Calculus II (MAC )

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1 Calculus II (MAC232-2) Tst 2 (25/6/25) Nam (PRINT): Plas show your work. An answr with no work rcivs no crdit. You may us th back of a pag if you nd mor spac for a problm. You may not us any calculators. Pag Points Scor Total: Pag of 6

2 Calculus II (MAC232-2) Pag 2 of 6. Circl Tru of Fals. (a) (3 points) TRUE Fals Th diffrntial quation dz dt z+t is sparabl. Solution: Not that dz dt z+t z t. (b) (3 points) TRUE Fals Th hydrostatic prssur at a point in a liquid is proportional to dpth and is indpndnt of dirction. Solution: Hydrostatic prssur P is givn by ρgd, whr d is th dpth of th point at which w want th prssur. It is indpndnt of dirction (.g., it is th sam in x, y, and z dirctions at any point). (c) (5 points) Tru FALSE 3 x (ln x )3 ln(2). Solution: This is an impropr intgral, sinc x is a vrtical asymptot in th intrval [, 3]. So, 3 x : lim t 3 x + lim s + s x, if both of th intgrals on th right-hand sid convrg. But lim t x lim (ln x )t t lim ln t. t (d) (5 points) Tru FALSE Lt a and b b ral numbrs. If a b, thn a 2 b 2. Solution: Th implication only holds if a b, sinc x 2 is an incrasing function for nonngativ ral numbrs. On th othr hand, 2 but ( 2) 2 4 ( ) 2. () (5 points) TRUE for all x. If Fals Suppos f and g ar continuous functions and f(x) g(x) g(x) is convrgnt, thn f(x) is convrgnt. Solution: This is th comparison tst for convrgnc of intgrals of nonngativ functions. If th intgral of th largr function convrgs, thn so dos that of th smallr function. 2. (5 points) Find th volum of th solid rsulting from rotating th rgion boundd by th curv x 2 + (y 6) 2 4 around th x axis. Hint: A thorm of Pappus (which conncts cntroids and volums of rvolution ) may hlp. Solution: Th quation x 2 + (y 6) rprsnts a circl with cntr at (, 6) and radius 2 (and ara A πr 2 π(2) 2 4π). Th cntroid of a circl is at its cntr (du to symmtry), which is at a distanc d 6 from th axis of rotation (th x axis). So, according to a thorm of Pappus, th volum of th solid rsulting from rotating th rgion boundd by this circl around th x axis is qual to th distanc travld by th cntroid of th circl (2πd) tims th ara of th circl (A), V 2πdA 2π(6)(4π) 48π 2.

3 Calculus II (MAC232-2) Pag 3 of 6 3. Considr th dfinit intgral I : x2. (a) ( points) Us Simpson s rul with 5 points to approximat I. Thr is no nd to valuat th final sum. Hint: Th intgral of a parabola with hights y start, y mid, and y nd at vnly-spacd points ovr an intrval of lngth l is l( 6 y start y mid + 6 y nd). Solution: With 5 points w hav n 4 subintrvals, so th points ar x 4 4 apart. S I 2(/4) ( ()2 + 4 (/4)2 + 2 (2/4)2 + 4 (3/4)2 + (4/4)2 ) 6 2 ( + 4 /6 + 2 / /6 + ) (b) ( points) Estimat th rror involvd in approximating I using th trapzoidal rul with 5 points. Hint: If f (x) K for all a x b, thn E T K(b a)3, whr n is th numbr of trapzoids. 2n 2 Solution: Th intgrand is f(x) : x2. So, f (x) 2x x2 and f (x) 2( x2 + x( 2x x2 )) 2( 2x 2 ) x2. To find a loos bound for f (x) 2( 2x 2 ) x2 2 2x 2 x2, not that 2x 2 + 2x 2 +2x 2, which is at most +2() 2 3 in [, ]. Also, x2 is at most () in that intrval. So, f (x) 2(3)() 6 : K and 6( )3 E T 2(4) W can find a tight bound for g(x) : f (x) in [, ] by finding its global maximum in that intrval. Th drivativ g (x) f (x) 4(2x 2 3)x x2 is zro at x and x 3/2. But 3/2 is not in [, ] (sinc 3/2 > 2/2 ). So, thr ar no critical points and hnc no xtrma (local or global) in th intrior of th intrval. At th nd points x and x w hav g() f () 2 and g() f () 2/. So, th global maximum of f (x) is 2 2 : K. W could hav sn this without diffrntiating g(x) f (x) too. Both ( 2x 2 ) and x2 ar dcrasing in [, ] (hnc, so is thir product). So, th maximum of f (x) 2( 2x 2 ) x2 must happn at on (or both) of th nd points of th intrval. In any cas, this tight bound givs a bttr rror stimat: E T 2( )3 2(4)

