Math 100 Sample Exam Answers

Math 100 Sample Exam Answers - mama Math 100 Sample Exam...

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Unformatted text preview: mama Math 100 Sample Exam Solutions 7 1. SHORT ANSWER QUESTIONS (a) A . $2+4mz5 . (x+5)(z—1)_1+5_6 mz2+5x46‘l§'i(z+6)(z—1) ‘ 1+6_ 7 (b) 1 1— 1+1 lim(x—\/:c2+:v)=lim [ — x2(1+l>]=limz<1— 1+—)=lim————z z—NDO z—mo a: 2—H» a: :c—ioo l/cc We“ a “3;”; .‘mnaie‘rwowaewww mwu " | Now we make a substitution y = l/z. So a: —> 00 ¢ y —> 0. We have: lim ———l — ' + y —) 9 u—io y 0 Thus we can use L’Hépital’s rule: _ 1 nm 14— VI” = lim Wu =_ 1 =1 y—w y y—w 1 2 1 + 0 2 So 1im(z — V22 +2) = —0.5. 2—)00 (c) At 3 =0, the limit takes the form 0/0 so we may apply L’Hépital’s rule: w- W -‘.n,_wmw~mm_ a . l—cosz . sina: 0 11m .— = hm +—— —> — z—ro msma: z—>0 s1nx+$cosz 0 So we apply L’Hépital again: _ sin a: _ cos a: 1 hm ———— = 11m 1 :—>0 sinac + zcosa: z—vo cosz — zsinx + cosa: _ 1 —- 0+ 1 _ 2 (d) The function is differentiable so it must be continuous at :1: = 1, and its derivative must also be continuous at a: = 1. So we have: A(1)2=13+B =>A=1+B 2A(1) = 3(1)2 => 2A = 3 Therefore A = 3/2 and B = A — 1 = 1/2. (e) 9(01) = 933M932) So we use the product and chain rules: g'(:1:) = 3m2h(z2) + msh’(a:2)(2m) = 39:2h(a:2) + 2z4h’(m2) At x=2, we get: g'(2) = 3(22)h(22) + 2(24)h’(22) = 12h(4) + 32h’(4) = 12(2) + 32(—2) = 24 — 64 = —-40 (f) We use the product rule to get f’ and f”: f’(m) = zcosz + sinx f”(9:) = x(— sinzv) + cosa: + cosx = 2cosa: — wsinm ) If f’ = —2 f , then f (9:) is an exponential function and we have: (s f (z) = Ae‘“ Atz=1,f=2sowehave: 2 = As"2 A = 262 f(0) = Ae° = A = 2e2 (11) Differentiate implicitly to get: ’69“1 + yea”-1 + e” + :ceyy’ = 0 y Plugin 2: = 1,1; :0: y’e°+Oe°+e°+le°y’=y’+1+y’=2y’+1=0 1 I— —— y ’ 2 (i) To find all inflection points, set y" = 0: y=zv+sinx y’ =1+cosx y" =-—sin:v er a: is a multiple of 7r. So the coordinates are (n1r,mr) where n is an integer. The sine function is 0 whenev (j) The next approximation is the a: intercept of the tangent line at the initial guess. Thus we set 3/ = 0 in the tangent ‘ line equation and get: 0 = 10x + 19 10:1: = —19 as = -1.9 (k) The speed of the object is :c’ (t) = 3t2 — 4. The rate of change of the speed is a)" (t) = 6t. At t = 1, the rate of change of speed is 6(1) = 6 > 0 and so it is INCREASING at 6 metres per second per second. (1) Using the quotient rule, we have: 2 .. _ f,(x) : 2:1:(3: + 2:12)$2 im:m)22)(2x + 2) f’(2)—WW_32‘12_ 5 - (22+2(2))2 - 64 —1-6. =width, l=length. Then the area A = wh = 40. (m) Let 10 Since the width and length are changing with respect to time, they are functions of time t. So we have: 40 l h t = —— ( ) w“) —40 —40w' I __ __ I __ h (t) - W2 (w (m — m? At w = 10, w' = +2. Therefore: —40(2) ——80 h, t = ——-— : ——-— :: —— . ( ) 102 100 0 8 Thus the height is decreasing at a rate of 0.8 cm/s. (11) Using the chain rule, we get: ftp) __ < 1 > (221272 + 1) — 2:1:(232 — 1)) 2 1 — ($371)? (“’2 + W 1 + x2 First let’s simplify the numerator of the first term: 223(z2 + 1) - 221(172 -— 1): 22:3 + 2:1: -— 29:3 + 2x = 4:1: Next we simplify the denominator of the first term: ( 1_(22—1)?)($2+1)2=(z2+1)2 W 2:2 + 1 (a:2 + 1)2 _ (2:2+1)2\/z4+2:52+1—:r4+2z2— 1 ‘ (22 + 1) = (x2 + mm = (9:2 + 1X22“) Thus our derivative becomes: Therefore, f (as) = constant. We put in a: = 1 and get f(1) = sin"1 0 — 2tan"1 1 = —27r/4 = -7r/2. Thus we have f(x) = -—7r/2. (0) Since we will be multiplying the Taylor series of cosine by an 23" term, we only need to expand cosine to 9:6. The Taylor Series for cosx is: So the series for cos(:z:/3) is: 2. (a) The function f is increasing when f’ > 0 and decreasing when f’ < 0. We have f’(:r) = Ke“'2/8(1 — (2:2/4». We know that K > 0 so Ke‘zz/8 > 0 for all 1:. Thus the sign of f’(:c) depends 