MA304 Differential Geometry

Size: px

Start display at page:

Download "MA304 Differential Geometry"

Katherine Adela McDaniel
5 years ago
Views:

1 MA34 Differential Geometr Homework solutions, Spring 8, 3% of the final mark Let I R, where I = (a, b) or I = [a, b] Let α : I R 3 be a regular parameterised differentiable curve (not necessaril b arc length), and let β : J R 3 be a reparameterisation of α(i) b the arc length s = s(t), measured from a Let t = t(s) be the inverse function of s and set dα = dt α, d α = α, d3 α = α Assume that α (t) and α (t) dt dt 3 for all t I Prove that (a) dt ds = α and d t ds = α α α 4 (b) The curvature of α at t I is (c) The torsion of α at t I is k(t) = α α α 3 τ(t) = (α α ) α α α Solution (a) dt ds = ds follows from the definition that α dt = α We have d t ds = d ds α = dt d ds dt (α α ) / = ( ) α (α α ) 3/ (α α α α ) = ) ( α α = α α α α 3 α 4 (b) Note that for an two vectors a, b R 3, we have a b = (a b) (a b) = a (b (a b)) = a ((b b)a (b a)b) = (a a)(b b) (a b) = a b (a b), () b the scalar and vector triple products (see homework, question ) Since β(s) = (α t)(s), β (s) = α (t(s))t (s) = α α, () β (s) = α (t(s))t (s) α (t(s))t (s) = α α (α α )α, (3) α 4 β (s) β (s) = α α (α α ) (α α ) (α α ) α 4 α 6 α 8 = α 6 ( α α (α α ) ) = α α α 6,

2 where we have applied () for the last equalit Therefore, the curvature of α is (c) Note that the torsion of α is k(t) = β (s) = α α α 3 (4) τ(t) = τ β (s) = (β (s) β (s)) β (s) k β (s), b homework, question 4, where k β (s) and τ β (s) are the curvature and torsion of β(s) We compute the expression on the right B () and (3), we have ( ) β (s) β (s) = α α α α (α α )α = α α, α 4 α 3 β (s) = d [ α (t(s))t (s) α (t(s))t (s) ] ds dt = α ds t (s) α t (s)t dt (s) α ds t (s) α t (s), (β (s) β (s)) β (s) = α α α 3 α Together with (4), we have dt ds t (s) = (α α ) α α 6 τ(t) = (β (s) β (s)) β (s) k β (s) = (α α ) α α α Find the curvature and torsion of the following curves For parts (b) and (c), find also all of the vertices (a) α : (, ) R 3, where α(t) = (cos t, sin t, log(t )) (b) α : [, π] R, where α(t) = (a cos t, b sin t) and < b < a (c) α : [, π] R, where α(t) = (( cos t) cos t, ( cos t) sin t) Solution We use the expressions for curvature and torsion from question (a) We have α (t) = ( ) sin t, cos t, t e e e 3 α α = sin t cos t t cos t sin t (t) ( cos t = (t ) sin t ) e t ( cos t α α = (t ) sin t ) t = (t ) 4 (t ), and α (t) = ( cos t, sin t, ) (t ) ( sin t (t ) cos t t ( sin t (t ) cos t ) t ) e (sin t cos t)e 3

3 and Therefore, the curvature of α is Also, (α α ) α = Therefore, the torsion of α is α = sin t cos t (t ) = (t ) k(t) = α α α 3 = ((t ) 4 (t ) ) / ((t ) ) 3/ α (t) = ( ) sin t, cos t,, (t ) 3 ( cos t (t ) sin t ) ( sin t sin t t (t ) cos t ) cos t t (t ) 3 = (t ) (t ) 3 τ(t) = (α α ) α α α = (t ) (t ) 3 (t ) 4 (t ) For parts (b) and (c), we note that the torsion of α should be, since a plane curve cannot have an twisting Indeed, α α is in the direction of e 3, and α has zero e 3 component, so (α α ) α =, which implies τ(t) = It remains to find the curvature and the vertices in parts (b) and (c) (b) We have and α (t) = ( a sin t, b cos t) and α (t) = ( a cos t, b sin t) Therefore, the curvature of α is We have e e e 3 α α = a sin t b cos t = abe 3, a cos t b sin t α = a sin t b cos t k(t) = α α α 3 = ab (a sin t b cos t) 3/ k (t) = 3 ab(a sin t b cos t) / (a sin t cos t b cos t sin t) = 3 ab(a sin t b cos t) / (a b ) sin t, 3

