1 question on the cycloid

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1 Answers to question sheet, qs.tex question on the cycloid cycloid The position ct of particle of mass m and electric charge e in combined electric and magnetic fields E and B according to Newton and Lorenz satisfies the differential equation m c ee + ċ B Assuming q e m and normal crossed fields E j and B k show that the equation becomes ẍ ÿ + ẏ ẋ z Show that, subject to initial conditions c ċ, the particle execute a cycloid in the plane z.. answer m c ee + ċ B c j + ċ k ẍ ÿ + z ẍ ÿ + z ẍ ÿ z ẏ ẋ i j k ẋ ẏ ż We solve the trivial differential equation z to obtain zt z +tżt + t using the initial conditions. The main differential equation now becomes dimensional ẍ ẏ ÿ ẋ Below we valiently solve this differential equation subject to the initial conditions but it suffices to check that a parameterization of the cycloid satisfies both the D.E. and the I.C. Recall, parameterization of the cycloid with velocity and acceleration is x t sint ẋ cost ẍ,, y cost ẏ sint ÿ ẏ ẋ sint cost Clearly the I.C. are satisfied x y ẋ ẏ. Also the D.E is satisfied ẍ sint ẏ and ÿ cost cost ẋ. Next we give a second method. We directly solve the D.E with I.C. Write the D.E. in the form ẍ ẋ + A where A ÿ ẏ

2 ẍ expat ÿ ẋ + AexpAt ẏ expat t where we multiply across by the integrating factor expat exp t d [ ] ẋ sint expat dt ẏ cost Z t [ ] Z d ẋ t sinp expap d p d p d p ẏ cosp [ ] t t ẋp cosp expap ẏp sinp p p ẋt cost expat on the LHS we used I.C. ẋ ẏ ẏt sint ẋt cost exp At ẏt sint ẋt cost sint cost ẏt sint cost sint ẋt cost ẏt sint Z t Z ẋp t cosp ẏp t xp yp xt yt d p p p sinp cosp t sint cost sinp t p d p on the LHS we used the I.C. x y cost sint sint cost The solution is the well known parameterization of the cycloid. question on cardioid Let C R be the cardioid curve obtained by rolling a circle of radius around the fixed circle x + y of radius and following the initial point iof contact. The parameterization is cost cost c : [,π] t ct R sint sint. Sketch the initial configuration. Sketch the cardioid heart shaped curve.. Explain how the parameterization is obtained from the physical description rolling circle above. 3. Compute the velocity and acceleration vectors and the speed scalar. 4. Compute the arc length function st, t π and the total arc length L. 5. Compute the Seret-Frenet frame. 6. Compute the curvature function. Prove that κt 3/[ 8 sint/]. 7. Parameterize the involute curve i.e. compute it. 8. Parameterize the evolute curve i.e. compute et. 9. Prove that the involute is another cardioid.. Prove that the evolute is also a cardioid.. Sketch the cardioid, its ivolute and its evolute together on one diagram

3 . answer Figure : i initial configuration of wheel rolling on wheel, ii cardioid involute and evolute See figure??.. answer The argument is similar to that for the nephroid which is given in full, see below. Since both circles have the same radius one might assume that a spoke on the rolling circle turns with angular speed. That is wrong! The spoke turns with angular speed. See the detailed treatment of the nephroid below..3 answer velocity ċt ẋt ẏt sint + sint cost cost cos3t/ 4sint/ sin3t/ acceleration speed scalar ẍt ct ÿt cost + 4cost sint + 4sint ċt ẋ + ẏ 4sin t/cos 3t/ + sin 3t/ 4sin t/ 4sint/ 3

4 Caution:- in the current parametrization domain t π the speed is always positive as it should be. In a larger parametrization domain we might have to put ċt sint/..4 answer arc-length and total arc length The total arc length is st Z t c p d p, Z t 4sinp/d p 8cos p t 8cost/ + 8cos 8[ cost/] L sπ 8[ cosπ] 8[ ] 6 Warning:- the arc length must be adjusted for parameterization of the involute, see below..5 answer Seret-Frenet Frame tt ċt ċt cos3t/ 4sint/ 4sint/ sin3t/ cos3t/ sin3t/ nt ttċt cos3t/ tt sin3t/ sin3t/ cos3t/ The Seret-Frenet Frame is.6 answer curvature {t,n} κt { cos3t/ sin3t/, sin3t/ cos3t/ } ẋÿ ẏẍ ẋ,ẏ 3/ t 4

5 sint + sint sint + 4sint cost cost cost + 4cost 4sint/ 3 4 sint + sint sint + sint cost cost cost + cost 64sin 3 t/ sin t 3sint sint + sin t cos t + 3cost cost cos t 64sin 3 t/ sin t + cos t 3cost cost + sint sint + sin t + cos t 3cost t + 64sin 3 t/ 4 3 cost 64sin 3 t/ sin t/ 64sin 3 t/ 4sin t/ 64sin 3 t/ 3 8sint/ 64sin 3 t/.7 answer Above we saw that ċt 4 sint/ and that t Z t 5 ċp d p 8 cost/.

