , e−t , sin πt⟩ dt =
0
D 3
E
t
1
−t
3 , −e , − π cos πt
|10 =
1
1
1
3, 1 − e, −π
.
√
√
Rπ√
4 + 1 = 5, L = 0 5 dt = π 5.
Rt
√
3. ⃗r′ (t) = ⟨−3 sin t, 3 cos t, 4⟩, |⃗r′ (t)| = 9 + 16 = 5, s(t) = 0 5 du = 5t,
t = 5s , ⃗r(s) = 3 cos 5s , 3 sin 5s , 4s
5 .
h √
i2
√
√
R2√
2
4. ⃗r′ (t) = ⟨2t, 1, 2⟩, |⃗r′ (t)| = 4t2 + 5, L = 0 4t2 + 5 dt = t 4t2 +5 + 45 ln(2t + 4t2 + 5) ≈
0
6.08.
D
E
R
4
⃗
5. ⟨cos 2t, t3 , e2t ⟩ dt = 12 sin 2t, t4 , 12 e2t + C.
2. ⃗r′ (t) = ⟨−2 sin t, 2 cos t, 1⟩, |⃗r′ (t)| =
2
√
Curvature of Space Curves
1. ⃗r′ (t) = ⟨2t, 3t2 , 1⟩, ⃗r′′√
(t) = ⟨2, 6t, 0⟩, ⃗r′ (1) × ⃗r′′ (1) = ⟨−6, 2, 6⟩, |⃗r′ (1) ×
′′
′
√ .
⃗r (1)| = 10, |⃗r (1)| = 7, κ(1) = (√107)3 = 710
7
2. ⃗r′ (t) = ⟨−3 sin t, 3 cos t⟩, |⃗r′ | = 3, T⃗ = ⟨− sin t, cos t⟩, T⃗ ′ = ⟨− cos t, − sin t⟩,
|T⃗ ′ | = 1, κ = 31 .
3. ⃗r′ (0) = ⟨2, 0, 0⟩, ⃗r′′ (0) = ⟨0, 2, 0⟩, |⃗r′ × ⃗r′′ | = 4, |⃗r′ | = 2, κ(0) = 84 = 12 .
√
4. ⃗r′ (t) = ⟨1, 4, 5⟩, |⃗r′ | = 42, ⃗r′′ (t) = ⃗0, κ = 0 (straight line).
5. ⃗r′ (t) = ⟨cos t − t sin t, sin t + t cos t, 1⟩, ⃗r′′ (t) = ⟨−2 sin t − t cos t,q2 cos t −
t sin t, 0⟩, at t = π2 , ⃗r′ = ⟨0, π2 , 1⟩, ⃗r′′ = ⟨−2, 0, 0⟩, |⃗r′ × ⃗r′′ | = 2
q
2
|⃗r′ | = π4 + 1, κ = π22 .
4
+1
3
π2
4 + 1,