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Unformatted text preview: Math 105—922 Solutions to Practice Midterm 2 Rflicltorm 2 on Vikxlnesday, July 14 in BUCH A101 1. Short Answer Questions 3’: ~— (28/3 —— 2
(a) Find / Wdﬂ?
‘ {I} '3::—:'8/3—2 ' r r r f r' 2
solution: / I i :3 (is: z ~i— 3:2” — 2$“2)dri: : 31n+ gala/‘3 + + C.
‘11.“ (h) For all values X} —1, define the function = 1—] {21; + 435m. Find 1" ( 8). Siinl‘JIii'y your answer.
,32/3 solution: Recalling the 17mniuirieiiinl '1711001‘011'1 oi Calculus, _["(3:) 2: .5?) "" ' Evalinlting at :1; :::= 8, If" 2:: 1 /100. (c) The daily revenue in dollars for Sally’s espresso stand is given by RH) = 500 + 8mm where
t is the minier of days since J urinary 1st.
W hat was Sally’s average revenue from the month of February (day 31 to 59)?
, 1 '5” 2 a
solution: average revenue = [,9 ‘31 560 + (:O'UHCH: : 2—8(5001’; + 100(30'0”§‘}J)
t){ *W I. . I 3] % $50} .57 per day. (cl) Find {lie consumer’s surplus for the (ilen'iuiltl curve p(:i:) :2 180 — 0.04:1;3 dollars per unit at
the series level :1: 5 units.
solution: At a; : 5, 10(5) z": 100 w U.{}4(5)3 m 95. The consumer surplus is F (100 "—0.0%;3 — 95cm: 2 5 s 0.04m3cim 2 5$~0.01‘r§ a $18.75
0 ( r "J o A. (e) If possible, (IOll’lPUiie /
o powwow1:: /  0 2
2:11:22 w~ Marine. Justify your 21.115wor. solution: Siruze the immlzion is continuous in [0,2]‘ we can apply the i'umlmneminl i;l‘ieorern. , l . . . . r (in .10 solve the ii’iizograi, use Si.ll)SiJLUi‘.l()l‘L Let u. 2: :21‘2 —— 4, so (in 2:: 21113: or c1571: :2 AL :1: r: 0,
:2: u :2: fl and 211'. :1: :1: 1, U. z: 8. Then, .2 '8
r r r 1
/ 221.132 + 4)"(£:L' : / 11.00111. : suﬁli = 43008
0 4 6 2 r,
(f) If possible, compute / O o’er. Justify your answer. 2
—2 'i’ 1)
solution: The function is discontinuous at x=~1, thus we cannot apply the fundamentai theorem of calcuius to compute this integral. Compute [323111.(m)d3;
3 solution." Using partial fraction, let ﬁre) 2 111(1‘) and g’(.’r:) = :1: .
Then, f’(:i:) : and g(:1r;) m fed. ' . l '1. l l l
.3 . ,i ,. ._ W11 . ,. __ W , m «I ,. :1: __,.'i . ,‘ ___ ,.’l M 7
jar, l.1](.t)CiJ, 4.1. limit) a; 4:1,. (ii. [1.1, 111(2)) 16st I C (h) Compute /7122(3$3(l05(8$3)d33. solutton: Use substitution, let u m 6"“ . Tl hen m— x 3113281 by chain rule, and do
(in: m 3 .
31826:” . ,1 c 1 ' l .
[332615 cos(e”‘3)drc : M / cos('u.)du =: sin('u.) + C = E); si1‘1(e"’d) + C .
‘ I (i) jOHHDiiUl/25/31?’t(22)dz. solution: Use integration by parts, Let = 111(22) and g’ =2 2””, so
f’(z) m g and x $28”. fzs/3ln(22)dz m 228/3ln(zg) — f g  gzs/sdz
' . 3 ' r  = 328/3 ln(z2) — 2"/3dz 3 ’8/3 I :w; ‘i 2:2 m [swamC
8 “( l 32 ' (i) Find the positive constant A: such that the area, under the graph 1 2: c"""’”’ with x going from
0 to inﬁnity is equal to 7r. solution: We need to ﬁnd it such that e‘“'k”"cl:i: r:— 7r. 0° w. 1”: i
/U(fkd$=—E€k‘lgo=g Thus, we need = 7r, or k = (k) Given a differentiable function Mm) such that 11(2) : 3 and 11(5) 2 —1 and ,f' 0 ‘LJ
/ l'i.(;i:)d:t 2: 8, find / :ti't’(:i:)d:i:
2 . 2
solution: Use integration by parts to solve: iet :L' and g’(:1;) 2r: li’(:1:) so that f’ (11;) 2: l. and M32). r. f0 wh’(:t)dq: = m V/o h(:t)d3: = (5 (—1)— 2 3) — 8 m ~19.
