Advanced Calculus (MAU11202), Homework Assignment 1 Solutions Straightforward 1. [3 points] Find the derivative of the vector-valued function 2 t −1 , sin(t) ln(t), cos(sin(t)) . r(t) = 2 t +2 What is r0 (1) equal to? Solution: To differentiate r(t) = (x(t), y(t), z(t)), we differentiate the three component functions separately. Using the Quotient Rule, x0 (t) = 2t3 + 4t − 2t3 + 2t 6t 2t(t2 + 2) − 2t(t2 − 1) = = 2 . 2 2 2 2 (t + 2) (t + 2) (t + 2)2 Using the Product Rule, y 0 (t) = cos t ln(t) + sin(t) · sin(t) 1 = cos(t) ln(t) + . t t Using the Chain Rule, z 0 (t) = cos(t)(− sin(sin(t))) = − cos(t) sin(sin(t)). Hence sin(t) 6t , cos(t) ln(t) + , − cos(t) sin(sin(t)) . r (t) = (t2 + 2)2 t Letting t = 1 in this expression, we obtain 6(1) sin(1) 0 r (1) = , cos(1) ln(1) + , − cos(1) sin(sin(1) (12 + 2)2 1 2 = , sin(1), − cos(1) sin(sin(1)) . 3 0 2. [3 points] Find the derivative of the vector-valued function 2 t −1 3 2 r(t) = (2t − 3t + 5t + 3) 2 , sin(t) ln(t), cos(sin(t)) . t +2 What is the domain of this function? Solution: Here we use Theorem 1.2.11(c) (Chapter 1 of course notes by Anthony Brown), the solution to Question 1, and the fact that the derivative of f (t) = 2t3 −3t2 +5t+3 is Fb(t) = 6t2 −6t+5, to obtain 6t sin(t) 0 3 2 r (t) =(2t − 3t + 5t + 3) , cos(t) ln(t) + , cos(t) cos(sin(t)) (t2 + 2)2 t 2 t −1 2 + (6t − 6t + 5) 2 , sin(t) ln(t), cos(sin(t)) . t +2 The domain of r is R ∩ (0, ∞) ∩ R = (0, ∞). 1 of 9 3. [3+1 points] Find an arc length parameterization of the curve h πi r(t) = cos3 (t)i + sin3 (t)j t ∈ 0, , 2 in the same direction as the original curve. Try to write your result without sin and cos. Solution: Here we are given our reference point t0 = 0, so we don’t have to calculate it. Next, let us calculate the arc length of the curve from t = 0 to t. Using the Chain Rule, we obtain r0 (t) = −3 sin(t) cos2 (t)i + 3 cos(t) sin2 (t)j. Hence q kr (t)k = 9 sin2 (t) cos4 (t) + 9 cos2 (t) sin4 (t) q = 9 sin2 (t) cos2 (t)(cos2 (t) + sin2 (t)) q = 9 sin2 (t) cos2 (t) 0 = 3 sin(t) cos(t). Thus the required arc length is t Z 0 Z kr (v)k dv = s(t) = 0 0 t t 3 2 3 sin (v) = sin2 (t), 3 sin(v) cos(v) dv = 2 2 0 where we have integrated ‘by inspection’ If you can’t spot the integral, then you can use the substitution u = sin(v). So we have s = 32 sin2 (t), and solving this for t we obtain 3 s = sin2 (t) 2 =⇒ r 2s sin2 (t) = 3 =⇒ sin t = 2s 3 r t = sin−1 =⇒ 2s 3 ! . Substituting this into r(t), we obtain r r(s) = cos3 sin−1 2s 3 !! r i + sin3 sin−1 2s 3 !! j Now, when t = 0, s = 0 and when t = π2 , s = 32 , so we obtain the arc length parametrization r !! r !! 3 2s 2s 3 −1 3 −1 r(s) = cos sin i + sin sin j s ∈ 0, . 3 3 2 The additional (+1) point goes to students trying to simplify this expression: Although this is a perfectly good answer, it doesn’t look particularly ‘nice’. We can get a ‘nicer’ answer (although of course this is a matter of opinion) by instead of solving s = 2 of 9 r 3 2 2 sin (t) for t, stopping when we got an expression for sin(t), that is sin t = 32 2s Then sin3 (t) = and 3 2s = 3 2s 3 12 . q 3 3 32 2s 2 2 2 cos (t) = 1 − sin (t) = 1 − sin (t) = 1 − . 3 3 So we obtain the parametrization r(s) = 2s 1− 3 32 i+ 2s 3 32 3 . j s ∈ 0, 2 t2 − 9 2 4. [3 points] Let r(t) = 2t + 1, , t . Find limt→3 r(t). Is r(t) continuous t = 3. t−3 Solution: From the lecture, we know that t2 − 9 2 lim r(t) = lim 2t + 1, lim , lim t = (7, 6, 9) t→3 t→3 t→3 t − 3 t→3 (1) It is not continuous at t = 3 because it is note defined there. √ R 5. [3 points] Find r(t) = r0 (t)dt, where r0 (t) = e2t , t, sin t , and r(π) = (0, 0, 0). Solution: From the lecture, we know that Z Z Z Z √ 1 2t 2 3 0 2t tdt, sin tdt = r (t)dt = e dt, e , t 2 , − cos t + c 2 3 (2) Evaluating at π yields the equation 1 2π 2 3 1 2π 2 3 e , π 2 , − cos π + c = e , π 2 , 1 + c = (0, 0, 0) 2 3 2 3 1 2π 2 3 which implies that c = − e , π2,1 . 