6 However use of these formulae in construction of it leads to a bizarre and uninteresting involute curve. As in the case of the cycloid we use st R t π ċp d p 8cost/. In effect we measure arc length from the mid not the start point of the cycloid. it ct sttt cost cost cos3t/ 8 cost/ sint sint sin3t/ cost cost cos3t/cost/ + 4 sint sint sin3t/ cost/ cost cost cost + cost + 4 sint sint sint + sint cost + cost 3 sint + sint Comparing it to the original ct we see that the involute is some kind? of cardioid scaled up by a factor 3. In fact we can manipulate the details and get it 3 Thus the involute is the original cardioid but scaled by 3 3ct π and reflected in the origin to explain the and time delayed by π.8 answer evolute et cost π cost π sint π sint π ct + nt κt cost cost sin3t/ + sint sint 3/8 sint/ cos3t/ cost cost + 8sint/ sin3t/ sint sint 3 cos3t/ cost cost + 4 sin3t/sint/ sint sint 3 cos3t/ sint/ cost cost + 4 cost cost sint sint 3 sint sint 6cost 3cost + 4cost 4cost 3 6sint 3sint 3 4sint 4sint cost + cost 3 sint + sint We see that et it/9 /3ct π. The description of the evolute is similar to that of the involute above but scaled,not by 3, but by /3. 6

7 .9 answer Sketch the involute,evolute and cardoid on one diagram. See figure?? 3 question on nephroid Let C R be the nephroid kidney shaped curve obtained by rolling a circle of radius around outside the fixed circle x + y 4 of radius and following the initial point iof contact. The parameterization is 3cost cos3t c : [,π] t ct R 3sint sin3t. Sketch the circles x +y 4,x +y 6 and x 3 +y and the nephroid curve together on one diagram.. Explain how the parameterization is obtained from the physical description rolling circle above. 3. Compute the velocity and acceleration vectors and the speed scalar. 4. Compute the arc length function st, t π and the total arc length L. 5. Compute the Seret-Frenet frame. 6. Compute the curvature function. Prove that κt /[ 3 sint ]. 7. Parameterize the involute curve i.e. compute it. 8. Parameterize the evolute curve i.e. compute et. 9. Prove that the involute is another nephroid.. Prove that the evolute is also a nephroid.. Sketch the nephroid, its involute and its evolute together in one diagram. 3. answer Figure : i initial configuration of wheel rolling on wheel, ii nephroid involute and evolute See figure??. 7

8 3. answer The center of the rolling circle travels on x + y 9, the circle of radius 3 centered at the origin. Initially this center is at position 3i and at time t its position is 3cos t at St3i. 3sin t One imagines a spoke on the rolling wheel extending from the center to the initial point of contact. Initially this spoke is represented by the vector b i. The position of the initial point of contact at time t is c a + b 3i i i. The small wheel rolls ACW, the crucial question is with what angular speed?. Since the ratio of radii is : one might think that this angular speed is. But that is wrong! The angular speed is in fact 3, see below for proof. Thus the position of the initial point of contact at time t is center + spoke 3cos t cos 3t ct at + bt St3i + S3t i 3sin t sin 3t We now give two arguments each of which justifies the assertion that the small wheel turns with angular speed 3. If the small wheel were fixed to the larger at its initial point of contact and both wheels rotated as a unit then the angular velocity of the spoke would be. If instead the small wheel rolls on the larger wheel it turns with additional angular speed. The total angular speed of the spoke is + 3. Make a diagram showing the small wheel in two positions, with center both at a and at aπ/. Draw in the spoke at position b i and position bπ/ j S3π/b. This shows that the spoke is turning at 3 times the angular speed of the center of the rolling circle. 3.3 answer velocity ċt ẋt ẏt 3cost cos3t The acceleration d dt 3sint sin3t sint + sin3t 3 cost cos3t use the trig formulae sin A sin B cos A + B cos A cos B sin A + B costsint 6 sint sint 6sint cost sint cost ct 6cost sint sin A B sin A B sint + sint cost There is no point in trying to simplify the acceleration as it will here be used only to compute a determinant en route to the curvature, see below. 8

