Name: Math 10550, Final Exam: December 15, 2007

Size: px

Start display at page:

Download "Name: Math 10550, Final Exam: December 15, 2007"

Norah Fowler
6 years ago
Views:

1 Math 55, Fial Exam: December 5, 7 Name: Be sure that you have all pages of the test. No calculators are to be used. The exam lasts for two hours. Whe told to begi, remove this aswer sheet ad keep it uder the rest of your test. Whe told to stop, had i just this oe page. The Hoor Code is i effect for this examiatio, icludig keepig your aswer sheet uder cover. PLEASE MARK YOUR ANSWERS WITH AN X, ot a circle!. (b) (c). (b) (c) 3. (b) (c) 4. (b) (c) 5. (b) (c) 6. (b) (c) 7. (b) (c) 8. (b) (c) 9. (b) (c). (b) (c). (b) (c). (b) (c) 3. (b) (c) 4. (b) (c) 5. (b) (c) 6. (b) (c) 7. (b) (c) 8. (b) (c) 9. (b) (c). (b) (c). (b) (c). (b) (c) 3. (b) (c) 4. (b) (c) 5. (b) (c)

2 Multiple Choice Name:.(6 pts.) Fid the limit lim x x +. x (b) (c) The limit does ot exist. 3 Solutio: x + lim x x ( x +) = lim ( + x +) x x ( + x +) (x +) = lim x x( + x +) x = lim x = lim x = + x( + x +) + x + =.(6 pts.) The fuctio f(x) = x + x 6 x 4 has a removable discotiuity at x =. We ca remove this discotiuity by defiig f() to be 3 (b) (c) 3 5 4

3 Solutio: Sice x + x 6 lim x x 4 (x +3) (x = lim ) x (x +) (x ) = 5 4, Thus defiig f() = 5 4 yields so f(x) is cotiuous at. lim f(x) =f(), x 3

4 3.(6 pts.) If the dr dθ = r = si θ +cosθ, (c) cos θ +cos θ si θ ( + cos θ) (b) +cosθ +cosθ cos θ ( + cos θ) cos θ +cos θ si θ ( + cos θ) Solutio: dr dθ = ( + cos θ)cosθ si θ( si θ) ( + cos θ) = cos θ +cos θ + si θ ( + cos θ) = +cosθ ( + cos θ) = +cosθ 4.(6 pts.) If the f (8) = f(x) = + +x, 4 (b) (c) 8 9 Solutio: Sice f (x) = ( + +x) ( + x) 4

5 pluggig i x = 8 gives f (8) = ( + +8) ( + 8) = ( + 9) = 4 3 (9) = 4. 5

6 5.(6 pts.) The secod derivative of is y =(x +)(x )(x +) 4x (b) x +x (c) x 4x 3 4x x + Solutio: Sice y =(x +)(x )(x +)=(x )(x +)=x 4, differetiatig yields y =4x 3 ad so y =x. 6.(6 pts.) A body travels alog a straight lie accordig to the law s = t 4 4t 3 +t, t. At what positio, after the motio gets started, does the body first come to rest? s = 3 (b) s =36 (c) s = s = s =4 Solutio: We first seek t> such that v(t) =, where v(t) =s (t) = 4t 3 t +4t. Sice factorig gives v(t) = 4t(t +3t ) = 4t(t )(t +5), we see that t = is the first time whe the body is at rest. The positio at t = is s() = =3. 6

7 7.(6 pts.) The equatio of the taget lie to the curve at x = is y = x 3 +6x +x +6 y = x (b) y = x (c) y = x + y = x + y = x Solutio: Sice pluggig i x = gives y =3x +x +, y () = 3 ( ) + ( ) + =. Whe x =, the y-coordiate o the give curve is Therefore the equatio of the taget is y =( ) 3 +6 ( ) + ( ) + 6 =. y = (x +) = y = x. 8.(6 pts.) Use the implicit differetiatio to fid the equatio of the taget lie to the curve 5x +9y =+xy + y at the poit (, ). y = 4 3 x + (b) y = 5 6 x (c) y = 3 x + y = 5 6 x + y = 3 x Solutio: Implicitly differetiatig both sides of the give equatio yields (5x +9y) / (5 + 9y )=y +xyy + y, 7

8 which after pluggig i x = ad y = simplifies to 6 (5 + 9y )=+y. Solvig for y gives y =. Therefore the equatio of the taget lie is 3 y = 3 x +. 8

9 9.(6 pts.) A cylider is carved out of ice ad the left i the su to melt. If the radius decreases at a rate of 3 iches per hour ad the height decreases at a rate of 6 iches per hour, how fast is the surface area of the cylider decreasig whe the cylider is at height 5 feet ad radius oe foot? (Hit: iches i a foot.) Aswer: The total surface area decreases at a rate of 3 4 ft /hr (b) 9 ft /hr ft /hr Solutio: The surface are is give by 5 4 ft /hr (c) A =r +rh. Differetiatig with respect to t the gives da dt =4rdr dt +(hdr dt + r dh dt ). Pluggig i the give values h =5,r =, dr = dt 4, ad dh dt = yields da dt =4 4 +(5 4 + )=9. 5 ft /hr.(6 pts.) Use liear approximatio to estimate (b) (c) Solutio: With f(x) =x, ad hece f (x) = x 3, the liear approximatio of f at x =4. is f(4.) = f(4) + f (4)(4. 4) = 79. =