2 3 (3) Intermediate 6. [3 points] Prove the Serret-Frenet formulae (a) dT ds = κ(s)N(s) Solution: By definition N(s) = dT ds dT | ds | 3 of 9 = dT ds κ(s) . (4) (b) dB ds = −τ (s)N(s), where τ is a scalar called the torsion of r(s). Solution: By definition dB dT dN dN = ×N+T× =T× . (5) ds ds ds ds Thus dB is normal to T. Since B is normal, dB is also normal to B. Thus, it is ds ds proportional to N. (c) dN ds = −κ(s)T(s) + τ (s)B Solution: We know that N = B × T, B = T × N, T = N × B. (6) Thus dN dB dT = ×T+B× = −τ (s)N × T + κ(s)B × N = −κ(s)T(s) + τ (s)B . (7) ds ds ds 7. [6 points] Consider the vector-valued function r(t) = (4 cos(t), 4 sin(t), 3t), t ∈ R. (a) Find an arc-length parameterization for the function with reference point t0 = 0. Solution: Here we are given our reference point t0 = 0, so we don’t have to calculate it. Next, let us calculate the arc length of the curve from t = 0 to t. We already did this in lectures, finding that s = 5t. s Thus t = and substituting this into r(t), we obtain the arc length parametrization 5 s s 3s r(s) = 4 cos , 4 sin , s ∈ R. 5 5 5 (b) Calculate the curvature κ of r, using the arc-length parameterization you found in Part (a) and the formula κ(s) = kT0 (s)k. Solution: Since we now have a unit speed parametrization, using Remark 1.4.2(b), s 4 s 3 4 0 T(s) = r (s) = − sin , cos , . 5 5 5 5 5 Hence 0 T (s) = s s 4 4 − cos , − sin ,0 . 25 5 25 5 Thus r 0 κ(s) = kT (s)k = s 42 42 2 s 2 cos + 2 sin = 252 5 25 5 r 42 4 = . 2 25 25 Note that the curvature is independent of the parametrization, as we would want for any reasonable measure of ‘curvature’. 4 of 9 (c) Calculate the torsion τ of r, using the arc-length parameterization you found in Part (a) and the formula τ (s) = −N(s) · B0 (s). Solution: We need to calculate N(s), B(s) and hence B0 (s). Firstly T0 (s) kT0 (s)k s s 1 4 4 = − cos , − sin 4/25 25 5 25 5 s s = − cos , − sin ,0 . 5 5 N(s) = Next, using B(s) = T(s) × N(s), i j k s 4 s 3 4 cos B(s) = − 5 sin 5 s 5 5s 5 − cos − sin 0 5 5 s 3 s 4 s 4 4 4 3 − sin − sin cos cos 5 5 +k 5 5 5 5s 5 − j 5s =i 5 s s − sin 0 − cos 0 − cos − sin 5 5 5 5 s s s 4 2 s 4 3 3 = sin i − cos j+ sin + cos2 k 5 5 5 5 5 5 5 5 s 3 s 4 3 = sin , − cos , . 5 5 5 5 5 Then 0 B (s) = s 3 s 3 cos , sin ,0 . 25 5 25 5 Finally, τ (s) = − N(s) · B0 (s) s 3 s 3 s s , − sin ,0 · cos , sin ,0 = − − cos 5 5 25 5 25 5 3 3 2 s 2 s = − − cos − sin 25 5 25 5 3 = . 25 Again, as we would want, the torsion does not depend on the parametrization. 5 of 9 8. [4 points] A curve C is the intersection of the cone z 2 = x2 + y 2 , with a plane defined through (a), (b), (c) or (d). Identify the curve, find a parametric representation. (a) z = 2 . Solution: It is a circle of radius 2: x = 2 cos t , y = 2 sin t , z = 2 (b) y = 0 . Solution: It is two intersecting lines: y = 0 , x = t , z = ±t (c) x + y = 1 . Solution: It is a hyperbola: x = 1 2 + 41 (t − 1t ) , y = 1 2 − 14 (t − 1t ) , z = (d) x + y + z = 1 . Solution: It is a hyperbola: x = 1 + t , y = 1 + 1 2t , z = −1 − t − 6 of 9 1 2t 1 (t 23/2 + 1t ) 9. [3 points] Consider the motion of a particle moving along a curve, described at time t by ds r(t). The velocity of the particle at time t is then v(t) = r0 (t) = T(t). dt (a) How do we find the acceleration a(t) of the particle at time t? Solution: First derivative is velocity v(t) = r0 (t), and the derivative of velocity is acceleration; so, a(t) = v0 (t) = r00 (t). (b) Using v(t), write an expression for calculating the distance traveled between times t1 and t2 . Solution: We know that distance traveled is the arc length between t1 and t2 which can be expressed as Z t2 Z t2 0 ||r (t)||dt = ||v(t)||dt (8) t1 t1 (c) Using v(t), write an expression to calculate the length of the displacement of the particle occurring from time t1 to time t2 , r(t2 ) − r(t1 ). Solution: Using R t the fundamental theorem of calculus for vector-valued functions, we know that || t12 v(t)dt|| = ||r(t2 ) − r(t1 )|| Challenge 10. [9 points] The parameterization x = α + a cosh t , y = β + b sinh t , −∞ < t < ∞ (9) of the hyperbola (x − α)2 (y − β)2 − = 1, a2 b2 represents only one branch of the hyperbola. a > 0, b > 0, Find a parameterization t = γ(p) which represents both branches of the hyperbola. Solution: 7 of 9 (10) Recall that using the definitions for sinh t and cosh t, we can rewrite the given parameterization as et − e−t et + e−t , y(r) = β + b , −∞ < t < ∞. (11) x(r) = α + a 2 2 Since 0 < et < ∞, we can reparameterize using r = et with 0 < r < ∞ yielding (with some algebraic manipulation) a 1 b 1 x=α+ r+ , y=β+ r− , 0 < r < ∞. (12) 2 r 2 r 1 a r+ , and it is not hard Lastly, observe that since −r < 0, we have that x(−r) = α − 2 r to show that for all negative values −r, y(−r) delivers the correct corresponding y-value, so that the full parameterization of both branches is 1 a x = α + (r + ) , 2 r b 1 y = β + (r − ) , 2 r −∞ < r < ∞ , r 6= 0 . (13) 11. [9 points] A curve C in the xy-plane is represented by the equation Ax2 + Bxy + Cy 2 + Dx + Ey + F = 0 . (14) In the uv-plane obtained by rotating the xy-plane through an angle φ u = x cos φ + y sin φ , v = −x sin φ + y cos φ , (15) the curve C is represented by a similar equation b 2 + Buv b + Cv b 2 + Du b + Ev b + Fb = 0 . Au (16) b B, b C, b D, b E, b Fb in terms of A, B, C, D, E, F and φ. (a) Express A, Solution: Expressing x, y in terms of u, v, one gets x = u cos φ − v sin φ , y = u sin φ + v cos φ . (17) Substituting these expressions in (14), one finds b = A cos2 (φ) + B sin(φ) cos(φ) + C sin2 (φ) , A b = −A sin(2φ) + B cos(2φ) + C sin(2φ) , B b = A sin2 (φ) − B sin(φ) cos(φ) + C cos2 (φ) , C b = D cos(φ) + E sin(φ) , D b = E cos(φ) − D sin(φ) , E (18) Fb = F . (b) Prove that if the angle φ satisfies cot 2φ = A−C , B (19) then the curve C is represented by the equation b 2 + Cv b 2 + Du b + Ev b + Fb = 0 , Au 8 of 9 (20) b = 0. i.e. B Solution: Solving the equation b = −A sin(2φ) + B cos(2φ) + C sin(2φ) = 0 B (21) for φ, one obviously finds (19). ! b2 E b2 D + . Give a parameterization b b A C 1 (c) Suppose now that (19) holds and that Fb < 4 of this curve. Solution: The key observation here is that if we complete the square correctly, we might have the b equation describing a circle. First, let us rewrite the equation (with B=0) as ! ! b b E D b v 2 + v + Fb = 0. b u2 + u + C (22) A b b A C We simply must complete the square in both parentheses, i.e., b b u2 + D u+ 1 A b 4 A b D b A !2 b b v 2 + E v+ 1 +C b 4 C b E b C !2 + Fb− 1 4 b2 D b A ! 1 − 4 b2 E b C ! = 0. (23) We can factor both polynomials now as perfect squares and move the compensation terms to the right, yielding b b u+ 1D A b 2A !2 b b v+ 1E +C b 2C !2 1 = 4 b2 b2 E D + b b A C ! − Fb. (24) This equation describes a circle in the new coordinate system, as long ! as the right2 2 b b 1 D E hand side is positive. This leads to the constraint that Fb < + . Thus, the b b 4 A C parameterization is x(t) = − y(t) = − b 1D + b 2A 1 4 b 1E + b 2C 1 4 9 of 9 b A ! cos t b2 E b2 D + − Fb b b A C b C ! sin t b2 E b2 D + − Fb b b A C