Math 501: Differential Geometry

Homework 1

Lawrence Tyler Rush

<me@tylerlogic.com>

January 26, 2013

http://coursework.tylerlogic.com/courses/math501/homework01

1 do Carmo pg. 5 exercise 2

Assume for later contradiction that α′(t₀) is not orthogonal to α(t₀). Since α′(t₀) is nonzero, then we are able to find a ϵ > 0 such that either |α(t₀ -ϵ)| < |α(t₀)| or |α(t₀ + ϵ)| < |α(t₀)|. However, either case contradicts the fact that α(t₀) is the point of the trace of α which is closest to the origin. Thus α′(t₀)⊥α(t₀) must be true.

2 do Carmo pg. 7 exercise 2

Figure 1: Cycloid at a Glance.

(a)

We will parametrize the cycloid by the angle of rotation, denote it by t. From the diagram in Figure 1 it can be seen that

1 θ = 2(π - t) π t φ = 2 - θ = 2

Thus the value of h is determined by the following sequence of equations.

h- = cosθ 2 h = 2cosθ( ) h = 2cos 1 (π - t) 2 t h = 2sin 2

With these things, we can finally derive the point p.

px = t- hcosφ t t px = t- 2sin 2 cos2 px = t- sin t

and

p = hsin φ y ( ) py = 2sin t sin t 2 2 py = 2sin2 t 2 py = 1- cos t

(b)

Let’s rotate from 0 to 2π, fixing t₀ at 0. By making use of the double angle formula for cos, we have the following.

∫ 2π ′ s(2π) = |α (t)| dt ∫0 2π = |(1- cost,sint| dt ∫0 ---------------- 2π ∘ 2 2 = 0 (1 - cost) + sin t dt ∫ 2π ∘--------------------- = 1 - 2cost+ cos2 t+ sin2t dt ∫0 2π = √2-- 2-cos-t dt 0 √-∫ 2π√ ------- = 2 1- cost dt ∫02π∘ ----(------------)- = √2 1 - cos2 t - sin2 t dt 0 2 2 √-∫ 2π∘ ---------------- = 2 1 - cos2 t+ sin2 t dt ∫0 ∘ ------- 2 2 √- 2π 2 t = 2 0 2 sin 2 dt ∫ 2π t = 2 sin - dt 0 |22π = - 4 cos t|| 2|0 = - 4cosπ + 4cos(0) = 8

3 do Carmo pg. 7 exercise 3

Figure 2: Cissoid of Diocles at a Glance.

(a)

For this problem we will use Figure 2 as a guide. Using some trigonometry, from the figure we get the following (remember the diameter is length 2a)

φ = 2θ OB = -OA- = -2a- cosθ cosθ OC = asin2φ-= a-sin2θ sinθ sinθ

These allow us to have the following

p = (OP cosθ,OP sinθ) = (CB cosθ,CB sinθ) = (O(B( - OC )cosθ,(O)B - OC( )sinθ) ) ) -2a- a-sin2θ -2a- asin2θ = cosθ - sinθ cosθ, cosθ - sin θ sin θ (( 2a 2a cosθ sinθ ) ( 2a 2acosθ sinθ) ) = ----- ----------- cosθ, ----- ----------- sin θ ( cosθ sinθ cosθ ) sinθ = 2a- 2a cos2θ, 2a-sinθ - 2acosθsinθ ( cosθ ) 2 2asin-θ--2acos2θsinθ- = 2asin θ, cosθ ( 2 2 2 ) = 2a-sin-θ2sec-θ, 2a-sinθ(1--cos-θ) ( sec θ cosθ ) --2asin2-θ- 2atan-θsin2θ-sec2θ = sec2θcos2θ, sec2θ ( 2atan2θ 2a tanθsin2 θ ) = ------2--,-------2----2-- (1 + tan2 θ (1+ tan3 θ))cosθ = -2atan-θ-, 2a-tan-θ- 1 + tan2 θ 1+ tan2θ

Hence letting t = tanθ we get

( 2 3 ) p = 2at--,-2at-- 1+ t2 1+ t2

(b)

Because

( 3 2 4 ) ( 2 2) α′(t) = -4at- - --4at--,-6at- - --4at--- = --4at--, 2at-(3+-t) 1 + t2 (1+ t2)2 1 + t2 (1+ t2)2 (1+ t2)2 (1 + t2)2

then α(0) = (0,0), and so α has a singular point at the origin (0,0).

