Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
MATHEMATICS 3202 ASSIGNMENT 3 - SOLUTION Due on Monday, February 1, 2016; Submitted to Department Exercise Box 33 Name M.U.N. Number 1. Describe the projections of the circle r(t) = hsin t, 0, 4 + cos ti onto the coordinate planes. Solution [p.728-Ex.12; SM-p.472] The projection onto the xy-plane is traced by hsin t, 0, 4 + cos ti, which is the segment [−1, 1] on the x-axis. The given circle is contained in the xz-plane, hence the projection on the sz-plane is the circle itself. We identify this circle. Since x = sin t and z = 4 + cos t, we have x2 + (z − 4)2 = sin2 t + cos2 t = 1. This is the circle of radius 1 centered at h0, 0, 4i. The projection onto the yz-plane is traced by h0, 0, 4 + cos ti, which is the segment [3, 5] on the z-axis. 2. i) Parametrize the intersection of the surfaces y 2 − z 2 = x − 2 and y 2 + z 2 = 9 using t = y as the parameter; ii) find a parametrization of the curve in i) using trigonometric functions. Solution [p.729-Ex.21,22; SM-pp.474-475] p i) Solve for z and x in terms of y. From y 2 + z 2 = 9 we get z = ± 9 − y 2 ; and from y 2 − z 2 = x − 2 we have √ x = 2 + y 2 − z 2 = 2y 2 − 7. Considering t = y as a parameter, we find z = ± 9 − t2 and x = 2t2 − 7, thereby getting the vector parametrization p r(t) = h2t2 − 7, t, ± 9 − t2 i for t ∈ [−3, 3]. ii) The curve in i) is the intersection of the surfaces y 2 − z 2 = x − 2 and y 2 + z 2 = 9. Note that the circle y 2 + z 2 = 9 is parametrized by y = 3 cos t, z = 3 sin t. Substituting in the first equation and using the identity cos2 t − sin2 t = cos 2t yields x = 2 + y 2 − z 2 = 2 + (3 cos t)2 − (3 sin t)2 = 2 + 9 cos 2t. Below is the trigonometric parametrization: r(t) = h2 + 9 cos 2t, 3 cos t, 3 sin ti. 3. Use hyperbolic functions to parametrize the intersection of the surfaces x2 − y 2 = 4 and z = xy. Solution [p.729-Ex.26; SM-p.477] x = 2 cosh t & y = 2 sinh t satisfy the equation of the cylinder since x2 − y 2 = 4(cosh2 t − sinh2 t) = 4. To find a parametrization of the curve of intersection, we substitute x = 2 cosh t & y = 2 sinh t in the equation of the surface z = xy and solve for z. This gives z = 4 cosh t sinh t. We obtain r(t) = h2 cosh t, 2 sin tt, 4 sinh t cosh ti. 4. In i) and ii) below evaluate the derivative by using the appropriate Product Rule, where r1 (t) = ht2 , t3 , ti and d 4 d r2 (t) = he3t , e2t , et i: i) dt (t r1 (t)); ii) dt (r1 (t) · r2 (t)) assuming that r01 (2) = h2, 1, 0i and r02 (2) = h1, 4, 3i. t=2 Solution [p.739-Ex.18,20; SM-p.489] i) d 4 dt4 (t r1 (t)) = t4 r01 (t) + r1 (t) = t4 h2t, 3t2 , ti + (4t3 )ht2 , t3 , ti = h6t5 , 7t6 , 5t4 i. dt dt ii) d d d (r1 (t) · r2 (t)) = ( (r1 (t))) · r2 (t) + ( r2 (t)) · r1 (t) t=2 t=2 t=2 dt dt dt 2 3t 2t t 2 3 = h2t, 3t , 1i · he , e , e i + ht , t , ti · h3e3t , 2e2t , et i t=2 6 4 t=2 2 = h2, 1, 0i · he , e , e i + h1, 4, 3i · h4, 8, 2i = 2e6 + e4 + 4 + 32 + 6 = 2e6 + e4 + 42. 