Math F412: Homework 3 Solutions February 14, Solution: By the Fundamental Theorem of Calculus applied to the coordinates of α we have

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1 1. Let k() be a mooth function on R. Let α() = ( θ() = co(θ(u)) du, k(u) du in(θ(u)) du). Show that α i a mooth unit peed curve with igned curvature κ p () = k(). By the Fundamental Theorem of Calculu applied to the coordinate of α we have Since α (t) = (co(θ(t)), in(θ(t))). co 2 (θ(t)) + in 2 (θ(t)) = 1, α (t) i a unit vector for each t T(t) = α (t). Now N p (t) = ( in(θ(t)), co(θ(t)) ince N p i obtained from T by rotating 9 degree. Since α i unit peed, Now κ p (t) = N p (t) α (t). α (t) = ( in(θ(t))θ (t), co(θ(t))θ (t)). Applying the Fundamental Theorem of Calculu to θ(t) we have Hence Since N p N p = 1, we conclude that θ (t) = k(t). α (t) = ( in(θ(t)), co(θ(t)))k(t) = N p (t)k(t). κ p (t) = k(t).

2 2. Let T R 2 R 2 be a tranlation (o T(x) = x + v for ome contant vector v). Let R R 2 R 2 be a rotation. Let S R 2 R 2 be the reflection S(x, y) = ( x, y). If α i a mooth plane curve, how that the igned curvature of T α R α are the ame a thoe of α, but the igned curvature of S α i the negative of the igned curvature of α. Without lo of generality we can aume that α i a unit peed curve. Let T be a tranlation, let β(t) = T α(t). So β(t) = α(t) + v for ome contant vector v. Then In particular we have for the unit tangent β (t) = α (t) β (t) = α (t). T β (t) = β (t) = α (t) = T α (t). Moreover, ince the planar normal N p,β N p,α are obtained from the tangent by rotating by 9 degree we have N p,β = N p,α. Finally, κ p,β = N p,β β = N p,α α = κ p,α. Let R θ be a rotation through angle θ let β = R θ α. Note that R θ can be repreented by the matrix co(θ), in(θ) R θ = ( in(θ), co(θ) ). Since R θ i a linear map, β (t) = R θ α (t) β (t) = R θ α (t). We need three propertie of rotation map, which we leave for the reader to verify. Firt, if x, y are vector, then (R θ x) (R θ y) = x y. (1) Second, if we apply the previou equation with y = x we have R θ x 2 = x 2 therefore Finally, for any two angle θ 1 θ 2 we have R θ x = x. (2) R θ1 R θ2 = R θ1 +θ 2. (3) From (2) we conclude that β (t) = R θ α (t) = α (t) = 1 2

3 o β i a unit peed curve T β = β = R θ α = R θ T α. By definition, N p,α = R /2 T α imilarly for β. So, applying our (3) we have N p,β = R /2 T β = R /2 R θ T α = R θ+/2 T α = R θ R /2 T α = R θ N p,α. Hence (applying (1)) we have κ p,β = β N p,β = R θ α R θ N p,α = α N p,α = κ p,α. Finally, let S be the given reflection let β = S α. Note that [S(a, b)] [S(c, d)] = ( a, b) ( c, d) = ( a)( c) + bd = ac + bd = (a, b) (c, d). Stated more uccinctly, if x y are vector, Sx Sy = x y. In particular, Sx = x. We have β = Sα hence β = Sα = α = 1. So β i unit peed. Alo, β = Sα. Let R be the rotation through 9 degree, o R(a, b) = ( b, a). Note that R(S(a, b)) = R( a, b) = ( b, a) S(R(a, b)) = S( b, a) = (b, a) = R(S(a, b)). Now T β = β = Sα = ST α therefore N p,β = RT β = RST α = SRT α = SN p,α = S( N p,α ). We conclude that κ p,β = β N p,β = Sα S( N p,α ) = α ( N p,α ) = κ p,α. 3

4 3. Suppoe α β are unit peed plane curve defined on the ame interval I = [a, b] uch that α(a) = β(a), α (a) = β (a) uch that their two curvature agree at every point in I. Show that α = β. Let T α N α be the tangent planar normal to α, ue imilar notation for β. Let D = T α T β + N α N β. From the Cauchy-Schwarz inequality applied to each term of D, uing the fact that all vector are unit vector, we ee that D 2 with equality if only if T α = T β N α = N β. Now D = T α T β + T α T β + N α N β + N α N β = κ p,α N p,α T β + κ p,β T α N p,β + κ p,α T α N β + κ p,β N α T β = (κ p,α κ p,α )N p,α T β + (κ p,β κ p,β )N p,β T α =. So D i contant D(a) = 2 ince the tangent ( hence normal) of α β agree at a. So D 2 the tangent normal of α β agree everywhere. Since the tangent agree on the interval, α = β on [a, b]. Since α(a) = β(a) we conclude from the Fundamental Theorem of Calculu applied to each coordinate function that α = β. 4. Ue Maple to plot the trace of a plane curve with igned curvature κ p () = 2 co(). Explain why α() lie on the x-axi. See workheet for plot. Note that where α(t) = ( Hence the y-coordinate of α() i θ(t) = co(θ(u)) du, t in(θ(u)) du) 2 co(u) du = 2 in(t). Now ubtituting u = v + we have in(2 in(u)) du. in(2 in(u)) du = 4 in(2 in(v + )) dv.