9 The speed scalar ċt ẋ + ẏ cos 6sint t + sin t 6sint 6 sint Caution:- speed is always positive so be careful since the norm of velocity always involves a square root. 3.4 answer total arc length is L Z π Z /pi Z π c p d p, 6 sinp d p note that sin ± sin according to the value ofp. 6sinpd p + Z π 6cosp π + 6cosp /pi π 6[ ] + 6[ ] 4 π 6sinpd p The following arclength function works only in the range t π st Z t 6sinpd p 6cost answer Seret-Frenet Frame tt ċt ċt sint sint cost sint The fraction introduces a factor ±, nπ < t < n + π or n + π < t < n + π The Seret-Frenet Frame is {t,n} nt tt sint cost sint sint sint sint sint cost sint { cost sint sint, sint cost } 9

10 3.6 answer Recall the formula for curvature κt detċ, c < ċ,ċ > 3/ ẋ ẍ det ẏ ÿ < ċ,ċ > 3/ 6sintcost 6costcost sintsint det 6 sint sint 6 cost sint + sint cost 6 3 sint 3 6sintcost sintsint det 6 sint sint sint cost 6 3 sint 3 by an E.C.Operation, col -> col - cott col cost sint 6sint sintdet sint cost 6 3 sint 3 6sint sint 6 3 sint 3 3 sint 3.7 answer Before we compute the parameterization of the involute let us rearrange for use below the parameterization of the nephroid using the triple angle formulae. An alternative parameterization of the nephroid is cos3t 4cos 3 t 3cost sin3t 3sint 4sin 3 t ct 3cost cos3t 3sint sin3t 3cost 4cos 3 t + 3cost 3sint 3sint + 4sin 3 t

11 6cost 4cos 3 t 4sin 3 t Furthermore we measure distance along the nephroid from the point cπ/: ie we take st 6 cost. The following is only strictly valid in the range t π but yields an interesting wind on-off sort of involute if taken in the range < t <. it ct sttt 3cost cos3t cost + 6cost 3sint sin3t sint 3cost cos3t + 6costcost 3sint sin3t + 6costsint next express this in terms of cost and sint only 3cost 4cos 3 t 3cost + 6costcos t 3sint 3sint 4sin 3 t + 6costsintcost 6cost 4cos 3 t + cos 3 t 6cost 4sin 3 t + sintcos t 8cos 3 t 4sin 3 t + sint sin t 8cos 3 t 8sin 3 t + sint Comparison of it with the alternate formula for ct suggests that the involute is some sort of another nephroid. We make this remark precise as follows it 4cos 3 t 6sin 3 t 4sin 3 t 4sin 3 t π/ 6cos 3 t π/ 4cos 3 t π/ 6cos 3 t π/ 4cos 3 t π/ 4sin 3 t π/ 6cos 3 t π/ 4cos 3 t π/ 4sin 3 t π/ Sπ/c t π/ We have π it S c t π We interpret this as follows the involute is the original nephroid but scaled by. rotated by π/ traversed parameterized in the opposite sense and out of phase by π/. 3.8 answer evolute et ct + nt κt

12 3cost cos3t + 3 sint sint sint 3sint sin3t sint cost 3cost cos3t sint + 3sint 3sint sin3t cost 3cost cos3t sint + 3sint 3sint sin3t cost 3cost cos3t 3sintsint 3sint sin3t + 3sintcost Observe that the first coord is an even and the second an odd function of t. It seems sensible to reduce these coords to be in terms of cos and sin respectively. We will make use of the triple angle formulae cos3t 4cos 3 t 3cost sin3t 3sint 4sin 3 t et 3cost [4cos 3 t 3cost] 3sint[sintcost] 3sint [3sint 4sin 3 t] + 3sint[ sin t] 6cost 4cos 3 t 6sin tcost 3sint sin 3 t 6cost 4cos 3 t 6 cos tcost 3sint sin 3 t cos 3 t 3sint sin 3 t Comparison with the alternate nephroid parameterisation see above 6cost 4cos ct 3 t 4sin 3 t suggests that the evolute is itself a nephroid. The clinching observation is et 4cos 3 t 6sint 4sin 3 t 4sin 3 t π/ 6cost π/ 4cos 3 t π/ 4sin 3 t π/ 6cos t π/ 4cos 3 t π/ 6cos t π/ 4cos 3 t π/ 4sin 3 t π/ Sπ/c t π/ Thus et Sπ c t π We interpret this as follows the evolute is the original nephroid but scaled by /. rotated by π/ traversed parameterized in the opposite sense and out of phase by π/.

