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- Michigan
- University of Michigan - Ann Arbor
- Mathematics
- Mathematics 255
- Waddle
- math255_hw5_sol.pdf

Arun G.

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MATH 255 Applied Honors Calculus III Winter 2008 Homework 5 Due: Friday, February 15th, 2008 Section 14.4, pg. 914: 6, 16, 22, 34, 40. Section 15.1, pg. 933: 6, 16, 32, 53{58, 60. Section 15.2, pg. 944: 6, 8, 28, 36, 42. Additional Problem: 1. Dave is riding the \corkscrew roller-coaster in an amusement park. he has a photo camera and wants to take a picture of his girlfriend who is waiting for him on the ground at the point P(12;8;11). Because of the restraints, Dave can only hold the camera looking forward, and can take pictures only in the direction of the motion of the roller-coaster. The motion of Dave as a function of time t is given by xD(t) = t2; yD(t) = 2t; zD(t) = 3t 1: Find the moment at which Dave has to take the picture of his girlfriend. Solutions 14.4: #6 Given r(t) =< sint;2 cost;0 >, the velocity is v(t) = r0(t) =< cost; 2 sint;0 >, the speed is v(t) = jv(t)j = pcos2t+ 4 sin2t and the acceleration is a(t) = r00(t) =< sint; 2 cost;0 >. Plugging in t = =6, we have v( =6) =, v( =6) =p7=2 and a( =6) =< 1=2; p3;0 > 14.4: #16 To nd the velocity, we write v(t) = v(0) + Z t 0 a(u)du =< 1;1; 1 > + Z t 0 < 0;0; 10 >du =< 1;1; 1 10t> Similarly, position vector of the particle is r(t) = r(0) + Z t 0 v(u)du =< 2;3;0 > + Z t 0 < 1;1; 1 10u>du =< 2 +t;3 +t; t 5t2 > See gure . 14.4: #22 If a particle moves with constant speed v(t) = C, then the velocity satis es v(t) v(t) = v2(t) = C2. Now if we di erentiate the dot product, we have v(t) v0(t) + v0(t) v(t) = (C2)0 = 0, and therefore 2v(t) v0(t) = 0. Finally recall that the dot product of two vectors is 0 means that the vectors are orthogonal. We have v(t) v0(t) = 0 and therefore v(t) is orthogonal to v0(t). 1 a y x 1 v Figure 1 14.4: #34 We use formulas aT = r 0(t) r00(t) jr0(t)j and aN = jr 0(t) r00(t)j jr0(t)j Since r0(t) =< 1;2t;3 >;r00(t) =< 0;2;0 >, we compute jr0(t)j = p10 + 4t2;r0(t) r00(t) = 4t and r0(t) r00(t) =< 6;0;2 >. Also jr0(t) r00(t)j=p36 + 4 = 2p10. Finally aT = 4tp10 + 4t2 and aN = 2 p10 p10 + 4t2 14.4: #40 (a) We have dv dt = 1 m dm dt ve To solve for v we integrate both sides of the equation v(t) v(0) = Z t 0 1 m dm duvedu The right hand side is Z t 0 1 m dm duvedu = ve Z t 0 m0(u) m(u)du = ve ln m(t) m(0) by substitution. Therefore, v(t) = v(0) + ve ln m(t)m(0) (b) We have jv(t)j= 2jvej and jv(0)j= 0. By part (a), v(t) = ve ln m(t)m(0) and therefore 2 = jv(t)jjv ej =jln m(t)m(0)j Mass is decreasing, so jln m(t)m(0)j= ln m(0)m(t) and nally m(t) = e 2m(0) Thus m(0) e 2m(0)m(0) is the fraction of the initial mass that is burned as fuel. 15.1: #6 Let f(x;y) = ln(x+y 1). a)f(1;1) = ln(1 + 1 1) = ln(1) = 0 b)f(e;1) = ln(e+ 1 1) = ln(e) = 1 c)Domain of f is a set of (x;y) so that x+y 1 > 0. It is the region in the xy-plane which is above the line y = 1 x. See gure ?? y=1−x y x Figure 2. Domain of f(x;y) d) The range of f is the range of ln function, which is all real numbers. 15.1: #16 For f(x;y) =py xln(y +x) the domain is D =f(x;y) : y x 0 andy +x> 0g It is the quarter of the xy plane which is above the lines y = x and y = x, including the line y = x but not the line y = x. See gure . 