5 But So Hence in(2 in(v + )) = in( 2 in(v)) = in(2 in(v)). in(2 in(u)) du = in(2 in(v)) dv = in(2 in(u)) du. in(2 in(u)) du = = =. in(2 in(u)) du + in(2 in(u)) du So the y-coordinate i zero α() lie on the x-axi. in(2 in(u)) du in(2 in(u)) du 5. Let α be a unit peed plane curve. It center of curvature i є() = α() + 1 κ p () N(). a) Show that the circle centered at є() i tangent to α at α() ha the ame curvature a α at that point. You hould ue fact you know about the curvature of a circle. b) The curve є() i called the evolute of α. Show that the unit tangent to є i N() the igned unit normal to є i T. c) Let v be the arclength parameter of є. Show that d) Compute the igned curvature of є(). dv d = p() κ κ 2 Solution, part a: Note that є() α() = 1/κ p ()N p () = 1/ κ p () = 1/κ(). So α() i on the circle centered at є() with radiu κ(). To how that it i tangent, it i enough to how that α () i perpendicular to the vector connecting α() to є(). But (є() α()) α () = But α () = T() T() N p () =. Hence 1 κ p () N p() α (). (є() α()) α () =. 5

6 Solution, part b: We note that But N p = κ p T α = T, o є () = α () + ( d d 1 κ p () ) N 1 p() + κ p () N p(). є () = T() ( κ p() κ p () ) N 1 p() 2 κ p () κ p()t() = κ p() κ p () N p() 2 Auming that κ () >, we have є ()/ є () = N p (). But rotating T by 9 degree obtain N p, o rotating N p by 9 degree obtain T, rotating N p by 9 degree obtain T. Letting T є N p,є be the tangent igned normal of є we have T є = N p Solution, part c: Note that If v i arclength for є, then N p,є = T. є () = κ p() κ p () 2 N p() = κ p() /κ p () 2. v () = є () = κ p() /κ p () 2. Solution, part d: Let ν = v (). Then arguing a we did in cla for non unit peed curve we have T є () = ν()κ p,є ()N p,є (). Now Hence T є () = N p() = κ p ()T() = κ p ()N p,є. ν()κ p,є () = κ p () κ p,є () = κ p() 3 κ p() 6

7 6. Exercie Let α be a unit peed curve which lie on the phere centered at p with radiu R. Since T(t), N(t), B(t) are an orthonormal frame, we can write Now α(t) p = [(α(t) p) T] T + [(α(t) p) N] N + [(α(t) p) B] B. (α(t) p) (α(t) p) = R 2. Taking a derivative with repect to t, dropping the (t) for brevity we have 2T (α p) =. Hence T (α p) =. Taking another derivative we have T (α p) + T T =. From the Frenet equation we know T = κn therefore Taking another derivative we have N (α p) = 1 κ. N (α p) + N T = ( 1 κ ). But then from the Frenet equation ( the fact that N T = ) we have We have already een that T (α p) =. Hence Hence ( κt + τb) (α p) = ( 1 κ ). B (α p) = 1 τ ( 1 κ ). α p = 1 κ N + 1 τ ( 1 κ ) B. Computing the norm of each ide of thi equation, uing the fact that N N = 1, B B = 1 N B = we have R 2 = ( 1 2 κ ) + 1 τ [( κ ) ] which i the deired equation. 7

8 7. (Thi problem to be done entirely with Maple.) Viviani curve i defined by α(t) = (co(t) 2 1/2, in(t) co(t), in(t)). a) Show that α lie on the phere of radiu 1 centered at ( 1/2,, ) on the cylinder x 2 + y 2 = 1/4. b) Make a plot in Maple to demontrate that α lie on thi phere. The comm plot[pacecurve], plottool[phere] plottool[diplay] might come in hy. Alo note that if you end a line in Maple with a colon rather than a emicolon, the output will be uppreed, which i hy for thing like the output of plottool[phere]. c) Compute the curvature torion of Viviani curve. d) Verify that the curvature torion of Viviani curve atify the formula from the previou problem. R 2 = (1/κ) 2 + ((1/κ) (1/τ)) 2 8