13 3.9 answer Sketch the involute,evolute and cardoid on one diagram. See figure?? 3. footnote cos3t + i sin3t exp3it expit 3 Taking real and imaginary parts [ cost + isint ] 3 4 question on nephroid caustic cos 3 t + 3icos tsint + 3i costsin t + i 3 sin 3 t cos 3 t + 3icos tsint 3costsin t isin 3 t [cos 3 t 3costsin t] + i[3cos tsint sin 3 t] [cos 3 t 3cost cos t] + i[3 sin tsint sin 3 t] [4cos 3 t 3cost] + i[3sint 4sin 3 t] cos3t 4cos 3 t 3cost sin3t 3sint 4sin 3 t Let cost c t 4 sint cos3t π, c t 4 sin3t π and ct 4 [3c t + c t]. Prove that the line segment l c t,c t meets the nephroid Γ see q3 tangentially at the point ct Γ.. Prove that the angle between the horizontal i and the radial r c t equals the angle between the radial r and the line segment l. 3. Prove that the rays of a horizontal beam of light reflect off the semicircular mirror { x x + y 4, x } to form a nephroid caustic. 4. See the nephroid formed by a bright light in your coffee cup. See figure??. 4. answer By definition ct 3 4 c t + 4 c t and therefore is a point on the line segment joining the point c t to the point c t, in fact it is the point dividing the line segment in parts with lengths in the ratio 3 :. Since ct 3 4 c t + 4 c t 3 cost cos3t π sint 4 4 sin3t π 3cost + cos3t π 3sint + sin3t π 3

14 Figure 3: envelop of reflected horizontal rays is a nephroid 3cost cos3t 3sint sin3t Thus ct is a point on the nephroid Γ see the standard parameterization of the nephroid. Thus indeed the line segment c t,c t meets the nephroid, i.e. at the common point ct. Next we will show that the tangent line at the point ct on the nephroid is parrallel to the line segment. This proves that as required the line segment meets the nephroid tangentially at the the point ct. The tangent vector has direction that of c 3sint + 3sin3t t 3cost 3cos3t while the line segment has direction that of 4cost 4cos3t π c t c t 4sint 4sin3t π These are parallel iff cost cos3t sint + sin3t iff cost cos3t cost + cos3t sint + sin3t cost + cos3t cost + cos3t sint + sin3t sint + sin3t sint + sin3t iff iff iff which is true. cos t cos 3t sin t + sin 3t cos t + sin t cos 3t + sin 3t 4

15 4. equiangles. The angle θ between the horizontal and the radial vector from to the point c t is of course θ t. The angle φ between the radial and the line segment is given by cosφ < c t,c t c t > c t c t c t / 4cost 4cost 4cos3t π <, > 4 4cost 4cos3t π 4sint 4sint 4sin3t π 4sint 4sin3t π / cost cost + cos3t cost + cos3t <, > sint sint + sin3t sint + sin3t + cos3tcost + sin3tsint + cos3tcost + sin3tsint + cos3t t + cos3t t + cost + cost Thus θ φ t. + cost + cos t cost 4.3 answer, nephroid from circular mirror See figure??. Figure 4: envelop of reflected horizontal rays is a nephroid 5

16 5 question and answer, conical helix Conical Helix with Darboux vector Let Γ be the conical helix curve with parametrization ct xti + ytj + ztk t costi + t sintj + tk.. Sketch Γ K S R 3 showing the curve as the intersection of cone and screw surfaces.. Compute tt, nt, bt. [Hint:- The computation is brutal here but: if you write vt costi + sint and wt vt and note that {v,w,k} is a P.O.N. triple with easily computed dot and cross products, then life becomes a lot easier. For example c is simply v +tw + k.] 3. Prove that κt t4 + 5t + 8 / t + 3/, τt t + 6 t 4 + 5t Let d : R t dt R 3 be the Darboux vector at the point ct Γ; i.e. dt is the eigenvector associated with eigenvalue of the curvature matrix, i.e. the angular velocity of the S.F. frame. Express dt in terms of the S.F. basis t,n,b. 5. Express dt in terms of the standard basis i,j,k. 6. Compute et S k tdt. [ S k, of course, denotes standard rotation about the z-axis]. 7. Compute dt et, the angular speed of the S.F. frame. Prove that lim t dt. dt 8. Prove that lim t dt k. 9. Explain, in your own words, what all this means. 5. answer See figures?? and??. Figure 5: i conical helix curve, ii intersection of cone and screw surfaces 6