2 y x 1 Figure 3. Domain of f(x;y) =py xln(y +x) 15.1: #32 If we start at the origin and move along the x-axis, the z-values of a cone centered at at the origin increase at a constant rate, so we would expect its level curves to be equally spaced. A paraboloid with vertex the origin, on the other hand, has z-values which change slowly near the origin and more quickly as we move farther away. Thus, we would expect its level curves near the origin to be spaced more widely apart than those farther from the origin. Therefore the contour map I corresponds to the paraboloid, and contour map II to the cone. 15.1: #53-58 53. The function is constant on any circle centered at the origin. a) B; b) III 54. The function is symmetric about the plane x = y. It is equal to 0 along the x- and y-axes. Moreover, it vanishes for x,y big enough. a) C; b) II 55. Level curves are ellipses, and for large x;y the the function becomes almost 0. a) F; b) V 56. Along the lines y = 1p3x and x = 0, this function is 0 a) A; b) VI 57. This function is periodic in both x and y, with period 2 in each variable. a) D; b) IV 58. This function is periodic along x-axis, and increases as jyj increases. a) E; b) I 15.1: #60 Level surfaces of this function have equations x2 + 3y2 + 5z2 = k These are ellipsoids for k> 0 and the origin for k = 0. 15.2: #6 Easy one, since the function is de ned and well behaved there. lim (x;y)!(6;3) xycos(x 2y) = 6 3 cos(6 6) = 18 15.2: #8 f(x;y) = x 2 + sin2y 2x2 +y2 Try two lines: x= 0 and y = 0. limx!0 x 2 + 0 2x2 + 0 = 1=2 limy!0f(x;y) = limy!0 0+sin2 y0+y2 = limy!0(sinyy )2 = (limy!0 sinyy )2 = (limy!0 cosy1 )2 = 16= 1=2 The limit doesn’t exist. 15.2: #28 This is a rational function, so it is continuous on the whole domain. The denominator is never zero, so the domain is all x, y. So, the function is continuous for all x and y. 15.2: #36 For this function to be continuous at (a,b), its limit must equal its value at (a,b). This is a rational function, so it is automatically continuous when the denominator polynomial isn’t equal to zero, which occurs only when (x,y) = 0. Let’s check if it is continuous at (0,0). f(x;y) = 0 if (x,y) = (0,0) xy x2+xy+y2 otherwise Following example 2, let’s take y = mx, for any slope m, and take the limit of x going to 0. limx!0f(x;mx) = limx!0 x(mx)x2 +x(mx) + (mx)2 = limx!0 mx 2 (1 +m+m2)x2 = limx!0 m(1 +m+m2) = m(1 +m+m2) As we can see, the limit depends on the slope of the line of approach we take. Therefore, it depends on our avenue of approach (whether or not it’s a line), and the limit doesn’t exist, much less equal 0. So f(x,y) is not continuous at (0,0), and is therefore continuous for all (x;y)6= (0;0). 15.2: #42 If f(x) is continuous, then limx!af(x) = f(a) for all a in Vn. So, we need the limit to exist, and to equal to f(a). We use the de nition of existence for the limit: jf(x) Lj< jx aj< We must nd a such that, given the second inequality, the rst inequality is also true. Let’s work on the rst inequality to try to get terms that look like the left hand side of the second. jf(x) Lj< jf(x) f(a)j< jc x c aj< jc (x a)j< jjcjjx ajcos j< Note that jcjjx ajcos jcjjx aj. If we can show that jcjjx aj< , then it will also be true for jcjjx ajcos . So, jcjjx aj< Give that jx aj< , what can we pick to make the inequality above true? Multiply the inequality by jcj on both sides to get: jcjjx aj= k< 2t;2;3 > Solving the system, we have 12 t2 = 2tk;8 2t = 2;12 3t = 3 Therefore k = 4 t, so we can plug it in the rst equation and get 12 t2 = 2t(4 t). There are two solutions: t = 2 and t = 6. Since k = 4 t and k has to be positive, t = 6 doesn’t work. Finally we conclude that the moment corresponds to t = 2.

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