5. Compute d dt (r1 (t) × r2 (t)) t=1 in two ways: i) Calculate r1 (t) × r2 (t) and differentiate; ii) Use the Product Rule. Solution [p.739-Ex.22; SM-p.490] i) since r1 (t) × r2 (t) = (et − 4t)i + (2t − t2 et )j + (2t2 − 1)k, it follows that d = (e − 4)i + (2 − 3e)j + 4k. = (et − 4)i + (2 − 2tet − t2 et )j + (4t)k (r1 (t) × r2 (t)) t=1 t=1 dt ii) Using the Product Rule we see d d d = r1 (t) × r2 (t) + (r1 (t)) × r2 (t) (r1 (t) × r2 (t)) t=1 t=1 t=1 dt dt dt and so d = (e − 4)i + (2 − 3e)j + 4k. (r1 (t) × r2 (t)) t=1 dt 6. Let v(s) = s2 i + 2sj + 9s−2 k. Evaluate d ds v(g(s)) at s = 4. assuming that g(4) = 3 and g 0 (4) = −9. Solution [p.739-Ex.28; SM-p.492] Applying the Chain Rule we have d 2/3k, we get ds v(s) s=4 d ds v(s) = g 0 (s)v0 (g(s)), and d ds v(s)s=4 = −9v0 (3). Also since v0 (3) = 6i + 2j − = −9v0 (3) = −54i − 18j + 6k. 7. In i)-ii)-iii)-iv), evaluate the integrals: i) 01 h(1 + s2 )−1 , s(1 + s2 )−1 i ds; ii) R1 Rt −2 −4 −5 2 1/2 hu , u , u i du; iv) 0 (3si + 6s j + 9k) ds. R R1 0 2 (te−t i + t(ln(1 + t2 ))j) dt; iii) Solution [p.739-Ex.40,42,44,46; SM-pp.495-497] i) Z 1 Z 1 h(1 + s2 )−1 , s(1 + s2 )−1 i ds = h 0 (1 + s2 )−1 ds, 0 Z 1 s(1 + s2 )−1 dsi = hπ/4, 2−1 ln 2i. 0 ii) Z 1 2 (te−t i + t(ln(1 + t2 ))j) dt = h2−1 (1 − e−1 ), −1/2 + ln 2i. 0 iii) Z 1 Z 1 hu−2 , u−4 , u−5 i du = h 1/2 u−2 du, 1/2 Z 1 u−4 du, 1/2 Z 1 u−5 duih1, 7/3, 15/4i. 1/2 iv) Z t 0 Z t (3si + 6s2 j + 9k) ds = ( Z t (3s)ds)i + ( 0 0 (6s2 )dsj + ( Z t 0 9ds)k) = (3t2 /2)i + (2t3 )j + (9t)k. 8. In i)-ii), compute the length of the curve over the given interval: i) r(t) = 2ti − 3tk, 11 ≤ t ≤ 15; ii) r(t) = (2t2 + 1)i + (2t2 − 1)j + t3 k, 0 ≤ t ≤ 2. Solution [p.744-Ex.2,4; SM-p.509-510] i) x(t) = 2t, y(t) = 0, z(t) = −3t gives s = R 15 11 √ kr0 (t)kdt = 4 13. ii) Since r0 (t) = h4t, 4t, 3t2 i, we use u = 32 + 9t2 to get Z 2 s= kr0 (t)k dt = Z 2 p t 32 + 9t2 dt = 0 0 Z 68 √ du u 32 18 ≈ 14.063. 9. Find an arc length parametrization of the cycloid with parametrization r(t) = ht − sin t, 1 − cos ti. Solution [p.745-Ex.26; SM-p.517-518] Step 1. Find the Rinverse of the arc length function. We are given r0 (t) = h1 − cos t, sin ti and kr0 k = 2 sin(t/2). This gives s(t) = 0t kr0 (u)k du = 4(1 − cos t/2). Then solve s(t) for t: t = 2 arccos(1 − s/4). Step 2. Reparametrize the curve using the t we just found. We will write this as t = 2 arccos(1 − s/4) and solving, let us first note that p sin t = 8−1 (4 − s) 8s − s2 & cos t = 2((4 − s)/4)2 − 1. So then, rewriting r(t) = ht − sin t, 1 − cos ti, and p r1 (t) = h2 arccos(1 − s/4) − 8−1 (4 − s) 8s − s2 , 2 − 8−1 (4 − s)2 i. 10. Express the arc length L of y = x3 for 0 ≤ x ≤ 8 as an integral in two ways, using the parametrizations r1 = ht, t3 i and r2 = ht3 , t9 i. Do not evaluate the integrals, but use substitution to show that they yield the same result. Solution [p.745-Ex.28; SM-p.518-519] √ √ For r1 = ht, t3 i we have r01 = h1, 3t2 i and hence kr01 k = 1 + 9t4 . Similarly, we have kr02 k = 9t4 + 81t16 . The length s may be computed using the two parametrizations by the following integrals (note that in the second parametrization 0 ≤ t3 ≤ 8 hence 0 ≤ t ≤ 2) Z 8 s= 0 kr01 k = Z 8p 0 1 + 9t4 dt & s = Z 2 0 kr02 k = Z 2p 0 1 + 9t12 3t2 dt = Z 8p 1 + 9u4 du. 0 This shows that the arc length is independent of the parametrization we choose for the curve.