# Math 100 Dec 03 Answers - Solutions for MATH 100/180...

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Unformatted text preview: Solutions for MATH 100/180 Sessional Examination — December 2003 1. (a) f’(:t) = 33:2 + 9 (b) Applying L’Hospital’s rule, we have . x—2 _ l 1 hm :lim—z— w~2x2~4 r~221~ 4 ((2) Applying L’Hospital’s rule, we have 1. sinﬂ l, 0056 l. 1 6 2 26 1 im : ______ z _ = _ ego tan 26 913(1) 2 3802 20 013(1) 2 COS cos 2 (d) g’(t) = e‘ sint+et cost, and hence g”(t) = e‘ sint+et cost+et cost—et sint : 26‘ cos t. (e) Since the graph equation is y = tan‘1(\$2), we have ,_ 2:: y _1+x“ Thus the equation of the tangent line to the graph at the point (1, is 7r 2*l( 1) 7r 1 __: \$_ __:._ y 4 1+14 y 4 r (f) We use Chain rule to differentiate 21/3 and the product rule to differentiate my: l' d 2x+y+;L‘(—y+6y2—g :0 (11‘ (1:1: So we have dy ~22: — 3/ all; 2 _:_2 _.: (\$+6y)dz (\$+y):>d\$ z+6y2 Thus the equation of the tangent line to the curve at the point (—2, l) is —2*(—2)—1 3 (g) Since the function is ﬂat) 2 sin‘1(ln 1‘), values of :3 should be _x>0 and ~1§lnx§1 Noting that lnI is an increasing function, we have 6’1 S a; g e. (h) L(s) = 65”” — s :> L’(8) = 6.5”” coss — 1 So the Newton iteration formula for approximating the roots of L is L(Su) esinsn _ S” : 5n S1L+l : 51L esinsn COS 8” __ 1 (i) P(:c) = 2:4 — 33:3 + 12:102 — 5x + 2 => P’(x) = 411:3 — 9x2 + 24\$ — 5. We know the critical points are the roots of P’(x), so compute p"(:I:) =12ac2 — 18:5 + 24 Thus the Newton iteration formula for approximating the critical points of P is P’(xn) 4x: — 92:31 + 24:1:n — 5 wn+1 = L; — ,, — —————— P (Inn) " 1220;"1 — 18:13" + 24 (j) We have x3+10922+3\$+1 x+41 2 —-——~———— => I = x + 10 — ﬁx) z2+4 f” \$2+4 Since lili] — (:1: + 10) = 111311 = 0, the slant asymptote of is I—‘ (X3 1“ 00 y = x + 10 (k) f(1.2) % f(1) + f’(1)(1.2 — 1) = 2 + 012 * \/5‘ z 2 + 0.2 * 2.236 m 2.447 (1) Since m) = f(0) + f’(0):c + = 5 + 32: + (0 < a < 1), f”(6:c) 2 1 4 1 <——=— 2137 162! 8 GT'T‘OT 2 Thus we can determine a value of M = 213. (m) We have f(w) = e” => f(0) =1 f’(2;) : 281' => f’(0) : 2 f”(\$) = 462* => f”(0) = 4 It yields P0 = f(0) : 1 P1: f(0) + f’(0)x = 1 + 2x P2 = “W + f’(0l\$ + f”;me :: 1+ 23; + 23:2 (n) We have f(1:) 2 xln(1 + 2x) => f(0) = 0 f’(:1:) =ln(1+ 2m) + 1 32\$ => f’(0) = 0 // _ ﬂ 2 2 ’l ._ f (\$)_ 1+2337L (1+2.1:)2 2” (Ol—4 III a: = I” _ f () (H202 (1+2x)3=>f (0)—~12 f////( ): + I/II (1+2\$)3 (1+2\$)4 :>f (0):“ It yields P0 = NO) 2 P1: (0) + f’(0)r : I ll 0 P2 =f(0)+f(0):c+ f; )x2—2112 I I, 0 III 133— no) + f (0) + f; )x2 + f3?le = 2x2 _ 213 I ll 0 III 0 [III P4 ‘f(0)+f(0) + f2(l )\$2+ fBS )I3+ f 41(0):):4 = 2332—2x3+ :14 2. (a) We have 2 — _ _1 ~~% lim ﬂ: lim 2 ﬂ: lim 21 = -3 :r-—54+ |x —4[ z—+4+ I ~4 1H4+ 1 4 2— _ _1 -% lim ﬁzlimQ ﬁzlim 2x :1 w—“I— [x — 4| x-4— 4 — a: :c—’4+ —l 4 Since the left— and right—hand limits are not equal, the limit does not exist. (b) We have f(x)=erl_2~2z=>f(0)=~1, f(1)= 1 —2z—0.6 Note that e“‘2 = 2 and 0 < ln2 < 1, so lim : —oo, lim ﬁx) 2 +00 IAIIiQ— I—'lll2+. Since f(a:) is a continuous function on the interval (ln2, 1) and f(1) < 0, there exists 3:“ E (ln2, 1) such that f(x*) = 0 which means has a root in the interval [0,1]. . According to Newton’s law of cooling and with A = 25, we have du (17 : —A~(u — A) = —lc(u . 