18.01 Practice Final, Fall 2003: Solutions

18.01 PRACTICE FINAL, FALL 2003 Problem 1 Find the following definite integral using integration by parts. � π 2 x sin(x)dx. 0 � � Solution Let u = x, dv = sin(x)dx. The du = dx, v = − cos(x). So udv = uv − vdu, i.e., π � π2 � π2 2 x sin(x)dx = (−x cos(x) | + 0 0 0 cos(x)dx = π (− π2 · 0 + 0 · 1) + (sin(x)|02 = 0 + (1 − 0) = 1. Problem 2 Find the following antiderivative using integration by parts. � x sin−1 (x)dx. Solution First substitute x = sin(θ), dx = cos(θ)dθ. Then the integral becomes, � θ sin(θ) cos(θ)dθ. Of course sin(θ) cos(θ) = 1 2 sin(2θ). Thus we need to compute, � 1 θ sin(2θ)dθ. 2 � � Set u = 21 θ, dv = sin(2θ)dθ. Then du = 12 dθ and v = − 21 cos(2θ). So udv = uv − vdu, i.e., � 1 � 1 1 cos(2θ)dθ = θ sin(2θ)dθ = − θ cos(2θ) + 2 4 4 − 14 θ cos(2θ) + 81 sin(2θ) + C. Using trigonometric formulas, this equals, � 1 1 2 − θ(1 − 2 sin (θ)) + sin(θ) 1 − sin2 (θ) + C. 4 4 Back­substituting, sin(θ) = x, gives the final answer, 1 1 � − (1 − 2x2 ) sin−1 (x) + x 1 − x2 + C. 4 4 Problem 3 Use L’Hospital’s rule to compute the following limits. x x (a) limx→0 a −b x , 0 < a < b. 3 (b) limx→1 4x −5x+1 . ln x Solution (a). As x approaches 0, both the numerator and denominator approach 0. The corre­ sponding derivatives are, d x d (a − bx ) = ln(a)ax − ln(b)bx , (x) = 1. dx dx Therefore, by L’Hospital’s rule, ln(a)ax − ln(b)bx ax − bx = lim = ln(a) − ln(b). lim x→0 x→0 x 1 Date: Fall 2003. 1 (b). As x approaches 1, the numerator approaches 4 − 5 + 1 = 0, and the denominator approaches ln(1) = 0. The corresponding derivatives are, d d 1 (4x3 − 5x + 1) = 12x2 − 5, ln(x) = . dx dx x Therefore, by L’Hospital’s rule, 12 − 5 12x2 − 5 4x3 − 5x + 1 = = 7. = lim 1 x→1 x→1 ln(x) 1 x lim Problem 4 Determine whether the following improper integral converges or diverges. � ∞ 2 e−x dx. 1 (Hint: Compare with another function.) Solution Because the integrand is nonnegative, the integral converges if and only if it is bounded. 2 Therefore the comparison test applies. For x > 1, x2 > x. Therefore −x2 < −x and 0 ≤ e−x < e−x . Integrating, � ∞ �∞ � e−x dx = −e−x �1 = (0 + e−1 ) = e−1 < ∞. 1 Therefore, also �∞ 1 2 e−x dx converges (and is bounded above by e−1 ). Problem 5 You wish to design a trash can that consists of a base that is a disk of radius r, cylindrical walls of height h and radius r, and the top consists of a hemispherical dome of radius r (there is no disk between the top of the walls and the bottom of the dome; the dome rests on the top of the walls). The surface area of the can is a fixed constant A. What ratio of h to r will give the maximum volume for the can? You may use the fact that the surface area of a hemisphere of radius r is 2πr2 , and the volume of a hemisphere is 32 πr3 . Solution The area of the base is πr2 . The area of the sides are 2πrh. The area of the dome is 2πr2 . Therefore we have the equation, A = πr2 + 2πrh + 2πr2 = πr(3r + 2h). A − 3r It follows that h = 2πr 2 . The volume of the cylindrical portion of the can is the area of the base 2 times the height, i.e., πr h. The area of the dome of the can is 32 πr3 . Therefore the total volume of the can is, � A � 2 3 − 3r V (r) = πr2 2πr 2 + 3 πr = Ar 5πr 3 2 − 6 . � A The endpoints for r are r = 0 and r = 3π . The critical points for r occur when, dV A 5πr2 = − = 0, dr 2 2 i.e., A = 5πr2 . Since A = 3πr2 + 2πrh, we conclude that 2πrh = A − 3πr2 = 2πr2 . Cancelling, we have that h = r. This is contained in the interval for r, moreover geometric reasoning (or the first derivative test) shows this is a maximum for V . Therefore the maximum volume is obtained when h = r. Problem 6 A point on the unit circle in the xy­plane moves counterclockwise at a fixed rate of radian 1 second . At the moment when the angle of the point is θ = π4 , what is the rate of change of the distance from the particle to the y­axis? 2 Solution The coordinates of the point are (cos(θ), sin(θ)). The distance from the y­axis is the absolute value of the x­coordinate. Since the point is in the 1st quadrant, this is just x = cos(θ). Therefore the rate of change of the distance is, dx dx dθ radian 1 radian = = − sin(θ) · 1 = −√ . · dt dθ dt second 2 second Problem 7 Compute the following integral using a trigonometric substitution. Don’t forget to back­substitute. � x2 √ dx. 1 − x2 Hint: Recall the half­angle formulas, cos2 (θ) = 12 (1 + cos(2θ)), sin2 (θ) = 12 (1 − cos(2θ)). Solution This integral calls for a trigonometric substitution, x = sin(θ), dx = cos(θ)dθ. The integral becomes, � � sin2 (θ) cos(θ)dθ = sin2 (θ)dθ. cos(θ) By the half­angle formulas, this is, � 1 1 1 (1 − cos(2θ))dθ = (θ − sin(2θ)) + C. 2 2 2 Using the double­angle formula, sin(2θ) = 2 sin(θ) cos(θ), and back­substituting sin(θ) = x yields, 1 1 � sin−1 (x) − x 1 − x2 + C. 2 2 Problem 8 Compute the volume of the solid of revolution obtained by rotating about the x­axis the region in the 1st quadrant of the xy­plane bounded by the axes and the curve x4 + r2 y 2 = r4 . Solution The curve intersects the y­axis when r2 y 2 = r4 , i.e. y = r. The curve intersects the x­axis when x4 = r4 , i.e. x = r. So the endpoints of the curve are (0, r) and (r, 0). Using the disk method, the volume of the solid is, � x=r πy 2 dx. x=0 Since y 2 = r2 − x4 , r2 the volume is, � V = x=r x=0 This evaluates to πr3 − 1 3 5 πr = πx4 πr − 2 dx = r 2 � �r πx5 �� πr x − 2 � . 5r 0 2 4 3 5 πr . Problem 9 Compute the area of the surface of revolution obtained by rotating about the y­axis the portion of the lemniscate r2 = 2a2 cos(2θ) in the 1st quadrant, i.e., 0 ≤ θ ≤ π4 . Solution The polar equation for arclength is ds2 = dr2 + r2 dθ2 , which is equivalent to r2 ds2 = r2 dr2 + r4 dθ2 . By implicit differentiation, 2rdr = −4a2 sin(2θ)dθ, r2 dr2 = 4a4 sin2 (2θ)dθ2 . Therefore, r2 ds2 = r2 dr2 + r4 dθ2 = 4a4 sin2 (2θ)dθ2 + 4a4 cos2 (2θ)dθ2 = 4a4 dθ2 . 