4 so k (t) = when sin t =, t =, π, π, 3π It follows that the vertices of α are (a, ), (, b), ( a, ) and (, b) (c) We have α(t) = ( cos t cos t, sin t cos t sin t) = ( cos t cos t, sin t sin t ), α (t) = ( sin t cos t sin t, cos t cos t) = ( sin t sin t, cos t cos t), α (t) = ( cos t cos t, sin t sin t), e e e 3 α α = sin t sin t cos t cos t cos t cos t sin t sin t and = [( sin t sin t)( sin t sin t) ( cos t cos t)( cos t cos t)]e 3 = (4 sin t sin t 6 sin t sin t 4 cos t cos t 6 cos t cos t)e 3 = (6 6 cos t)e 3 α = ( sin t sin t) ( cos t cos t) = 4 sin t sin t 4 sin t sin t 4 cos t cos t 4 cos t cos t = 4 cos t Therefore, the curvature of α is We have k(t) = α α α 3 = 6 6 cos t ( 4 cos t) 3/ k (t) = 6 sin t( 4 cos t)3/ (6 6 cos t) 3 ( 4 cos t)/ ( 4 sin t) ( 4 cos t) 3 Since 4 cos t >, we have k (t) = when the numerator is equal to The numerator simplifies to 6( 4 cos t) / sin t( cos t), which is when sin t = or cos t =, giving t =, π, π, 4π It follows that the vertices of α are (3, ), ( 3, 3 3), (, ) and 3 3 ( 3, 3 3 ) 3 Determine if the following surfaces in xz-space are regular, and explain our reasoning carefull For those surfaces that are regular, give a local parameterisation and show how to cover the surface with several such parameterisations You ma use an of the propositions that were proved in class (a) The surface T, obtained b rotating the circle (x a) z = b in the xz-plane about the z-axis b π, where < b < a 4

5 (b) The surface C given b z = x, where x > and < x (c) The surface S given b { x z = for z, x = for < z Solution (a) The surface T is a torus, and is given b the equation ( x a) z = b B letting f(x,, z) = ( x a) z, we see that T is the set f (b ) (x, ) (, ), we have For f x = x( x a), x f = ( x a), x f z = z, so the three partial derivatives vanish simultaneousl at (x,, ) such that x = a These points do not belong to f (b ), b is a regular value of f B Proposition 3, T is a regular surface Finall, T can be covered b three parameterisations For example, x (u, v) = ((a b cos u) cos v, (a b cos u) sin v, b sin u), < u, v < π, x (ũ, ṽ) = ( (a b cos ũ) cos ṽ, (a b cos ũ) sin ṽ, b sin ũ), < ũ, ṽ < π, x 3 (û, ˆv) = ((a b sin û) sin ˆv, (a b sin û) cos ˆv, b cos û), < û, ˆv < π The parameters u, v, ũ, etc can be chosen, x (u, v) covers T except for the circles C = {(x,, z) : x = (ab), z = } and C = {(x,, z) : (x a) z = b, = }; x (ũ, ṽ) covers T except for the circles C = {(x,, z) : x = (a b), z = } and C = {(x,, z) : (x a) z = b, = }; and x 3 (û, ˆv) covers T except for the circles C 3 = {(x,, z) : x = a, z = b} and C 3 = {(x,, z) : ( a) z = b, x = } It is eas to see that ever point of T is covered b at least one of the three parameterisations (b) Since C is the graph of f(x, ) = x on the open set U = {(x, ) R : x >, < x}, C is regular b Proposition Clearl, the single parameterisation x(x, ) = (x,, x ), for (x, ) U covers the whole of C (c) We show that S is not a regular surface Otherwise if S is regular, then b Proposition, there is a neighbourhood V S of (,, ) which is the graph of a differentiable function in one of the forms z = f(x, ), = g(x, z) or x = h(, z) The first two cases are not possible since the projections of V onto the x-plane and the xz-plane are not one-to-one Indeed, for the first case we would have both (,, ), (,, ε) V being projected to (, ), for some small ε > For the second case we would have both ( ε, ( ε), ), ( ε, ( ε), ) V being projected to ( ε, ), for some small ε > In the final case, we have { z x = h(, z) = for z, for < z

