1. A convergent, B divergent. 2. both series divergent A divergent, B convergent

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1 Version Homework grandi () This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. CalCcExam. points Let h be a continuous, positive, decreasing function on [, ). Compare the values of the series and the integral A = 5 n= h(n) Clearly from this figure we see that a = h() < a 5 = h(5) < while a 6 = h(6) < a 7 = h(7) < and so on. Consequently, h(z) dz, h(z)dz, h(z) dz, h(z)dz, B = 5 h(z)dz. A < B.. A = B. A > B. A < B correct In the figure keywords: Szyszko CalCcs. points Determine the convergence or divergence of the series (A) , and (B) k= ke k.. A convergent, B divergent a a 5 a 6 a both series divergent. A divergent, B convergent. both series convergent correct the bold line is the graph of h on [, ) and the areas of the rectangles the terms in the series a n, a n = h(n). n= (A) The given series has the form = n= n.

2 Version Homework grandi () This is a p-series with p = >, so the series converges. (B) The given series has the form with f defined by k= f(k) f(x) = xe x. Notefirstthatf iscontinuous andpositiveon [, ); in addition, since f (x) = e x ( x ) < for x >, f is decreasing on [, ). Thus we can use the Integral Test. Now, by substitution, and so t xe x dx = [ e x ] t xe x dx = e. Since the integral converges, the series converges. CalCcexam. points Which, if any, of the following series converge? (A) n= n(ln(n)), (A) The given series is of the form with f(x) = n= f(x) x(ln(x)), x. Since f is positive and decreasing on [, ), Integral Test ensures that the given series is convergent if the improper integral I = f(x)dx converges. But, after the substitution u = ln(x) we see that I = lim [ ] t t u ln() Consequently, the series (B) The function converges. f(x) = ln(x) x = ln(). is continous and positive on [, ); in addition, since ( ) ln(x) f (x) = < x (B) m= ln(m) m on [, ), f is also decreasing on this interval. This suggests applying the Integral Test. Now,after substitutionu = ln(x),weseethat. A only correct. A and B. neither A nor B. B only t f(x)dx = Thus the improper integral [(ln(x)) ] t f(x)dx.

3 Version Homework grandi () does not converge, and so the Integral Test ensures that the given series keywords: diverges. CalCc6a. points If the improper integral x p dx converges, which of the following statements is (are) always true? (A) n p converges; (B) (C) (D) (E) n n n n n. A, D and E diverges; np+ converges; np diverges; np converges. np+. A and E only correct. A only. A, C and E only 5. B and D only To apply the Integral test we need to start with a function f which is positive, continuous and decreasing on [, ). Then the integral test says that the improper integral f(x)dx converges if and only if the infinite series n= converges. In the given example f(n) f(x) = x p, which is a function both continuous and positive on [, ). It will also be decreasing on [, ) if f (x) < for all x >. But f (x) = p x p+, so f will be decreasing provided p >. On the other hand, the improper integral converges if and only if exists. But n x p dx = lim n n x p dx x p dx. [ ] n px p = ( ) p n p. Consequently, the improper integral x p dx converges if and only if p >. Hence by the Integral test, the infinite series n= x p converges if and only if p >. Now we can check which of the statements is (are) always true. (A) This is always true because of the Integral test. (B), (E) Since p > = p+ >, the Integral test ensures that n n converges. p+ Thus (B) is false and (E) is true.

4 Version Homework grandi () (C), (D) The series n n converges if p and only if p >, i.e., when p >. Since the convergence of the improper integral x p dx guarantees only that p >, we see that statements(c) and(d) aretruefor somevaluesof p and false for others. Consequently, of the statements, are always true. only A and E CalC6b7a 5. points Find the volume, V, of the solid obtained by rotating the bounded region in the first quadrant enclosed by the graphs of about the x-axis. y = x 5, x = y. V = 7 π cu. units. V = π cu. units 5. V = 7 cu. units. V = π cu. units correct 5. V = cu. units Thus the volume of the solid of revolution generated by rotating this region about the x-axis is given by { } V = π (x / ) (x 5 ) dx { } = π x x 5 dx Consequently, ( V = π 5 6) [ = π 5 x5 6 x6 ] CalC6bexam 6. points The shaded region in y = π cu. units.. 6. V = 5 cu. units Since the graphs of x y = x 5, x = y intersect at (, ) and at (, ) the bounded region in the first quadrant enclosed by their graphs is the shaded area shown in is bounded by the graphs of y = x+, y =, x =,

5 Version Homework grandi () 5 (not drawn to scale). Find the volume of the solid obtained by rotating this shaded region about the dotted line y =.. volume = π cu. units correct 6. volume = 5 π cu. units. volume = π cu. units. volume = π cu. units 6 5. volume = π cu. units When a region { } (x, y) : g(x) y f(x), a x b is rotated about a line y = c to form a hollow solid of revolution, then the volume of this solid is given by the integral b π ((f(x) c) (g(x) c) )dx. a In the figure above, f(x) = x+, g(x) =, c =, where x. Thus the solid has volume π Consequently, ( ( ) x+) dx ( = π x+ ) x dx. [ volume = π x + x/] = 6 π. keywords: definite integral, volume of revolution, hollow solid, square root function, CalCb9a 7. points Which one of the following integrals gives the length of the parametric curve x(t) = t, y(t) = t, t.. I = t +dt. I =. I =. I = 5. I = t + dt t + dt t + dt t +dt 6. I = t +dt correct The arc length of the parametric curve (x(t), y(t)), is given by the integral I = But when we see that b a a t b (x (t)) +(y (t)) dt. x(t) = t, y = t, x (t) = t, y (t) =. Consequently, the curve has arc length = CalCa5a 8. points t +dt.

