Reparametrize the Curve

Problem Set 5 Section 14.3: 5) Reparametrize the curve   2 2t r(t) = 2 −1 i+ 2 j t +1 t +1 with respect to arc length measured from the point (1, 0) in the direction of t. Express the reparametrization in the simplest form. What can you conclude about the curve? Solution At (1, 0), t = 0. 0 r (t) = |r0 (t)| = = = 2 − 2t2 −4t i+ 2 j (t2 + 1)2 (t + 1)2  4 1/2
1/2 4t + 8t2 + 4 2 4 2 = t + 2t + 1 (t2 + 1)4 (t2 + 1)2 2 (t2 + 1) 2 2 (t + 1) 2 t2 + 1 Then, the arc length function s is Z t Z t 2 0 s= |r (u)|du = du = 2tan−1 (t) 2 0 0 u +1 =⇒ t = tan(s/2) The reparametrization with respect to arc length be1 comes  r(t(s)) = = = = =  2 2tan(s/2) −1 i+ 2 tan (s/2) + 1 tan2 (s/2) + 1   2tan(s/2) 2 − 1 i + j sec2 (s/2) sec2 (s/2) (2cos2 (s/2) − 1)i + 2sin(s/2)cos(s/2)j     1 + cos(s) 2 − 1 i + sin(s)j 2 cos(s)i + sin(s)j with s ≥ 0. The curve is a circle. 6) At what point does the curve y = lnx have maximum curvature? What happens to the curvature as x → ∞? Solution curvature_lnx.pdf The curvature To find the maximum of the curvature, we should 1/x2 x |f 00 (x)| check the critical points. The end= = κ(x) =the endpoints and 2 ]3/2 [1 + (f 0 (x)) [1 + 1/x2κ(x) ]3/2 →[x02 as + seen 1]3/2 points are x→ 0 and x→∞ where in the plot above. We can show this by using limits: x lim κ(x) = lim = 0. x→0 x→0 [x2 + 1]3/2 lim κ(x) = lim x→∞ x→∞ x [x2 + 1]3/2 = lim x→∞ 3 2 1 [x2 + 1]1/2 2x The critical points occur where (x2 + 1)3/2 − x · 32 (x2 + 1)1/2 · 2x κ (x) = =0 [x2 + 1]3 0 2 = 0. Then (x2 + 1)3/2 − x · 23 (x2 + 1)1/2 · 2x = 0, [x2 + 1]3 =⇒ −2x2 + 1 = 0, 1 x = ±√ , 2 1 x = √ , 2 since ln(x) takes in only positive arguments. Then, the maximum curvature   2 1 = √ , κ √ 2 3 3    and κ(x) → 0 as x → ∞. occurs at √12 , ln √12 7) Find the unit tangent, T, the unit normal, N, and the binormal vector, B, to r(t) = het , et sint, et costi at the point (1, 0, 1). Solution At (1, 0, 1), t = 0. T(t) = r0 (t) = |r0 (t)| = = r0 (t) |r0 (t)| het , et sint + et cost, et cost − et sinti [e2t + e2t (sint + cost)2 + e2t (cost − sint)2 ]1/2 √ t 3e Then, 1 T(t) = √ h1, sint + cost, cost − sinti, 3 3 so, 1 T(0) = √ h1, 1, 1i, 3 N(t) = T0 (t) = |T0 (t)| = = T0 (t) |T(t)| 1 √ h0, cost − sint, −sint − costi 3 1 √ [cos2 t − 2costsint + sin2 t + sin2 t + 2costsint + cos2 t]1/2 r3 2 3 Then, 1 N(t) = √ h0, cost − sint, −sint − costi, 2 so 1 N(0) = √ h0, 1, −1i 2 i j k 1 1 B(0) = T(0) × N(0) = √ 1 1 1 = √ h−2, 1, 1i 6 0 1 −1 6 4 Section 14.4 8) Use a computer to graph the path of the particle with acceleration a(t) = ti + t2 j + cos2tk, initial velocity v(0) = i + k and initial position r(0) = j. Solution Z v(t) = Z a(t)dt = t3 sin2t t2 k + C1 [ti + t j + cos2tk]dt = i + j + 2 3 2 2 v(0) = i + k = C1 Then  v(t) =    t2 t3 sin2t +1 i+ j+ +1 k 2 3 2     t2 t3 