2 So ds = 2ar dθ. The area of the surface of revolution is given by, � θ= π � � π �π � 4 4 2a2 dθ = 2πxds = 2πr cos(θ) 4πa2 cos(θ)dθ = 4πa2 sin(θ)�04 . r θ=0 0 3 Therefore the surface area is 4π 2 √ a . 2 Problem 10 Compute the area of the lune that is the region in the 1st and 3rd quadrants contained inside the circle with polar equation r = 2a cos(θ) and outside the circle with polar equation r = a. Solution Setting 2a cos(θ) equal to a, the points of intersection occur when 2 cos(θ) = 1, i.e. θ = − π3 and θ = + π3 . So the lune is the region between the two graphs for − π3 ≤ θ ≤ π3 . The outer curve is ro = 2a cos(θ) and the inner curve is ri = a. The formula for the area between two polar curves is � 1 2 Area = (r − ri2 )dθ. 2 o In this case, the area is, � π 3 − π3 1 (4a2 cos2 (θ) − a2 )dθ = a2 2 � π 3 (4 cos2 (θ) − 1)dθ. 0 To evaluate this, use the half­angle formula, cos2 (θ) = 21 (1 + cos(2θ)). The integral becomes, � π π 3 2 a (2 + 2 cos(2θ) − 1)dθ = a2 (θ + sin(2θ)|03 . Therefore the area is a2 ( π3 + 0 √ 3 2 ), i.e. √ 2π+3 3 2 a . 6 Problem 11 Find the equation of every tangent line to the hyperbola C with equation y 2 − x2 = 1, that contains the point (0, 21 ). Solution By implicit differentiation, 2y dy − 2x = 0, dx dy x = . dx y Therefore, the slope of the tangent line to C at (x0 , y0 ) is xy00 . So the equation of the tangent line to C at (x0 , y0 ) is, x0 (y − y0 ) = (x − x0 ). y0 If the tangent line contains the point (0, 21 ), then (x0 , y0 ) satisfies the equation, x0 1 ( − y0 ) = (0 − x0 ), 2 y0 y0 − y02 = −x20 . 2 Of course also y02 − x20 = 1, therefore y20 = y02 − x02 = 1. So y0 = 2. The two solutions of x0 are √ √ x0 = 3 and x0 = − 3. The equations of the corresponding tangent lines are, � √ √ 3 (y − 2) = (x − 3), √2 √ (y − 2) = − 23 (x + 3). Problem 12 Compute each of the following integrals. � (a) � sec3 (θ) tan(θ)dθ. x−1 (b) x(x+1) 2 dx. � 2x−1 (c) 2x2 −2x+3 dx. �√ e3x dx. (d) 4 Solution (a). Substituting u = sec(θ), du = sec(θ) tan(θ)dθ, the integral becomes, � 1 1 u2 du = u3 + C = sec3 (θ) + C. 3 3 (b). This is a proper rational function. Use a partial fractions expansion, x−1 A B C = + + . 2 x(x + 1) x x + 1 (x + 1)2 By the Heaviside cover­up method, A = −1 and C = −2 −1 = 2. This only leaves B to compute. Plug in x = 1 to get, −1 B 2 B 1 0= + + 2 = − , B = 1. 1 2 2 2 2 So the partial fraction decomposition is, x−1 −1 1 2 = + + . x(x + 1)2 x x + 1 (x + 1)2 Thus the antiderivative is, � −1 1 2 2 + + + C �. dx = − ln(x) + ln(x + 1) − 2 x x + 1 (x + 1) (x + 1) (c). The derivative of the denominator is 4x − 2. This is twice the numerator. Substituting u = 2x2 − 2x + 3, du = (4x − 2)dx, the integral becomes, � � � 1 1 1 1 du = ln(u) + C = ln(2x2 − 2x + 3) + C. u 2 2 2 √ 3 (d). Of course e3x = e 2 x . Therefore the antiderivative is, � 3 2 3 e 2 x dx = e 2 x + C. 3 5