6 Version Homework grandi () 6 What are the coordinates of the projection of the point Q( 5,, ) onto the xy-plane?. ( 5,, ). ( 5,, ) correct. (,, ). ( 5,, ) 5. (5,, ) 6. ( 5,, ) The xy-plane consists of all points P(x, y, z) whose z-coordinate is given by z =, and the projection of P(x, y, z) onto the xy-plane has coordinates (x, y, ). Consequently, the projection of Q( 5,, ) onto the xy-plane has coordinates ( 5,, ). keywords: coordinate plane, projection, point, coordinate, -space, CalCa5b 9. points Determine the distance of the point Q(, 5, ) from the xy-coordinate plane.. distance =. distance = correct. distance = 6. distance = 7 5. distance = 6. distance = 5 7. distance = Since the distance of a point P(x, y, z) from the xy-, yz-, and zx-coordinate planes is given respectively by z, x and y, the point Q(, 5, ) has from the xy-plane. distance = keywords: coordinate plane, projection, point, distance, -space CalCaa. points Find an equation for the sphere having center at (,, ) and radius.. x +y z +6 =. x +y +z +6x+y 6z =. x +y +z 6x y +6z =. x +y z = 6 5. x +y +z +6x+y 6z + = 6. x + y + z 6x y + 6z + = correct The sphere consists of all points P(x,y,z) such that dist ((x, y, z),(,, )) =. Thus, by the distance formula in -space, (x ) +(y ) +(z +) = 6, which after expansion becomes x +y +z 6x y +6z + =. CalCca. points

7 Determine the dot product of the vectors Version Homework grandi () 7 a =,,, b =,,.. a b = 6. a b =. a b =. a b = correct 5. a b = 8 The dot product, a b, of vectors a = a,a,a, b = b,b,b is defined by a b = a b +a b +a b. Consequently, when a =,,, b =,,, we see that 6. angle = 5π 6 Since the dot product of vectors a and b can be written as a.b = a b cosθ, θ π, where θ is the angle between the vectors, we see that cosθ = But for the given vectors, while a.b a b, θ π. a b = ()(5)+( )( ) =, Consequently, a =, b = 5. cosθ = = where θ π. Thus a b =. angle = π. CalCc5a. points Find the angle between the vectors a =,, b = 5,. CalCcb. points Find the scalar projectionof b onto a when b = i+ j k, a = i j k.. angle = π 6. angle = π. scalar projection =. scalar projection = correct. angle = π. angle = π 5. angle = π correct. scalar projection = 8. scalar projection = 5. scalar projection = 7

8 Version Homework grandi () 8 Thescalar projectionofbonto aisgivenin terms of the dot product by Now when comp a b = a b a b = i+ j k, a = i j k,. Now PQ =, 5,, But then PQ PR = PR =,,. i j k 5 = 5 k. we see that a b = 9, a = Consequently, () +( ) +( ). Consequently, P QR has area =. keywords: comp a b =. CalCd9a. points Find the area of the triangle having vertices P(, ), Q(, ), R(, ).. area = 8. area = 9. area = 9. area = 7 5. area = correct To use vectors we shall identify a line segment with the corresponding directed line segment. Since the area of the parallelogram having adjacent edges PQ and PR is given by PQ PR, PQR has area = PQ PR. keywords: vectors, cross product area, triangle, parallelogram CrossProductTFa 5. points Ifu, vandwarenon-zerovectorssuchthat (u v) w =, then u and v must be parallel. True or False?. TRUE. FALSE correct If w, then (u v) w = if and only if u v = (in which case u and v are parallel) or w and u v are parallel. But if w is orthogonal to both u and v, then w is parallel to u v. For example, when u = i, v = j, w = k, then u is perpendicular, not parallel, to v, while (u v) w = (i j) k = k k =.

9 Version Homework grandi () 9 Consequently, the statement is FALSE. CalCea 6. points Find parametric equations for the line passingthroughthepointp(,,)andperpendicular to the plane x+y z = 6.. x = + t, y = + t, z = t correct. x = +t, y = +t, z = t. x = +t, y = t, z = +t. x = t, y = +t, z = +t 5. x = t, y = t, z = t 6. x = +t, y = +t, z = t A line passing through a point P(a, b, c) and having direction vector v is given parametrically by r(t) = a+tv, a = a, b, c. Now for the given line, its direction vector will be parallel to the normal to the plane Thus and so a =,,,, Consequently, x+y z = 6. v =,,, r(t) = +t, +t, +t. x = +t, y = +t, z = +t are parametric equations for the line. CalCe5a 7. points Find the point of intersection, P, of the lines x 5 x 5 = y 6 = y 5. P(9, 9, 9) correct. P(9, 9, 6). P(9, 9, 7). P(9, 9, 9) 5. P( 9, 9, 9) = z 7 = z 5 To determine where the lines intesect it is convenient first to convert the lines from equations in symmetric form to ones in parametric form: x = 5+t, y = 6+t, z = 7+t, x = +5s, y = 5+s, z = 5+s. For then the lines intersect when the equations 5+t = +5s, 6+t = 5+s, 7+t = 5+s, are satisfied simultaneously. Solving the first two equations gives t =, s = and a check shows that these values then satisfy the third equation. Consequently, the lines intersect when s = t =, i.e., at the point P(9,9,9). CalCea 8. points,.

10 Version Homework grandi () Find the point at which the line x = +t, y =, z = t, intersects the plane x+y z =.. P(,, 9). P(,, 9). P(,, 9) correct. P(,, 9) 5. P(,, 9) 6. P(,, 9) The line x = +t, y = z = t intersects the plane when x+y z = (+t)+( ) (t) =, i.e. when t =. Thus the point of intersection occurs at t =, which corresponds to the point P(,, 9).