MA34 Geometry and Motion Selected answers to Sections A and C Dwight Barkley 26 Example Sheet d n+ = d n cot θ n r θ n r = Θθ n i. 2. 3.
4. Possible answers include: and with opposite orientation: 5.. sin 2, cos 2, 2 and sin 2, cos 2, 2. 2. Yes. No. rt = + 3 cos t, 2 + 3 sin t, t [, 2π] rt =, 2 + 3 cos t, 3 sin t, t R rt =, 2 + 3 cos2πt, 3 sin2πt, t [, ] rt =, 2 + 3 cos2πt 2, 3 sin2πt 2, t [, ] rt =, 2 + 3 sin t, 3 cos t, t [, 2π] rt =, 2 + 3 cos t, 3 sin t, t [, 2π] rt =, 2 + 3 cos t, 3 sin t, t [, 2π] rt =, 2 + 3 sin t, 3 cos t, t [, 2π] rt = 2, 2 + 2 cos t, sin t, t [, 2π]. vt = r t = t, 2t 3/2, 2t 2 Example Sheet 2 c = vt = t 2 + 4t 3 + 4t 4 /2 = t + 4t + 4t 2 /2 = t + 2t at = v t =, 3t /2, 4t Length of path lc = 4 r t = 4 ct = 4 t + 2t 2 = 8 + 28/3. 2. From the relationship between Cartesian and polar coordinates, the polar parametrisation gives a Cartesian parametrisation Differentiating gives xt = rt cos θt, yt = rt sin θt. x t = r t cos θt rt sin θt θ t y t = r t sin θt + rt cos θt θ t The formula for arc length of a curve in the plane in Cartesian coordinates is lc = b a x 2 t + y 2 t /2 substituting the above and simplifying gives the desired result lc = b a [ r t 2 + rt 2 θ t 2 ] /2
3. lc = π r t 2 + r 2 tθ t 2 π /2 = 4 sin 2 t + 4 cos 2 t π /2 = 2 = 2π. 5. Start from rt 2 = rt rt so where in the last step rt. d rt 2 = d rt rt 2 rt d rt = r t rt + rt r t = 2rt r t d rt = rt r t rt. 2. parametrisation: rt = t, t 2 /2, t 3 /6. Length = 42. 3. 3 sin, 4, 3 cos. Example Sheet 3. a κt = 2/4t 2 + 3/2 and b κt = / + 2t 2 3/2 2. T = / 2, / 2, N = / 2, / 2, κt = 2e t. for sketch, radius of osculating circle is ρ = 2 and centre is at, + ρn =, + 2 / 2, / 2 =, 3. Ts = 4 5, 3 5 cos s 5, 3 5 sin s s, Ns =, sin 5 5, cos s 3, Bs = 5 5, 4 5 cos s 5, 4 5 sin s 5 8. x, y = 2 ln 2, / 2. κ = 3/25, ρ = 25/3, τ = 4/25.
. a b c 2. a b 3. a Example Sheet 4 x = 3y x = πe t sin πx = 2y/x + y2 x y = 5y4 3x y f = e xy + xy, x 2 t = e t cos πx f = y x + y x, = 2x/x + y2 dg = 2x + y, 2y + x cos t, sin t = 2 sin t + cos t cos t 2 cos t + sin t sin t = cos 2t Verify: 4. By definition fx, y = y2 x 2, 2y x f, 2 = 4, 4 gt = + 2 sin2t. dg = cos 2t. D u f, 2 = f, 2 u = 4, 4 2/3, 5/3 = 8 3 + 4 5 3 D u f, 2 = lim f + 2h/3, 2 + 5h/3 f, 2 h h 2 + 5h/3 2 = lim 4 h h + 2h/3 4 5h/3 8h/3 + 5h 2 /9 = lim h h + 2h/3 4 = lim h 5/3 8/3 + 5h/9 + 2h/3 = 4 3 5 2 = 4 5/3 8/3 = 4 3 5 2
5. x 2 = y 2 = xey sin πz z 2 = π2 xe y sin πz x y = ey sin πz y z = πxey cos πz x z = πey cos πz 4. ω 2 /k 2 = a 2. a Example Sheet 5 and similarly for b. fx, y = 2x + 3 4y + f, + x, x +, y = 3 + 2x 2y y 3. This elliptic paraboloid is the zero level set of fx, y, z = 2x 2 + y 2 z. f,, 3 is normal to the surface at,, 3 and hence normal to tangent plane through the point,, 3. Thus the equation for the tangent plane is f,, 3 x, y, z 3 = 4x + 2y z 3 = Note, the above f is natural to use, but it is not unique. For example fx, y, z = z 2x 2 y 2 might have been used giving an equally valid i.e. equivalent equation for the tangent plane 4x 2y + z 3 = 4. We need linear approximations to f and g based at,. Evaluating the partial derivatives gives f, =, g, =,, = x g g, = x, = y, = y so in the linear approximation the ODEs become ẋ = y ẏ = x Solution with x =., y = is xt =. cost yt =. sint 5. x = 3x2 y = gives y = x 2, y = 3y2 x = gives x = y 2. Solving by plugging first into second gives x = x 4, so x = or = x 3 giving x =. Using y = x 2 these give y = and y =. Hence critical points are, and,. = 6x, 2 f = 6y, 2 f x 2 y 2 x y = 3, so D = 36xy 9. At,, D < so saddle. At,, D = 25 > and = 6 > so minimum. x 2 x = y = /2 is between saddle and minimum so initial condition will roll down hill to the minimum. Hence as t the xt, yt,.