2 (17 only on the sign of 1 — I. 2 2 :1: 2: 1—— 0 1 — 4 2 4> <=> >4<2> >zc ¢>2>zor —2<:c So f(z) is INCREASING on —2 < a: < 2. For decreasing, we just reverse all the inequalities and get -—2 > a: and a; > 2. For critical points, we change all the inequalities to equal signs and get a: = :l:2. Since f is decreasing on the left of a: = —2 and increasing on the right of a; = —2, the point a; = —2 is a local minimum. Similarly f is increasing on the left of a: = 2 and decreasing on the right of a: = 2 so this point is a local maximum. Since the function approaches 2 as a: approaches 00, f (—2) = 0 < 2 is an absolute minimum. Similarly f (2) = 4 > 2 is an absolute maximum. (b) For concavity we must find the second derivative f”(2:). Using the product rule we get: ._ x 2 $2 2 _ W) = Keg—Xe” “5 (1 — I) + KW “(5-) f"(2) > 0, and is concave DOWN if f”(:v) < 0. 3 ... As in part a, the sign of f"($) depends only on the sign of 17‘ 1612z' The function is concave UP if 2:3 —— 122: 16 <=>a:(a:2-—12)>0 <=> bothx>0ANDz2—12>00rbothx<0AND:c2—12<0 <=>m>\/fior0>:c>-\/T2_ >0<=>x3—12:z:>0 Thus the function is concave UP when 0 > a: > —\/1_2 or x > . To get concave DOWN, reverse the inequalities. So f (9:) is concave down on —-\/—1-2 > a: and 0 < a: < \fl-2,. The inflection points are when f”(a:) = 0 and that occurs at :1: = mix/1‘2. ‘ e have the following picture: information from parts a and b, and the asymptotes f —+ 2 as a: —-> :too, w U6- .(c) Given the 3. Let a: be the length of a side of a corner that is removed. Since we are removing a corner from each side, the length of the box will be L = 4— 22:, and the width will be W = 2 - 22;. The height of the box will be equal to 1. To maximize volume we take the derivative and set it to zero: V = Lch = (4 - 21:)(2 — 22:)(13) = (8 — 8x — 42: + 49:2)(93) = 8m — 121:2 + 42:3 f} V’(:c)=8—24w+12x2=1222—24a:+8=0 y 95 H’ g Dividing through by 4, we get: . 3x2—6z+2=0 be, x=6d=‘/36-4(3)(2)=1i_\/_T—3: ‘ w:2»gx (2X3) 6 X < y 0423,1577 Now we check second derivativeTo/determine whether the critical Thus we have two critical points a; n points are max or min: f7 V"($)=24:v—24=24(z——1) / X;’_ 3, We see that we get a max if x < 1. So the max volume occurs when a: = 1 — J35 m 0.423. 3 So the maximum volume is 42:3 — 12:1:2 + 8a: N 1.54 cubic metres. Milena-E 4. The speed of the ant’s shadow is 2—3:, where y(t) is the coordinate of the ant’s shadow on the y axis. Because light travels in a straight line, the position of the shadow on the y-axis will be the y — intercept of the line joining the ant and the light source. Given an arbitrary point along the ant’s path, the slope of the line will be: 1/0 ~ 0 _ x0 — 3 (gab) Mm? Note that $0 is the w-coordinate of the position of the ant and is a function of time since it is moving. The equation of the line is y — 0 _ 29:0 2 — 3 _ 2:0 — 3 $ — 2% (:1: 3) y — 2:0 — 3 Now, the y-intercept occurs at a: = 0, so we have: 21230 —6£Eo = —3 : g 11:0 — 3( ) 2:0 — 3 The rate of change, then is: ,(t) = —6$6(.'B0 — — (—6220X116) y (930 — 3)2 At 2:0 = 1,2:6 = —1, we have: 3,, _ —6(—1)(1~ 3) —(—6>(1)(—1) —12 — 6 _ \ (1 — 3)2 4 So the ant is moving downwards at 4.5 units per minute. _—9 ‘2 5. According to Newton’s Law of Cooling, if T(t)=object temp. at time t (t in minutes), Ta=ambient temp., T0=initial temp., then we have: for some constant k. We are given: Solving for k, we have: 50 = 45 + (400 — 45)e-'=/2 = 45 + 3555“2 i = i = —k/2 355 71 7l=ek/2 k: 21n71 Now we plug this in and solve for T at t = 5. T(5) = 45 + 3555mm)5 = 43 + 355(e’n71r10 = 45 + 355(71-10) = 45 + 1.09 x 10—16 So T is VERY close to 45. End of Solutions. ...
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Math 100 Sample Exam Answers - mama Math 100 Sample Exam...

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