sin2t v(t)dt = +1 i+ j+ + 1 k dt 2 3 2     3 t4 −cos2t t +t i+ j+ + t k + C2 6 12 4 −1 k + C2 j= 4 1 j+ 4 Z r(t) = = r(0) = C2 = Z  Thus,  r(t) =   4    t3 t 1 − cos2t +t i+ +1 j+ +t k 6 12 4 which gives the path 5 9) Find the tangential and normal components of the acceleration vector for the position vector r(t) = ti + cos2 tj + sin2 tk. Solution The velocity vector is v(t) = r0 (t) = i − sin(2t)j + sin(2t)k a(t) = r00 (t) = −2cos(2t)j + 2cos(2t)k a = aT T + aN N v 2sin(2t)cos(2t) + 2sin(2t)cos(2t) 2sin(4t) p p = = |v| 1 + 2sin2 (2t) 1 + 2sin2 (2t)   i j k   |r0 (t) × r00 (t)| 1 1 −sin(2t) sin(2t) = =p |r0 (t)| 1 + 2sin2 (2t)  0 −2cos(2t) 2cos(2t)  √ | − 2cos(2t)j − 2cos(2t)k| 2 2cos(2t) p = =p 1 + 2sin2 (2t) 1 + 2sin2 (2t) aT = a · aN Section 15.1 9) Let f (x, y) = ln(x + y − 1). (a) Evaluate f (1, 1). (b) Evaluate f (e, 1). (c) Find and sketch the domain of f . (c) Find the range of f . 6 Solution (a) f (1, 1) = ln(1 + 1 − 1) = ln(1) = 0 (b) f (e, 1) = ln(e + 1 − 1) = ln(e) = 1 (d) The range of f is the range of the natural log which is −∞ < f (x, y) < ∞ (c) Domain of f is x + y − 1 > 0 or y > 1 − x. 10) Draw a contour map of the function f (x, y) = y/(x2 + y2) showing at least four level curves. 11) Describe the level surfaces of (a) f (x, y, z) = x2 + 3y2 + 5z2 (b) f (x, y, z) = x2 − y2 Solution (a) The level suraces of f (x, y, z) = x2 + 3y 2 + 5z 2 are k = x2 + 3y 2 + 5z 2 , where k ≥ 0. They form a family of ellipsoids. (b) The level surfaces of f (x, y, z) = x2 − y 2 are k = x2 − y 2 where k is any real number. The level surfaces are a family of hyperbolic cylinders with axis the z-axis. Section 15.2 7 12) Find the limit, if it exists, or show that the limit does not exist for (a) lim xycos(x − 2y) (x,y)→(6,3) x2 + sin2 y (b) lim (x,y)→(0,0) 2x2 + y 2 Solution (a) xycos(x − 2y) = (6)(3)cos(6 − 2(3)) = 9 lim (x,y)→(6,3) (b) Approach along x-axis. =⇒ y = 0. Then, x2 1 x2 + sin2 y = lim 2 = lim 2 2 x→0 2x 2 (x,y)→(0,0) 2x + y Approach along y-axis. =⇒ x = 0. Then, x2 + sin2 y sin2 y 2sinycosy lim = lim = lim y→0 y 2 y→0 2y (x,y)→(0,0) 2x2 + y 2 sin2y 2cos2y = lim = lim =1 y→0 2y y→0 2 The function has two different limits along two different lines, thus the limit does not exist. 13) Use polar coordinates to find the limit lim (x2 + y 2 )ln(x2 + y 2 ). (x,y)→(0,0) [Note that if (r, θ) are the polar coordinates of the point (x, y) with r ≥ 0, then r → 0+ as (x, y) → (0, 0).] 8 Solution lim (x2 + y 2 )ln(x2 + y 2 ) = lim+ r2 lnr2 = 2 lim+ r2 lnr r→0 (x,y)→(0,0) r→0 lnr 1/r = 2 lim r→0 1/r 2 r→0+ −2/r 3 = − lim+ r2 = 0. = 2 lim+ r→0 14) Use spherical coordinates to find xyz . lim (x,y,z)→(0,0,0) x2 + y 2 + z 2 Solution xyz ρ3 sin2 φcosφsinθcosθ lim = lim+ ρ→0 ρ2 (x,y,z)→(0,0,0) x2 + y 2 + z 2 = lim+ ρsin2 φcosφsinθcosθ = 0 ρ→0 9