10 .(6 pts.) The maximum ad miimum values of f(x) = x x +, o the iterval [,] are M =,m= (b) M =,m= (c) M =,m= 3 5 M = 5,m= m = is a miimum; there is o maximum. Solutio: The critical poits are where f (x) = x + x(x) (x +) = x ( + x ) equals zero ad the edpoits x =,x =. Sice f (x) = if ad oly if x = ±, we take x = as our third critical poits. Sice f() =, f() =,adf() =,wesee 5 that M = ad m =..(6 pts.) Determie the umber of solutios of the equatio x 3 5x += i the iterval [, ]. The umber of solutios is (b) (c) 3 4 Solutio: Set f(x) =x 3 5x +, so that f (x) =3x 5. Sice f( ) = 3 ad f() =, the itermediate value theorem guaratees that f has at least oe root i [, ]. Because x < 5forx [, ], it follows that 3x < 5 ad hece f (x) =3x 5 < forx [, ]. Thus f is strictly decreasig o [, ] ad hece caot have more tha oe zero o [, ]. Therefore f has exactly oe root o [, ].

11 3.(6 pts.) Cosider the fuctio f(x) = x +3 x. Oe of the followig statemets is true. Which oe? The lie y = x + is a slat asymptote of f, ad the lie x = is a vertical asymptote of f. (b) f has o horizotal or slat asymptotes, ad the lie x = is a vertical asymptote. (c) The lie y = is a horizotal asymptote of f, ad the lie x = is a vertical asymptote of f. The lie y = x + is a slat asymptote of f, ad the lie f has o vertical asymptotes. The lie y = x is a slat asymptote of f ad the lie x = is a vertical asymptote of f. Solutio: Sice (as log divisio easily verifies) x +3 x = x ++ 4 x, the slat asymptote is y = x +. Thus there is o horizotal asymptote. Because the deomiator is udefied at x = ad x is ot a factor of the umerator, x = is a vertical asymptote. 4.(6 pts.) Cosider the fuctio f(x) = x +3 x. Oe of the followig statemets is true. Which oe? f is icreasig o the iterval (, 3). (b) f has a local miimum at x =. (c) f is decreasig o the itervals (, ) ad (, 3). f is icreasig o the itervals (, ) ad (, 3). f has a local miimum at x =. Solutio: From the previous problem, we kow f(x) =x ++ 4 x,

12 so that The for x = f (x) = f (x) > > Name: 4 (x ). 4 (x ) (x ) > 4 (x 3)(x +)> x< orx>3. Thus f is icreasig o (, ) ad (3, ) ad decreasig everywhere else (i.e. o (, ) ad (, 3)). Clearly x = is ot a local miimum sice f has a vertical asymptote there. Although f ( ) =, this is actually because of a local maximum. Ideed f 8 (x) = (x ), 3 so f ( ) <.

13 5.(6 pts.) Cosider the fuctio 9x6 x f(x) = x 3 +. Oe of the followig statemets is true. Which oe? (b) y = 3 is a horizotal asymptote of f, ad y = 3 is ot a horizotal asymptote. f has o horizotal asymptotes. (c) y = ad y = 3 are both horizotal asymptotes of f. y = ±3 are both horizotal asymptotes of f. y = is a horizotal asymptote of f. Solutio: Sice the problem oly asks about horizotal asymptotes, we compute limits as x ± : 9x6 x lim f(x) = lim x 6 x x x 3 + = lim x = =3, 9 + ad similarly (sice x 3 = x 6 whe x<) lim f(x) = x lim x = lim x = = 3, 9 + so y = ±3 are both horizotal asymptotes of f. 9 x 5 + x 3 9x6 x x x 5 + x 3 x 3 x 6 x 3 3

14 6.(6 pts.) The fuctio f(x) =(x +) 4 4x +5x is cocave dow o which of the followig itervals? (, ) (b) (, ) (c) (, ) (, ) (, ) ad Solutio: Sice f (x) =4(x +) 3 48x +5=8(x +) 3 48x +5 f (x) =4(x +) 48 = 48((x +) a b )=48(x)(x +)=9x(x +), fidig where f (x) < amouts to solvig x(x +) <. The curve x(x + ) is a upward-opeig parabola with roots at x = ad x =, ad whece is egative whe <x<. Therefore f is cocave dow o (, ). 4