2 2 . _ 2 2. i incl the men lj>enn(:leci by the curves y r: (:1; ~§~ 2)2 and y 2: “A :1: + 5:1: ~l 7.
sol'utio'n:
Te graini tile ein‘ves, note the following ' 'U I (73 "3“ 2P is a, iimi‘aiJDizL with a. zero and 21380 a minimum point at :1: : ~2.
At :1: =: 0, y 1",: 4 o y 1,: mg:2 + 5.1: + 7 is another parabola but it’s concave down with zeros at 1i: 53 H “1.14;, 6.14 and at maxin'lnm point at a: = 5/2. At x = O, y m 7. o The ii‘itersection points are (:2: l 2)2 “:32 + 5:1: + 7
(1:2 i 11.7: "in 4 "—1132 — 5:1: —1— 7
2:1:2 — :2: 3 3:: i)
(:1: ~i 1,)(22: 3) = i}
3 ——> :1; : ~], — ’ 2 “WW ’lihen, the men is o3/2 I r g
/ 3:2 —i— 5:3 + 7) — (:2: + 2)3dm :2
 ‘ ,, ii. A modern cm'nping tent is (ionic simixrd whose vertical cross section is <ic35criimd by the equation
y :2 “332:2 i 5. Find the volin'ne of the tent. solution:
(16 c, 1 f 2
Volume 2 / 27¢ (73:3 ~i~ 5) (in:
. 0 3
i "m 1 i 2‘71" ‘/ 3:4 m 53:2 + 25dr1:
,F x; . o 4'
e” 10 3 W N 9,16%”sz 41. 30:1’11mte the average mine of the function f(:1:) 2: :2sin(:::) over the ii‘itervel [0, mr].
solution: Use integration by parts with f (:12) r: a: and ‘q’ 2 Sin(2:} SO that f’(:L') : 1 and
g(:1:) 2: cos(r1:). Average :
TNT “'1'”
m 0 msin{m)dm
1 “(HI '
2: W? [at cos.(:1;)i3‘7T + / cos(:1:)d:z:J
‘ {3 TL 7T 3?? )”('.'i.7r} «in Si]1(£l.‘) 37'} :: (.__1)'”"'] 7171' since sin(7'm) ::= 0 and C()8(‘l'.'.?r) 2: 1)" for any integer n. 5. Compute the area bounded by the graphs 3; 3 3:2 w 10:1? — 5 and y =2 3:1:2 17:1: — 0.
solution: To graph, use the lollowing: o y : 1‘2 m 10:1: w 5 is a parabola with zeros at 5 :l: \/ 30 a: “0.48, 10.48. AL 3; 0. y 2: _ l7 :i': V297 N w—— W  y 3.1:2 u 172; — .9 is another paraboia that‘s concave up with zeros at
—0.49 ,6.15. At :1: x 0,11 = “9. o The intersection point is found by equating 6 2:2_10:L' —5 = 32:2 17.’L‘— 9
“2:232 + 73: + 4 = 0
#(22311M1L'm 4) =0 so the intm‘seetiorl is at :1: : m% and 4. Them «1
area = f ((132 — 3.0.1: A 5) — (3.7;:2 "r 17:1: w 9)d:1: —1/2
’1 Hm“
I, u“ m / w2952 + 7mg} 143953: E .3
. m1/2 a. 8 6. Compute the solid of revolution obtained by rotating about the runaxis the region under the graph
of y r: m2 H over the interval [1, 3]. solution:
.E I 1 2
volume = / 7r (3:2 1— :2 \ﬂ > (I I
1
‘3 . 1
2271"] 3:41— Una—4.7;ch 4
_W EE+2$W2+EE a
”" 5 7 8 1 7. A cargo ship accumulates harnacles on its hull at an instantanemis rate of (f) 1000 )0 1 )ml ( , ::= WWW77 ' runs 301‘ on '. '1 .1 (t $74)3/2 i 3 where t n'reasured in terms 01' mo1’1ths. At the beginning, the ship already has 3000 [moods of
barnacles on its hull. How many pounds of barnaeles does the ship have on its hull at the end of
twelve months. solution:
With the rate of accumulation given by g(t), the total of barnacles aecunmlated in 12 months can 12
be computed by / g(a:)d.2:.