25) We separate the variables and integrate / du :/kdt 25—11 —ln(25 ~ u) = kt+ C 3 25 _ u : e—(kt+C) : Be—kt where B = e‘C. Next, u(0) = 65 implies that B = 25 — 65 = —40, so u(t) = 25 + 40646t We also know that u = 37 when t : 15. Substituting these values then yields So we ﬁnally solve the equation u(t) = 50 = 25 + 40500803t t__ 1 150—25 _ 0.0803n 40 Thus it was at 50°C after 5.8531 minutes in the room. 2 5.8531 1.2 I2+1 : O (ii):c—intercept: y : = x = 0 => 2: = 0 12+] (iii) We have (i)y-intercept: :L' = 0 => y = f(2:) = f’(a:) = It yields f/(as) : 0 only if IL‘ 2 0 f'(:r) > 0 when (E > 0 f’(\$) < 0 when 1" < 0 Since f(m) 2 0 for all real numbers, the critical point a: = 0 is a minimum. increasing function when 1: > 0 and f(IL‘) is decreasing function when 1: < 0. (iv) We have ,,, _ _ 2—6x2 fill—W f"(:r)=0:>6x2=2:>1:::t13-§ f"(;r)>0 when —%§<x<€ x/3 ﬂ f"(:r) < 0 when .7; < —?or whenz > ~3— So the inflection points are 1‘ 2 ﬁg. is concave upward when —§ < 1‘ < is concave downward when x < ——3‘6 and when 1' > fix) is V75 and (v) Since ﬂit) = l' = 1 — l' = I2 + 1 f(x) has horizontal asymtote y = 1. 1 5. We have 3/ = 100135“, so that y' = 1006—,” — lOOkme‘kx = 0 => k2: = 1 Since at most. 12000 cars can travel the road in one hour and :1: = is maxinia, we have 1006—1 1 = 12000 [C = ‘ z . k => 1206 0 0031 ght circular cone of height Then we can set up an equation according to ghﬁt) = 10*10‘3*t—0.01*7r*h2(t) *t When h = 0.2m we have ’ x3 1 Z 7r*0.23 2 1—0an” M _ 10_2~0.01>i<7r>l<0.22 A} Next using the chain rule to differentiate 113(t) and the product rule to differentiate h2(t)t, we have 1m and top radius 1m, we have h = 7“. conditions dh(t) (It what)? = 0.01 — 0.01 * 7r * h2(t) — 0.02 * 7r * h(t) * t * So we can get the rate for water level rising when h = 0.2m dh(t) 0.01— 001* 7r * 0.04 0.01— 001* 7T * 0.04 dt 7r*0.04+0.02*7r*0.2*t z7r“104+0.02*7r*0.2*1 (6) Ifhz 1m, we have a: 0.0633 7r t: ———3 10-2 — 001* 7r dh(t) (ll <0 It is impossible. Actually, when have the highest water level = 0 that means the water level stops rising, so we. ~ 0.01 H 001* 7r h2(t) => h(t) 2 Therefore we know this tank does not overflow. Q1 ...
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## This note was uploaded on 01/30/2011 for the course MATH 100 taught by Professor Lamb during the Fall '08 term at UBC.

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Math 100 Dec 03 Answers - Solutions for MATH 100/180...

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