17 5. answer First write, for convenience, vt costi + sintj and wt sinti + costj Note that v, w, k is a positively oriented orthonormal triple: all inner products and all cross products amongst these three vectors are known trivial. Also v w and w v c tv +tk v +tv + k vt +tw + k c v +tw + k v + w +tw + k w + w tv + w tv We compute s t 5.3 answer, S.F. Frame s t < c,c > / c c w tv w v tv v v tw 3v tw < vt +twt + k,vt +twt + k > / +t using orthonormality 6 times +t To compute t,n,b we first need the following quantities < c,c >,< c,c >,c c,< c c,c c > and < c c,c >. < c,c > < vt +twt + k,vt +twt + k > +t < c,c > < vt +twt + k, tvt + w > t 7

18 c c v +tw + k w tv v w tv v + tw w t w v + k w tk v k t + t +t k v tw k +t k v tw v tw + +t k < c c,c c > < v tw + +t k, v tw + +t k > 4 +t + +t 8 + 5t +t 4 < c c,c > < v tw + +t k, 3v tw > t + 6 Now we are ready to compute t,n,b,κ,τ. The unit binormal vector is The unit normal is n b t v tw + t + k v +tw + k 8 + 5t +t 4 t + t c c v +tw + k +t +t b c c c c cost t sint sint +t cost see expressions for v,w v tw + t + k 8 + 5t +t t +t 4 cost +t sint sint t cost t + see expressions for v,w 8

19 v v tv w v k tw v t w w tw k + t + k v + t + tk w + t + k k 8 + 5t +t 4 t + tk + w +tk t tv + t + w t + tv + t t +t 4 t + t3 3tv + t + 4w tk 8 + 5t +t 4 t t +t 4 t + Finally the S.F. frame is { } t,n,b t3 3tcost t + 4sint t 3 3tsint + t + 4cost t +t cost t sint sint +t cost, 8 + 5t +t 4 t + t 3 3tcost t + 4sint t 3 3tsint + t + 4cost t, 8 + 5t +t 4 cost +t sint sint t cost t answer, curvature and torsion 5.5 answer, Darboux vct w.r.t S.F.basis κ < c c,c c > / < c,c > 3/ t4 + 5t + 8 / t + 3/ τ < c c,c > < c c,c c > t + 6 t 4 + 5t + 8 dt is by definition the angular velocity converted from matrix dt κ κ τ τ τt κt where κ and τ are as above. But this column vector is w.r.t. to the S.F frame. In other words 5.6 answer, Darboux vct w.r.t. standard basis dt τttt + κtbt 6 +t 8 3t +t 4 8 3t +t 4 Note the following which we will use below dt τttt + κtbt cost t sint sint +t cost dt 6 +t 8 3t +t 4 8 3t +t 4 S kt S k t 6 +t 8 3t +t 4 8 3t +t 4 t + + t + form to vector form and is 8 + 5t +t t +t 4 S kt 8 + 5t +t 4 cost +t sint sint t cost t + t t + t t + 9

20 5.7 answer, computing et From the last remark, since S k t S k t S k I. 5.8 answer, dt Note that et 6 +t 8 3t +t 4 8 3t +t 4 t t +t 4 dt S k tet et t t + We can compute dt with a hack slog from the preceeding explicit equation for e. But a clever way is to use Pythagoras theorem on the expression d τt + κb So that d τ + κ t + 6 t 4 + 5t t4 + 5t + 8 t + 3 It is clear without working details that dt like /t, ie that t dt as t 5.9 answer, computing limiting direction of Darboux vct It is trivial to check that Thus dt lim t dt lim tκt, lim tτt t t tdt lim t t dt lim t tdt lim t t dt lim t tdt lim tdt t lim tτttt + tκtbt t lim tκtbt t since limtκt and t is a unit vector

21 tt 4 + 5t + 8 / lim t t + 3/ 8 + 5t +t 4 cost +t sint sint t cost t + t lim t t + 3/ cost +t sint sint t cost t + k The last step follows by taking the straightforward limit of each of 3 vector components; in particular 5. answer, explain tt + lim zt lim t t t + 3/ The speed with which the SF frame spins, i.e. d tends to zero over time but the direction of spin d/ d settles down to the vertical k, compare with the situation for the standard helix where d is constant at k. The vector e could have been put to use in the above computations but proved unnecessary in the method used here.