0 12
1 0
amount of barnacies :— 3000 + / WQQT—dt
. o (i '1" 413/2 2000 r
“)2 :: ‘3500 pounds
2'; + 4 2: 300 0 — 8. A certain annuity provides nieonie [or a period of 20 years. it includes a. yearly cost oi' living
increase so that at time I... measured in years. from the date of pnrel‘u‘ise, it will provide income at
an iiistantaneons rate of iii] :12 ii(}(l(}[)~£rU.Ufit per year. Assuming tl'iat interest rates remain fixed
at 400 arnmally over 20 year period, what would be the present value of the annuity if purchased today.
solution:
20
Present Value == f(i':)e"e'm{'dt
. 0
20 20 ~20 f (3000 4— 0.(}6t)e"'0‘0’“‘dt / 30005“”’”dr + / ().06te"”0'0’”'dt
0 . 0 . 0 5000* )"““”“‘lm l 0.00 {tiff 9” — / ( e 0 (W (a
"M004 ' 0 M004 0 ‘0 0.04
using integration by parts with 21:: t and g’(t)e“0'{m
:: (ﬁnish the algebraic: steps)
a ﬁlial13010.40
w .. . . 2
9. 1<1nd the area of the finite region borinded by the yams, the curve 1; = a: and the curve y = MIT.
:1: v it wili be useful to sketch the region before attempting any caleuiation. solution: The intersection between :1; :2 (r: and y : 1 1 is
:1: u
2 _. :1:
1 'i l Hwm2:0 (1 *‘ll{w i2):: )
1 t 1 ,w2 #1 The only relevant intersection for this problem is at m = 1. Then, /l 2 i
area : — w Italy:
I 0 :1; —{— 1
r r 1 2 1 r i 1
:: 2111(5): + i) w 5:12 [0 :3: 2 111(2) E): 10. Suppose that the marginal revenue function [’or a conniany producing q units of a product is
[\di’i’tq) r— ]i’.’(q) 400 ~— 3qg. Find the additional revenue received from doubling production if
Currently 10 units are being produced. 23 20
additional revenue 2 / R’(q)dq : / 400 —— 3.920%;
. 10 . 10 x 400g — (131%3 = ~3000 11. A tire manufacturer estimates that (1 thousand radial tires will be demanded by wholr—rsalers when
the price is p = D(q) m —0.1q2 + 90 dollars per tire
and that the same number of tires will be supplied when the price is
p 2:: S(q) 3: 0.2g2 + q + 50 dollars per the (a) Find the equilibrium price and the quantity supplied and demanded at that price.
solution: The equilibrium price is found when 13((1) :: qu), 41.162 + 90 2 6.2g2 ~i~ q + 50 0.3qgulq—4Om0w(0.3q+4)(q"10l=0—lqﬁmgﬁ ,10 Since q represents the number of tires, only q a: 10 a valid solution. ’l‘herefore, 10 is the
quantity supplied / demanded and the equilibrium price is
p 2.». one) x —0.1(102)+ 90 x $80 W (b) Determine the consumers and producers surmuses at the equililfiriun'i price
Cgmfglj NF“ serrate solution:
S‘ .10 66
‘ consumer surplus m D(q) ~ D(10ldq = / (—0.lq2 «IA 90) —— 80dq
“a.”  0 i 0
\ 10
r ’* ——1 . 200
a i) =f ~0qu +10dq z —iﬁ + 16(1130 : ‘ m $66.67
i G 39 3
\ @%uii'i'rxiis‘ﬁ ‘ 1 0 m
f “"5” producer surplus m / 8(10) — S(q)dq = f 80 — (0.2g2 + q + 50mg;
0 0
Vm‘l‘r‘” ‘0 1 , 1 5" )
SUVWUS =3 / ~0.2(12 q  30dq r: 3f)(] :~;q‘l m Tr/QW :: ):J( *3: $183.33
. 0 ii 2 d 12. An oil well that yields 900 barrels of crude oil per month will run dry in 3 years. T he price of
crude oil is currently at $40 per barrel. If the oil is sold as soon as it is extracted from the ground
and the money invested at 5% per year compounded continuously, what will be the total. future
value of the revenue from the well over its 3 year lifetime.
solution: a rSl‘he production rate per year is 900 barrels/niontli X 12 months / year 2 1.0800 barrels/ year.
a The annual income rate is then $IlU/barrel x 10800 barrels/ year I $432000 /year. 0 The future value of revenue when the oil runs out is ‘3
/ 432fl0060'05l3"l'lclt = $1, 398, 247.86
(J ...
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This note was uploaded on 01/14/2012 for the course MATH 105 taught by Professor Malabikapramanik during the Fall '10 term at UBC.
 Fall '10
 MalabikaPramanik
 Calculus

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