(c)

( -2at2- -2at3-) lt→im∞ α(t) = tli→m∞ 1+ t2,1+ t2 ( 2 3 ) = 2a lim--t-2,2a lim --t-2 t→∞ 1+ t t→∞ 1 + t = (2a,∞ )

( 2 2 ) lim α′(t) = lim ---4at2-2, 2at(3-+2t2) t→ ∞ t→( ∞ (1 +t ) (1+ t ) ) = 4a lim ---t---,2a lim t2(3+-t2) t→ ∞ (1 + t2)2 t→∞ (1 + t2)2 = (0,2a)

4 do Carmo pg. 11 exercise 10

(a)

Left Equation For α(t) = (x(t),y(t),z(t)) the Fundamental Theorem of Calculus gives us that there is a c ∈ I such that

′ ′ ′ ′ (q - p)⋅v = (q1 - p1)v1 + (q2 - p2)v2 + (q3 - p3)v3 = x(c)v1 + y(c)v2 + z(c)v3 = α (c)⋅v

but since v is a vector of magnitude 1, then

′ ∫ b ′ α (c) ⋅v = α (t)⋅v dt a

∫ b (q- p)⋅v = α′(t)⋅v dt a

Right Inequality By the definition of dot product

∫ b ∫ b ∫ b α′(t)⋅v dt = |α ′(t)||v|cosθ dt = |α ′(t)|cosθ dt a a a

Now since cosθ ranges between -1 and 1 then

∫ b ′ ∫ b ′ a |α (t)|cosθ dt ≤ a |α (t)| dt

(b)

(a)

To show that α(t) is parametrized by arc lenght, we must show |α′(t)| = 1 for all t. So proceeding that way, we have

′ -1-- x (t) = 1 - t+ 1 √2t y′(t) = t+-1 1 z′(t) = ---- t+ 1

giving us

( 1 √2t- 1 ) ( t √2t 1 ) α′(t) = 1- ---- ,----,---- = ----,----,---- . t+ 1 t+ 1 t+ 1 t+ 1 t+ 1 t +1

Therefore |α′(t)| is

┌│ (-----)---(-√---)2---(-----)-- ∘------------------------- ∘ ------- │∘ --t- 2 --2t -1-- 2 --t2--- --2t--- --1---- (t+-1)2- t +1 + t+ 1 + t+ 1 = (t+ 1)2 + (t +1)2 + (t+ 1)2 = (t+ 1)2 = 1

Hence α is parametrized by arc length.

(b)

Curvature. The curvature, κ, is |α′′(t)|. Using what we know from the previous part,

( √ -- ) ( ) α′′(t) = -1--- --t---,√---1----- ---2t-,----1-- = ---1---,√--1--t---,--- 1-- t+ 1 (t+ 1)2 2t(t+ 1) (t+ 1)2 (t+ 1)2 (t+ 1)2 2t(t+ 1)2 (t+ 1)2

then

∘ (-------)----(----------)----(-------)-- ∘ ------------------ ∘ ----------- ∘ --------- ---1--- 2 --1---t--- 2 --- 1-- 2 ---2--- -(1--t)2- 4t+-(1--t)2- -(t+-1)2- κ(t) = (t+ 1)2 + √2t(t+ 1)2 + (t+ 1)2 = (t+ 1)4 + 2t(t+ 1)4 = 2t(t+ 1)4 = 2t(t+ 1)4 1 = √-------- 2t(t+ 1)

Torsion. The torsion of α, τ, is the magnitude of the derivative of the binormal vector, b, as given by the Frenet equation b′ = τn. The binomial vector is

b(t) = t(t′)∧ n(t) - ′′ b(t) = α( (t)∧ (κ(t)) 1α (t)) t √2t- 1 √ -- ( 1 1- t - 1 ) b(t) = t+-1,t+-1,t+-1 ∧ 2t(t+ 1) (t+-1)2,√2t(t+-1)2,(t+-1)2- ( √ -- ) ( √ -- √--) -t-- --2t --1- --2t 1--t ---2t b(t) = t+ 1,t+ 1,t+ 1 ∧ t+ 1,t+ 1, t+ 1 [( √-- ) (√ -- √ --)] b(t) = ---1--2 t, 2t,1 ∧ 2t,1- t,- 2t (t+ 1) ( ) b(t) = ---1--- 3t- 1,(t- 1)√2t,- t2 - t (t+ 1)2

