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**Unformatted text preview: **Name: Math 115 — F2011: Quiz 1 - September 6th 2011 Read the instructions to each question carefully. Provide all required reasoning, and show calculations.
The Honor Code is in effect. Calculator use is NOT allowed. fry} 2 0.5 - C 4'} (1) Find the sinusoidal function 9(m) Whose graph is shown below:
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Q___ .L- V.__—__,___,.___ _,. WE Slim} 3 03902; [x] “l” 1‘ Li} ”x, 15(2) Give the formula for a sinusoidal function h(t) which has these propertieS: h(%) = _1 is the N/ minimum value, h(—%) = 3 is the maximum value. ‘ - i %~(-0 "5+1 1
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' lt, s,— 1 ﬂ ti} r (/6) SU’UJL Mammy. it, ﬁt; ﬁrst: 0 =9 put) :7 (1, gm U213) Jr‘s: qr 3)th H l L/ (5) A mow Lows moves Wlm velocnzy ME) 2 4 — o -.-- IILBBel‘S/ILOUI', w1m v 111 Hours. 0 Find 11(0), v(10), v(20).
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3 Math 115 — F2011: Quiz 1 - September 7th 2011 Name: _ V1 [5% ' if; 7Q * Read the instructions to each question carefully. Provide all required reasoning, and show calculations.
The Honor Code is in effect. Calculator use is NOT allowed. W (1) Find the sinusoidal function 9(33) whose graph is Shown below: For the sinusoidal function 9(0) is not the
max, so g(a:) : asin(bm) + k. The max is O and the min is -4, so the
amplitude is a = (0 — (—4))/2 : 2L1) The average value is the k: = (0 + (—4))/2 = —2. C1) midline, The period is p = 277 so I) : 15% a,
{WM/‘4 Therefore, 9(33) 2 2 sin(m) — 2. C49 ,g/‘\_ (2) Give the formula for a sinusoidal function h(t) which has these properties: Mg) = 1 is the mini— :’ “a, E «ESE , mum value, 71(0) 2 3 is the maximum value. “‘ng 2 It is simplest to take the period p = T” The amplitude a is half the difference between the max and the min, so a. = (3 — 1) / 2 = 1. The midline is the average of the max and the min, so k = (1 + 3) / 2 = 2. Finally, h(t) is a cosine function since its max is at t = 0. [remember — we don’t use phase shifts!]. a” 1 Thus, h(t) = cos(2t) + 2. (LE) 0 The period of h(t) is: [email protected]} a”) o The amplitude of h(t) is: 1. 1,) o The midline of h(t) is: 2. CV (3) A raccoon moves With velocity v(t) = 5 (5 — 2—0“) meters/ hour, with t in hours. {2“ c Find 12(0), v(10)@£
1/ 11(0) 2 5(5 — 2—0”) = 5(5 — 1) = 20. 11110) = 5(5 — 2-01-10) = 5(5 — 1/2) = 45/2. ﬁg 0 At What time does the raccoon reach a velocity of %? / Solve 11(231—5 for t 11
(1)—97fk<=”15(5 20-1t)=%1§<=>5— 20:“ 13:15—13:20” :> 2_E'i)=2—O.lt :> 211-22—0'1t ﬂ 2—222—0.1t; Taking log2 on each side —2 : —% 4:» t f 20.
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Li 0 For very large values of t What is the approximate value of 1205)? f? _‘ e/ For 75 very large 2 0 11__2__01H is very small close to 0; So, v(t ) is approximately 5 - 5 : 25.
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