MATH 155, EXAM 3 SOLUTIONS 1. I fidig a approximate solutio to the equatio x 3 +x 4 = usig Newto s method with iitial approximatio x 1 = 1, what is x? Solutio. Recall that x +1 = x f(x ) f (x ). Hece, x = 1 (1)3 + (1) 4 3(1) + (1) = 1 1 5 = 6 5. A box with a square base ad ope top must have a volume of 3 cm 3. Fid the dimesios of the box that miimizes the surface area of the box. Solutio. Let the edge legth of the base of the box be s, ad the Figure 1 height of the box be h. The volume of the box is give by V = hs ad the surface area is give by A = s + 4sh. Sice we require 3 = V = hs, we get that h = 3, ad thus A = s + 18s 1. To fid s the miimum of A(s) - the surface area, we eed to fid the critical umbers of the fuctio A(s). Solvig the equatio = A = s 18 18 = s 3 64 = s 3 s = 4, s gives the critical poit s = 4. Aother critical poit occurs at s =, but we are oly iterested i s o the iterval (, ). Furthermore, A (1) = 16 < ad A (8) = 14 >, ad so by the First Derivative Test, s = 4 is a local miimum. Sice A(s) is cotiuous ad differetiable
EXAM 3 SOLUTIONS o (, ) ad s = 4 is the oly local extrema, it must be a global miimum. Thus the box will have miimal surface area whe s = 4. Hece, the dimesios of the box that miimize its surface area are 4 4. 3. Calculate the followig idefiite itegral Solutio. x x x x x dx = x dx = (x x 1/ ) dx = x dx x 1/ dx = 1 x 4 x + C. 4. Calculate the followig defiite itegral π/ π/ si x dx = Solutio. Recall that si x < whe π x < ad si x whe x π. Thus, { si x if π si x = x < si x if x π Therefore, we have that π/ π/ si x dx = = π/ ( si x) dx + π/ [ cos x [ + cos x π/ ( ( π cos() cos = = (1 ) + ( + 1) =. π/ 5. What is the idefiite itegral cos x dx =? (si x 1) si x dx ) ) + ( cos ( π ) ) + cos()
EXAM 3 SOLUTIONS 3 Solutio. Use substitutio. Let u=si x 1; the du = cos x dx. Thus we have that cos x 1 (si x 1) dx = u du = u du = u 1 +C = (si x 1) 1 +C. 6. The equatio of the slat asymptote of the curve y = x3 + x + 3 x + 1 is: Solutio. Usig log divisio to divide x +1 ito x 3 + x +3, we get x 3 +x +3 = x + 1 x+. We the have that x3 +x +3 x + 1 = x+. x +1 x +1 x +1 x +1 x+ Sice lim x =, we kow that the slat asymptote is y = x+1. x +1 7. A table of values for a fuctio f is give below. t 4 6 f(t) 1 4 5 Use 3 rectagles ad right edpoits to estimate the value of the itegral 6 f(t) dt. Solutio. Each of the 3 rectagles will have width 6 3 =. Sice we are usig right edpoits, we have 6 8. Let g(x) = f(t) dt (f() + f(4) + f(6)) = ( + 4 + 5) =. (si x) 3 1 + t dt. Fid g (x). Solutio. Lettig u = (si x) 3 we have, g(x) = (si x) 3 1 + t dt = u 1 + t dt. Thus, by the Chai Rule ad The Fudametal Theorem of Calculus Part I dg dx = dg du du dx = d ( u 1 + t dt ) d(si x) 3 du dx = 1 + u 3(si x) cos x = 1 + (si x) 6 3(si x) cos x. 9. Calculate the itegral 3x x 3 + 1 dx
