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Unformatted text preview: MATHEMATICS 1LS3 TEST 2 Day Class .E. Clements, R. Eftimie, M. Lovric Duration of Examination: 60 minutes
McMaster University, 12 October 2010 FIRST NAME (please print): M FAMILY NAME (please print):
Student No: THIS TEST HAS 8 PAGES AND 7 QUESTIONS. YOU ARE RESPONSIBLE FOR EN—
SURING THAT YOUR COPY OF THE PAPER IS COMPLETE. Total number of points is 40. Marks are indicated next to the problem number. Any Casio
fx991 (or lower, nonegraphing) calculator is allowed. USE PEN TO WRITE YOUR TEST. IF YOU USE A PENCIL YOUR TEST WILL NOT
BE ACCEPTED FOR REMARKING (IF NEEDED). You need to Show work to receive full credit. Problem Points Mark Continued on next page MATH lLSB * Test 2 * 12 October 2010 Name:
Student N0: 1. (a)[2] Find 311:1: such that 1:1(3: + 5) = 21m.
9"“(“*‘° )= “m? A ‘MG all
XQx 2%
Cx—%3LX+13=O f
$23—19 $.0me Um MM‘i (>9ka QM mm M (Mi) X :7. —’2.—P “0'? SgkUwa (NJ? ivx oKQMMCm 3Q— KM‘L)
(b)[2] Solve the equation 261+”: : 7. I x1: MGMx
x = t {MGML 9:. 1:06 (c)[2] Given that tan 9 = 2 (where 0 < 6 < 7r/2), ﬁnd cos 6. 4.
CD89 = a
$3 2 _ \PE
3'
’1.
(d)[2] If arm+1 : mt + 12 and m0 = 7, what is W100?
“Mum +9.5; +11 0V1
_ U$Q QN'qulq
W‘Qrm'L+ u: 1L+u+ u = re +u 2
mph: mLk—L’Z t; 2?. +12. 3 W: l
i W‘Loo : ¥+ ’Lrl' LOO 1‘ 120? Continued 011 next page MATH lLSB * Test 2 * 12 October 2010 Name:
Student N0.: 2. A radioactive substance got reduced to 2/5 of its mass in 100 years.
(a)[2] Determine (Without actually calculating it — that Will be in part Whether the
halfelife of the substanCe is more or less than 100 years. Explain your reasoning. LtOo/o 0Q (that gubelnwxcr ls 0\ t“ “30 L103. s—> mthth 4 lost“; Am 9. {AIM
Ema W451 KN m subswam, ta mule
500,0 0Q it": OVC'Qtw‘SL ameuW'lx‘ (b)[3] Calculate the half—life of the substance, thus checking your answer in Note that
you are asked to calculate the halfelife — so you cannot use the ready—made formula for it. “Q‘WWG‘: “9 mutt/Suzerwm 90:}: E (Q) io'oogzt \MAQ on; 0.5 €§®o°¥lt QMQfg._:_0.009”,t at: W055 “0.00259. WWW Wk mhuw w \s mwvgvj \ézonooi ——a t: #201 \L=~00032.—§1 \L= —o.00°516 —s £27436? . It= 0 Mama's v t=ﬂ_‘&___— Continued on next page w‘ﬁﬁs
W5 MATH ILSB * Test 2 * 12 October 2010 Name:
'Student No.2' 3. (a)[3] Sketch the graph of the function = —_l.2 sin($ — 71') + 2.8. e‘m Lx—w)... we»: my: w udrs (b)[2] What is the range of the function from (a)? [L9H] (c)[2] Identify the Inaximum and the average values of f WM = L's
Woe; :: 948 Continued on next page MATH 1LS3 * Test 2 * 12 October 2010 Name: 
Student No.: 4. It has been determined that a population of green—striped ants (endangered species)
behaves according to Pt+1 : 0.65Pt, where P; is the number of ants in millions and t is the
time in months. (a)[2] The constant 0.65 is called the per capita production rate. Explain what it means for
the population of ants. I (b)[3] If the present population is P0 : 34,000,000 (i.e., 34 million) ants, What will the population be two years from now? [Be careful about the time units!]
' l:
945:35000‘000 O‘GS 3c“; Mouths
. ' on Continued on next page MATH 1LS3 * Test 2 * 12 October 2010 Name:
Student No: 5‘ A population of birds is given by ﬁt) 2 4000 — 30006_t, Where t 2 0 is time in days.
(a)[3] Sketch the graph of H000  "5600 E)“ _ (1))[1] Describe the longsterrn behaviour of ﬁt); i.e.J what happens as t keeps increasing? in vvgwiadﬁw .QUr) @Vwaxms Lmm Continued on next page MATH 1LS3 * Test * 12 October 2010 Name:
Student No: 6. Consider the system mt+1 : —0.5mti+ 6, where m0 = 1.
(a)[l] Find the equilibrium point(s) of the system. m* =40.‘5M*+Q 45 \m* = 6
Nip: H (b)[3] Starting with m0 = 1, cobweb for three steps; i.e., in your diagram, show m3. Also,
indicate the equilibrium point(s) that you calculated in (a). 1; UNA: 0 8mg 6 . (c)[1] Calculate the value of mg algebraically and compare with your diagram in mo=l
m: OS~’L+Q= ‘55 mlzOS 55 +Q '= 3'25
“‘5” "05415 +6: HERE Continued on next fage MATH 1LS3 * Test 2 * 12 October 2010‘ Name:
Student No.: 7. Consider the continuous model for bacterial growth. It islknown that the population
doubles every 8 hours anEl that the initial count is 600 bacteria. (a)[2] What will the population be in 24 hours? [Hint: think! You don’t need part (b) for
this] (500 19.0 o ZHBO L300
Mg m J '\_Jﬁ _
8 E H
(6mm (mm CM;th “3
(b)[3] Find the formula for the population POE) of bacteria as a function of 15.
EU?) = €00 apt
' — 81L ,
17290 —— em) e
8L=QMZ$ L=R£§E REUOS’GQWBH
QM'Z.
BUG): 600?. = Geog (c)[1] When will the population reach 12000? ' 0.08€C'c
Liege 2: Goo a 0‘0 =e0.09€£’c
__ M10 ~
'lr." #00966 ~ 3R5?) lQMS K: onset $505 We. DQENBS IL= 0.081 15an ON V»
VJ: 0t B%~.3%1g its0,0869% ._. ’o‘mﬁé}
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This note was uploaded on 11/07/2010 for the course 115 cs taught by Professor Kuzak during the Spring '10 term at Waterloo.
 Spring '10
 kuzak

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