MATH 227 MIDTERM 1 – SOLUTIONS Wednesday, February 4, 2009 1. (a) Sketch the flow lines of the vector field F(x, y, z) = x2 y x i− 2 j. 2 +y x + y2 The flow lines are circles centered at the origin, oriented clockwise. (b) Calculate the curl of the vector field in (a). i j k −(x2 + y 2 ) + x(2x) (x2 + y 2 ) − y(2y) ∂y ∂z = ∇ × F = ∂x − k (x2 + y 2 )2 (x2 + y 2 )2 2 y 2 2−x 2 0 x +y x +y −2(x2 + y 2 ) + 2x2 + 2y 2 k = 0. = (x2 + y 2 )2 2. Prove that if f and g are two C 2 functions on an open set X ⊂ R3 , then ∇ · (∇f × ∇g) = 0. We first compute i j k ∇f × ∇g = ∂x f ∂y f ∂z f = (fy gz − fz gy )i − (fx gz − fz gx )j+)fx gy − fy gx )k, ∂x g ∂y g ∂z g so that ∇ · (∇f × ∇g) = (fxy gz + fy gxz − fxz gy − fz gxy ) − (fxy gz + fx gyz − fyz gx − fz gxy ) + (fxz gy + fx gyz − fyz gx − fy gxz ) = 0. 3. The length of a curve x(t), t ∈ [0, ∞), is defined as the limit of the length of x(t) between t = 0 and t = R as R → ∞. Prove that the curve x(t) = (e−at cos t, e−at sin t), t ∈ [0, ∞), in R2 has finite length for any a > 0, and find all values of a for which the curve has length 2. We have x0 (t) = (−ae−at cos t − e−at sin t, −ae−at sin t + e−at cos t) = e−at (−a cos t − sin t, −a sin t + cos t), so that 1/2 kx0 (t)k = e−at a2 cos2 t − 2a cos t sin t + sin2 t + a2 sin2 t + 2a cos t sin t + cos2 t √ = e−at a2 + 1, and the length of the curve is Z R Z √ −at 2 L(a) = lim e a + 1dt = R→∞ 0 0 ∞ √ √ e−at ∞ −at 2 2 e a + 1dt = a + 1 = −a 0 √ a2 + 1 , a which is finite. We used here that limx→∞ e−at = 0 if a > 0. √ √ Suppose that√L(a) = 2, then a2 + 1 = 2a, a2 + 1 = 4a2 , 3a2 = 1, a = 1/ 3. (We discard the negative root a = −1/ 3 because we assumed that a > 0. Also, when a < 0, the curve has infinite length.) 4. Find the unit tangent and principal normal vectors, and the curvature at a general point on the curve 3 2 x(t) = ti − t2 j + t3 k. √ We have x0 (t) = (1, −t, t2 ) and s0 (t) = kx0 (t)k = 1 + t2 + t4 . Hence 1 1 1 1 2 T= √ (1, t, t ) = √ ,√ ,√ , 1 + t2 + t4 1 + t2 + t4 t−2 + 1 + t2 t−4 + t−2 + 1 −4t3 − 2t) −2t−3 + 2t 4t−5 + 2t−3 , , T (t) = 2(1 + t2 + t4 )3/2 2(t−2 + 1 + t2 )3/2 2(t−4 + t−2 + 1)3/2 −2t3 − t −1 + t4 2t + t3 = , , (1 + t2 + t4 )3/2 (1 + t2 + t4 )3/2 (1 + t2 + t4 )3/2 0 T0 (t) = κN = T (s) = 0 s (t) 0 κ = kT0 (s)k = −1 + t4 2t + t3 −2t3 − t , , (1 + t2 + t4 )2 (1 + t2 + t4 )2 (1 + t2 + t4 )2 1 4t6 + 4t4 + t2 + t8 − 2t4 + 1 + t6 + 4t4 + 4t2 , 1/2 (1 + t2 + t4 )2 1/2 1 t8 + 5t6 + 6t4 + 5t2 + 1 = 2 4 2 (1 + t + t ) 1/2 (1 + t2 + t4 )(t4 + 4t2 + 1) = (1 + t2 + t4 )2 (t4 + 4t2 + 1)1/2 = . (1 + t2 + t4 )3/2 5. Recall the Frenet-Serret formulas: T0 (s) = κN, N0 (s) = −κT + τ B, B0 (s) = −τ N. Find a vector W such that these formulas can be written as T0 (s) = W × T, N0 (s) = W × N, B0 (s) = W × B. Let W = aT + bN + cB, where a, b, c are unknown parameters to be found. Then W × T = bN × T + cB × T = −bB + cN = κN, W × N = aT × N + cB × N = aB − cT = −κT + τ B, W × B = aT × B + cN × B = −aN + bT = −τ N, so that a = τ , b = 0, c = κ, W = τ T + κB.