4 EXAM 3 SOLUTIONS Solutio. Let u = x 3 + 1, ad thus du = 3x dx. The at x =, we have u = 3 + 1 = 1, a dat x =, we have u = 3 + 1 = 9. Makig these substitutios, we have 3x 9 9 x 3 + 1 dx = u du = 1 3 u 3 = ( ) 9 3 3 1 1 3 = ( 3 3 1 ) = 3 3 (6) = 5 3. 1. Fid a Riema sum correspodig to the itegral 1 Solutio. x = 1 ad x i = i. Thus, the Reima sum is f(x i ) x = ( i cos ) 1 = 1 ( i cos. ) cos xdx 11. (a) Evaluate the defiite itegral x3 dx usig the defiitio of the defiite itegral. Hit: [ ] i3 = (+1) Solutio. We have x = x i = a + i x = + i x 3 dx = lim 16 = lim 4 16 = i. Thus f(x i ) x = lim =, ad usig right-had edpoits, i 3 4 = lim 4 = lim ( + 1) = lim 4 4 4 ( ) + 1 = 4 lim = 4 lim ( ) 3 i = lim [ ] ( + 1) ( + 1) ( 1 + 1 ) = 4. 8i 3 3 (b) Verify your result usig the Fudametal Theorem of Calculus. Solutio. Usig the secod part of the FTC, we obtai: x 3 dx = 1 4 x4 = 1 4 ()4 1 4 ()4 = 4. 1. Fid the poits o the ellipse 4x + y = 4 that are farthest away from the poit (1, ). (Note that there may be more tha oe!) Solutio.
EXAM 3 SOLUTIONS 5 For ay poit (x, y) o the ellipse, the distace from (1, ) is give by the fuctio D = (x 1) + (y ). Further, if (x, y) is o the ellipse, we have y = 4 4x, ad thus D = (x 1) + (4 4x ). To maximize D, we fid the critical poits. Takig the derivative of D with respect to x, we have D = 1 1 ((x 1) 8x). (x 1) +4 4x This does ot exist whe D = (x 1) + (4 4x ) = which is a miimum for D, ot a max. D is zero whe (x 1) 8x = 6x = 6x = x = 1/3. Pluggig i D () = ( 1)( 1 5 )( ) <, ad D ( 1) = ( 1)( 1 )(4) >. 4 So by the First Derivative Test, x = 1/3 is a local max, ad sice D is cotiuous ad x = 1/3 is the oly critical poit o [1, 1) it must be a global max. We solve for y usig y = 4 4x = 4 4 = 3 ad obtai y = 4 9 9 3 or y = 3 4. Hece, the farthest poits are ( 1, 4 3 3 ) ad ( 1, 4 3 3 ). 13. Fid the idefiite itegral (x 3 (x + 1) + cos x si x)dx. Solutio. We split up the itegral to have x 3 (x + 1) dx + cos x si x dx. Thus for the first half we get (1) x 3 (x + 1) dx = (x 7 + x 5 + x 3 ) dx = x8 8 + x6 6 + x4 4 + C 1. For the secod half, we ca use u substitutio. Lettig u = si x ad thus du = cos x dx, we get () Thus, we get cos x si x dx = udu = u + C = si x + C. x 8 8 + x6 6 + x4 4 + si x + C as a fial aswer (where C = C 1 + C ).
6 EXAM 3 SOLUTIONS A alterate way of doig the first half is usig substitutio with u = x + 1 ad thus du = xdx. The x = u 1, ad x 3 (x + 1) dx = x (x + 1) xdx = (u 1)u 1 du = 1 (u 3 u )du = 1 ( ) u 4 4 u3 + C 1 3 = 1 ( ) (x + 1) 4 (x + 1) 3 (3) + C 1. 4 3 Notice that Equatio (1) ad (3) just differ by a costat that is absorbed ito the C 1. A alterate way of doig the secod half is usig substitutio with u = cos x ad thus du = si x dx to get (4) cos x si x dx = cos x + C. However, these aswers () ad (4) are the same, sice si x + cos x = 1 is ab- ad so si x = cos x + 1. Sice we already have +C at the ed of the solutio, the 1 sorbed ito the C. Thus, both of our aswers are correct.