4 Calculus II (MAC232-2) Pag 4 of 6 4. ( points) Find th valus of p for which th intgral x(ln x) p convrgs. Solution: t : lim x(ln x) p t lim t x(ln x) p x (ln x) p lim (ln x) p d(ln x) (substituting u : ln x on th fly ) ( ) lim (ln x) p+ t t, p p+ lim t (ln(ln x)) t {, p p lim t ((ln t) p ), p, p. t Sinc lim t ln t, th rmaining limit is finit (and th givn intgral is convrgnt) only if th powr p of lnt is ngativ (so that ln t gos to th dnominator). This givs p <, i.., p >, for which x(ln x) : p p lim ( (ln t) p ) ( ) t p p. 5. ( points) Find th lngth of th curv y ln(sc x) whn x π/4. Hint: An antidrivativ for sc x is ln sc x + tan x. Solution: W hav dy (sc x tan x) sc x tan x. L : π/4 π/4 π/4 π/4 π/4 π/4 ds + ( dy )2 + tan 2 x sc2 x sc x sc x (sinc sc x > for x π/4) (ln sc x + tan x ) π/4 ln /(/ 2) + ln + ln( + 2)

5 Calculus II (MAC232-2) Pag 5 of 6 6. (8 points) Vrify that th function y(t) t cos t t is a solution of th initial-valu problm t dy dt y + t2 sin t, y(π). Solution: First, not that th givn function satisfis th initial condition, y(π), sinc y(π) π cos(π) π π π. To s if it also satisfid th rst of th diffrntial quation, w find dy dt (cos t + t( sin t)) cos t + t sin t. So, th lft hand sid is t dy dt t cos t + t2 sin t t, which is idntical to th right-hand sid y + t 2 sin t t cos t t + t 2 sin t. 7. (6 points) For what valus of k dos th function y(t) cos kt satisfy th diffrntial quation 4 d2 y dt 2 25y? Solution: W hav d2 y dt d dy 2 dt dt d dt in th diffrntial quation w gt d(cos kt) dt d( k sin kt) dt k 2 cos kt. Substituting 4 d2 y dt 2 25y 4k 2 cos kt 25 cos kt, that is, (4k 2 25) cos kt, for which to hold for all t w nd (4k 2 25) (2k 5)(2k+5), giving k ±5/2.

6 Calculus II (MAC232-2) Pag 6 of 6 8. ( points) Th arc of th parabola y x 2 from (, ) to (2, 4) is rotatd about th y axis. Find th ara of th rsulting surfac. Solution: W hav dy 2x. So, S 2π 2π 8 2π 8 2π x ds 2π 2 x 8x + 4x 2 + ( dy )2 2π 2 x + (2x) 2 ( + 4x 2 ) /2 d( + 4x 2 ) (substituting u : 4x 2 on th fly ) 2 3 ( + 4x2 ) 3/2 2 π 6 (73/2 5 3/2 ) π 6 ( ) 9. ( points) Solv th initial-valu problm dl dt L2 ln t, L() 8. Hint: Intgration by parts may hlp with on of th intgrals. Solution: This is a sparabl diffrntial quation: dl ln t dt L2 dl L ln t dt 2 L 2 dl ln t dt, intgrating th right-hand sid by parts (with u : ln t and dv : dt, so that du dt t v t) w gt, L uv v du t ln t t dt t L t ln t t + C t ln t t + C. L From th initial condition L() 8, w hav and that is, C 9/8. So, L(t) 8 () ln() + C, t ln t t + 9/8 t t ln t 9/8.

Solution: APPM 1360 Final (150 pts) Spring (60 pts total) The following parts are not related, justify your answers:

Solution: APPM 1360 Final (150 pts) Spring (60 pts total) The following parts are not related, justify your answers: APPM 6 Final 5 pts) Spring 4. 6 pts total) Th following parts ar not rlatd, justify your answrs: a) Considr th curv rprsntd by th paramtric quations, t and y t + for t. i) 6 pts) Writ down th corrsponding