15 7.(6 pts.) A ope box is to be made from a square of side oe by cuttig four idetical squares ear the vertices. The box with the largest volume has a height of 6 (b) Solutio: If the height of the box is h (which is also the side legth of the cutout square), the the volume is give by Thus (c) V = h( h) = h 4h +4h 3. V = 8h +h =(4h )(3h ), so that V = whe h = or h =. 6 I order to make a box, h must be i the iterval (, /). Because V is a upwardopeig parabola, it must switch from positive to egative at h = ad be egative util 6 h =,soh = is gives a maximum o (, /) (6 pts.) Whe applyig Newto s method to approximate a root of the equatio x 3 x + =, with iitial guess x =, the value of x is:.5 (b).5 (c) 3 Solutio: With f(x) =x 3 x +, we have Thus f (x) =3x. x = x f(x ) f (x ) = f() f () = =. 5

16 9.(6 pts.) Which of the followig is a Riema sum correspodig to the itegral 3 x 4 dx? ( + i )4 (b) i= ( + i )4 (c) i= ( i )4 i= i= ( +i )4 ( i )4 i= Solutio: With f(x) =x 4 ad x = 3 =, the Riema sum i this case is f( + i x) x = ( + i )4 = ( + i )4. i= i= i=.(6 pts.) A fuctio f(x) defied o the iterval [, ] has a atiderivative F (x). Assume that F ( ) = 8 ad F () = 7. Which oe of the statemets below is true? (b) (c) f(x)dx =. F (x) is a icreasig fuctio. f(x) ca be a odd fuctio. f(x)dx =. f(x)dx =. Solutio: If f is ot assumed cotiuous, the f might ot be itegrable, so that oe of the choices are correct. Thus we add the hypothesis that f is cotiuous. By the fudametal theorem of calculus, f(x) dx = F () F ( ) = 7 8=. Note that f(x) caot be a odd fuctio sice if it were, the cotrary to the calculatio above. f(x) dx =, 6

17 .(6 pts.) Calculate the itegral 3 si x dx. (b) (c) Solutio: Sice si x< oly o (, 3/), 3/ / si x dx = / si xdx+ 3/ = cos x / +cosx 3/ =. si xdx.(6 pts.) The volume of the solid obtaied by rotatig the regio give by x + y =, x ad y, about the lie y = is (b) (c) ( x )dx [ x + x ]dx x[ x + x ]dx x x dx ( + x ) dx 7

18 Solutio: The outer radius is x + ad the ier is, so V = = = ((outer radius) (ier radius) ) dx (( x +) ) dx [ x x ] dx. 8

19 3.(6 pts.) Fid the volume of the solid obtaied by rotatig about the y-axis the regio betwee y = x ad y = x 4. 6 (b) (c) Solutio: The curves itersect whe x = ad x = (ad x =, but sice the solid is obtaied by rotatig aroud the y-axis, this itersectio poit is irrelevat). Thus V = x[x x 4 ] dx x 4 = 4 x6 6 = 4 6 = (6 pts.) Fid the average of f(x) = si (x) cos(x) over [, ]. 3 (b) Solutio: The average is give by / / 3 si (x) cos(x) dx = u du = 3. (c) u du 3 9

20 5.(6 pts.) A (vertical) cylidrical tak has a height meter ad base radius meter. It is filled full with a liquid with a desity kg/m 3. Fid the work required to empty the tak by pumpig all of the liquid to the top of the tak. 5 kg-m (b) kg-m (c) kg-m kg-m 5 kg-m Solutio: Noe of the give solutios are correct. We cosider the cylider sliced ito slabs of equal height x so that the work doe o the i th slice is W i = F i x i =( V i 9.8)x i =( () x o.8)x i =(98 x)x i, where x i is a poit i the i th slab. The the total work is W = lim W i i= =98 lim =98 =98/ =49. x i x i= xdx

21 Math 55, Fial Exam: December 5, 7 Name: ANSWERS Be sure that you have all pages of the test. No calculators are to be used. The exam lasts for two hours. Whe told to begi, remove this aswer sheet ad keep it uder the rest of your test. Whe told to stop, had i just this oe page. The Hoor Code is i effect for this examiatio, icludig keepig your aswer sheet uder cover. PLEASE MARK YOUR ANSWERS WITH AN X, ot a circle!. (b) (c) ( ). (b) (c) ( ) 3. ( ) (c) 4. ( ) (b) (c) 5. (b) ( ) 6. ( ) (b) (c) 7. ( ) (c) 8. (b) ( ) 9. (b) (c) ( ). (b) (c) ( ). ( ) (b) (c). (b) ( ) 3. ( ) (b) (c) 4. (b) ( ) 5. (b) (c) ( ) 6. (b) (c) ( ) 7. ( ) (b) (c) 8. (b) ( ) 9. ( ) (c). (b) (c) ( ). (b) (c) ( ). ( ) (c) 3. ( ) (b) (c) 4. (b) ( ) 5. (b) (c) ( )

MATH 10550, EXAM 3 SOLUTIONS

MATH 10550, EXAM 3 SOLUTIONS 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,