. a Example Sheet 6 3 + 4xy dx dy = 3 [ x + 2x 2 y ] 3 dy = + 2y dy = b or, since it separates 2 π/2 2 π/2 x sin y dy dx = 2 [ 2 x sin y dy dx = [ x cos y] π/2 dx = 2 + x dx = 2 ] [ ] π/2 x dx sin y dy = 2 = 2 2. It was assumed that the rectangle is in the x-y plane. Probably the question should have stated that explicitly. The upper boundary of solid is the value of z, which is a function f of x, y z = fx, y = x2 4 y2 9 So volume under this function, above the rectangle is 2 V = fda = x2 4 y2 dx dy = 66/27 9 R 2 This could also be done as a triple integral V Ω = where Ω dv Ω = {x, y, z x, 2 y 2, z x2 4 y2 9 } V Ω = 2 2 x 2 4 y2 9 dz dx dy = 2 2 [ z] x 2 4 y2 2 9 dx dy = x2 2 4 y2 dx dy 9 so it reduces to the previous case. 3. 4/3. 4. a and b e 3 /3. 3. 8/3e 4a i α/2 and ii α. a z = r 2, b r = 2 cos θ. Example Sheet 7
3. Use cylindrical coordinates. Region in cylindrical coords: Ω = {r, θ, z r 4, θ 2π, 5 z 4} Function in cylindrical coords: x 2 + y 2 = r 4 2π 4 Hence: x 2 + y 2 dv = r rdr dθ dz = 9 2π 4 3 /3 = 384π. 32a 3 /9.. a 3 9π 6/36. Ω 5 2. dv = abc r 2 sin φ dr dθ dφ. 3. a b e 2 /2 8. 3/2 sin 9. e /e Example Sheet 8 Example Sheet 9. For disk, question did not state radius or orientation. For a radius R in the x y plane use parametrisation Then For cylinder use parametrisation rr, θ = r cos θ, r sin θ,, r, θ [, R] [, 2π] r r r θ = r ds = r dr dθ rθ, z = R cos θ, R sin θ, z, θ, z [, 2π] R Here I use z for parameter since it does correspond to z in Cartesian coordinates. One could use other parameter names as in the notes. Also, one could limit the height, but not important for element of surface area ds. For sphere use parametrisation r θ r z = R ds = R dθ dz rθ, φ = R cos θ sin φ, R sin θ sin φ, R cos φ, θ, φ [, 2π] [, π] r θ r φ = R2 sin φ ds = R 2 sin φ dθ dφ 2. For above parametrisation AS = S ds = π/2 2π sin φ dθ dφ = 2π
For other parametrisation evaluating ds gives ds = u 2 v da 2 where region Ω in u, v coordinates is u 2 + v 2. AS = S ds = Ω u 2 v 2 da This double integral is evaluated by changing to polar coordinates AS = 2π r dr dθ = 2π r 2 3. Many parametrisations are possible. A natural one would be or equivalently with different notation. rr, θ = r cos θ, r sin θ, r, r, θ [, ] [, 2π] rθ, z = z cos θ, z sin θ, z, θ, z [, 2π] [, ] 4. Uinsg parametrisation for disk from Q Flux is For this parametrisation, r r r θ ds = 2r dr dθ v =,, x 2 + y 2 =,, r 2 AS = 2π r ± v Ω r r drdθ θ r r r =,, r θ points upward so use + sign, so flux is 2π,, r 2,, r dr dθ = π 2 8. 8π. 9. 2π 2.. Hint: in this problem the region in Ω is not a rectangle. I get AS = 8.. π/2 3. 2π/5 The tangent plane is given by the parameterised surface: ph, k = 2 + / 2,, / 2 + h 2,, 2 + k, 2 + 2, with h, k R. Redefining h and k this can be simplified to ph, k = 2 + / 2,, / 2 + h,, + k,, Example Sheet. 24π and 8π 2. 5π 3. π 4. log8/5 6. By inspection F = f where fx, y, z = cos x cos y cos z. 7. Many ways to show this, e.g. compute line integral around the closed curve square with vertices,,, π/2,,, π/2, π/2,,, π/2, and show result is not zero. 8. 2πα in both cases.