6 We have { x ( z = z ) 3/ z ( z ) / for z, for < z, x z = (,z)=(,) { for z, for < z We see that x = h(, z) is not differentiable at (, z) = (, ), which is a contradiction 4 Verif the Cauch-Crofton formula for the following curves (a) x = where x (b) (Optional, for bonus) = x where x Solution In both parts, let f(p, ) denote the multiplicit of the line with parameters (p, ), with respect to the curve in consideration Let R = {(p, ) R : p and < π, or p > and π < π} (a) The curve is a semi-circle with radius, so its arc length is π We must show that f(p, ) dp d = π We note that, since the curve is smmetric about the x- R axis, we have f(p, ) = f(p, π ) for π < < π Thus, it suffices to show that π f(p, ) dp d = π (a) (b) d x x d Figure For < π, we have d = sin in Figure (a), f(p, ) = { for p < sin or p =, for sin p < For π < π, we have d = cos( π ) = sin in Figure (b), f(p, ) = for p sin Therefore, f(p, ) dp d = / (sin ( sin )) d = [ cos ] π/ [ cos ] π π/ = π π/ sin d 6

7 (b) We first find the arc length of the curve We can parameterise the curve b α(t) = (t, t ) for t, the arc length is α (t) dt = (t) ( ) / dt = ( u ) / du [ = 4 u( u ) / 4 log u ( u ) / ] = 4 log( ) 4 log( ) = ( ) 4 log = 4 log [ ( ) ] = log( ) We now simplif the task slightl We ma instead consider the curve x = for, which has the same arc length and is smmetric about the x-axis, so that f(p, ) = f(p, π ) for π < < π Similar to part (a), it suffices to show that f(p, ) dp d = log( ) Furthermore, it suffices to determine f(p, ) for < < π For the lines corresponding to, let m denote their common gradient, so that m = cot (a) (b) d 3 (c) L 4 d x x d 4 x L d L (d) d L 6 d 6 x (e) d 7 x Figure Let < π In Figure (a), the equation of the line L 4 is = cot (x ), the x-intercept is tan, and d = ( tan ) cos = cos sin The equation of the line L is = cot (x ), the x-intercept is tan, and d = ( tan ) cos = cos sin Therefore, { for cos sin < p cos sin, f(p, ) = for p cos sin 7

8 Let π < α < 3π be the angle such that tan α = For π < π, we have d 3 = cos sin in Figure (b), as before For π < α, in Figure (c), the x-intercept of the line L 4 is again tan <, d 4 = (tan ) cos(π ) = cos sin Since the tangent line to the curve at (, ) has gradient, we conclude that for π < α, 4 f(p, ) = for p cos sin Let α < 3π In Figure (d), we have d 4 = cos sin as before We compute the value of d 6 The line L 6 has gradient m and is tangent to the curve At the point of contact, we have x / = m, which gives x = and = x / = 4m m The equation of L 6 is therefore m = m (x be 4m < Therefore, d 6 = 4m f(p, ) = ) Setting =, we obtain the x-intercept of L 4m 6 to cos(π ) = 4 tan cos = cos sec, and { for p < cos sin or p = cos sec, for cos sin p < cos sec Let 3π 4 < < π We have d 7 = cos sec in Figure (e), as before Therefore, { for p = 4 f(p, ) = cos sec, 4 for p < cos sec Note that tan α = implies sin α =, cos α =, and sec α = We have as required f(p, ) dp d = = /4 /4 ( sin (cos sin )) d 3π/4 α α π/4 (cos sin ) d ( cos sin ( 34 cos 4 )) sec sin d ( 3π/ cos ) 4 sec d 3π/4 cos d (cos sin ) d ( ) sec d 3π/4 α π/4 ( 3 cos sin ) d 3π/4 α cos d [ = [ sin ] π/4 [sin cos ] 3π/4 π/4 ] π log sec tan α [ 3 ] 3π/4 sin cos α [ ] π sin 3π/4 = log log 3 3 = log( ), 8

Find the indicated derivative. 1) Find y(4) if y = 3 sin x. A) y(4) = 3 cos x B) y(4) = 3 sin x C) y(4) = - 3 cos x D) y(4) = - 3 sin x

Assignment 5 Name Find the indicated derivative. ) Find y(4) if y = sin x. ) A) y(4) = cos x B) y(4) = sin x y(4) = - cos x y(4) = - sin x ) y = (csc x + cot x)(csc x - cot x) ) A) y = 0 B) y = y = - csc