( √ -- ) ( ) b′(t) = --- 2-3 3t- 1,(t- 1) 2t,- t2 - t + --1--2- 3, 3t√--1,- 2t - 1 (t+ 1) (t+ 1) ( 2t ) ′ --1----( √ -- 2 ) ---1--- 3t2 -√2t--1 2 b(t) = (t+ 1)3 2- 6t,(2 - 2t) 2t,2t + 2t + (t+ 1)3 3t+ 3, 2t ,- 2t - 3t - 1 1 ( - (t- 1)2 ) b′(t) = (t+-1)3- - 3t - 5,-√---- ,t- 1 2t

and thus

τ(t) = |b′(t)| ∘ -------------------------------- ( 2)2 τ(t) = --1--3- (- 3t - 5)2 + - (t√--1) + (t- 1)2 (t+ 1)∘ -----------------2t--- --1---- 2 (t--1)4 τ(t) = (t+ 1)3 10t + 28t+ 26+ 2t

6 do Carmo pg. 22 exercise 1

The following are for

( ) α(s) = acos s,asin s,bs c c c

(a) The parameter s is the arc length

( - a s a s b) α′(s) = ---sin -,--cos -,- c c c c c

∘ ---------------------- ∘ ------- ∘ ------- ∘ --- ′ a2 2 s a2 s b2 a2 b2 a2 + b2 c2 |α (s)| = -c2 sin c + c2 cos2 c + c2 = c2 + c2-= --c2-- = c2 = 1

Therefore s is the arc length.

(b) Curvature and torsion.

′′ ( -a s -a s ) α (s) = -c2 cos c,-c2 sinc ,0

∘ (-------)2--(-------)2 k(s) = |α′′(s)| = a-cos s + a-sin s = a- c2 c c2 c c2

b(s) = α′(s)∧(k(s)-1)α ′′(s) ( - a s a s b) c2 ( a s a s ) = ---sin -,--cos -,- ∧ -- - -2 cos-,- 2-sin-,0 ( c c c c c) (a c c c) c = - a-sin s, a-cos s, b ∧ - cos s ,- sin s,0 c c c c c c c ( b s b s a- 2 s a- 2 s ) = c sin c,c cos c,c sin c + c cos c ( b s b s a) = -sin -,- cos -,-- c c c c c

( b s b s ) b′(s) = -2 cos ,- -2 sin-,0 c c c c

∘ (-------)2---(-------)2 ∘-2---------2------ ∘ -2- τ(s) = |b′(s)| = b2 cos s + b2-sin s = b4-cos2 s+ b4-sin2 s = b4 = b2 c c c c c c c c c c

(c)

The osculating plane of α at s = s₀ is defined by the following point, normal vector tuple

(( s0 s0 s0) ( b s0 b s0 a )) (α(s0),b(s0)) = a cos c-,asin c-,bc- , c sin c-,c cos-c ,c

the equations for which are given and determined in the previous problem.

(d)

The angle between the z-axis and a line passing through α(s) and containing n(s) will be the same as the angle between n(s) and any vector contain by the z-axis, say (0,0,1). And that angle will be the inverse cosine of their dot product, since both vectors have magnitude of one.

Since n(s)(k(s))^-1α′′(s) = ( s s )
- cosc,- sinc,0 , then the dot product of it and (0,0,1) is zero, implying that the cosine of the angle between them is zero, which in turn confirms that the angle between them is π
2 .

(e)

This is similar to the previous problem except that we will use t(s) in place of n(s). Because

( ) t(s)⋅(0,0,1) = --asin s, acos s, b ⋅(0,0,1) = b c c c c c c

i.e. the cosine of the angle between them is a constant, then the angle bewteen t and (0,0,1) is constant, since they both have a magnitude of one.

7 do Carmo pg. 25 exercise 12

(a)

Since

∫ t s(t) = |α′(t)|dt t0

then by the fundamental theorem of calculus we have ds∕dt = |α′(t)|, which implies dt∕ds = 1∕|α′(t)|.

Taking the derivative of both sides of the above equation leads to the following sequence of equations.

d2t∕d2s = d---1′-- ds|α (t)| = d-√--1--- ds α′ ⋅α′ - (α′′ ⋅α′ddts + α′ ⋅α ′′ddts) = ------∘---′--′3------ 2 (α ⋅α ) = - (α′′ ⋅α′)ddts |α′|3 - (α′′ ⋅α′) = --|α′|4---

(b)

(c)

(d)

Assuming that the yet-to-be-proven part (b) of this problem is true, then the signed curvature is

α′ ∧-α′′ k(t) = |α′|3 (x′,y′)∧ (x ′′,y′′ = -----′--′-3--- |(x′ ,′′y)|′′ ′ = (--x-y---x-y-)-- ∘ (x′)2-+(y′)2-3 = --x′y′′ --x′′y′- ((x′)2 + (y′)2)3∕2