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 . $2+4mz5 . (x+5)(zβ€”1)_1+5_6 mz2+5x46β€˜lΒ§'i(z+6)(zβ€”1) β€˜ 1+6_ 7 (b) 1 1β€” 1+1 lim(xβ€”\/:c2+:v)=lim [ β€” x2(1+l>]=limz<1β€” 1+β€”)=limβ€”β€”β€”β€”z zβ€”NDO zβ€”mo a: 2β€”HΒ» a: :cβ€”ioo l/cc Weβ€œ a β€œ3;”; .β€˜mnaieβ€˜rwowaewww mwu " | Now we make a substitution y = l/z. So a: β€”> 00 Β’ y β€”> 0. We have: lim β€”β€”β€”l β€” ' + y β€”) 9 uβ€”io y 0 Thus we can use L’HΓ©pital’s rule: _ 1 nm 14β€” VI” = lim Wu =_ 1 =1 yβ€”w y yβ€”w 1 2 1 + 0 2 So 1im(z β€” V22 +2) = β€”0.5. 2β€”)00 (c) At 3 =0, the limit takes the form 0/0 so we may apply L’HΓ©pital’s rule: w- W -β€˜.n,_wmw~mm_ a . lβ€”cosz . sina: 0 11m .β€” = hm +β€”β€” β€”> β€” zβ€”ro msma: zβ€”>0 s1nx+$cosz 0 So we apply L’HΓ©pital again: _ sin a: _ cos a: 1 hm β€”β€”β€”β€” = 11m 1 :β€”>0 sinac + zcosa: zβ€”vo cosz β€” zsinx + cosa: _ 1 β€”- 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 β€” 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(β€”2) = 24 β€” 64 = β€”-40 (f) We use the product rule to get f’ and f”: f’(m) = zcosz + sinx f”(9:) = x(β€” sinzv) + cosa: + cosx = 2cosa: β€” wsinm ) If f’ = β€”2 f , then f (9:) is an exponential function and we have: (s f (z) = Aeβ€˜β€œ Atz=1,f=2sowehave: 2 = As"2 A = 262 f(0) = AeΒ° = A = 2e2 (11) Differentiate implicitly to get: ’69β€œ1 + yea”-1 + e” + :ceyy’ = 0 y Plugin 2: = 1,1; :0: y’eΒ°+OeΒ°+eΒ°+leΒ°y’=y’+1+y’=2y’+1=0 1 Iβ€” β€”β€” y ’ 2 (i) To find all inflection points, set y" = 0: y=zv+sinx y’ =1+cosx y" =-β€”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 β€˜ line equation and get: 0 = 10x + 19 10:1: = β€”19 as = -1.9 (k) The speed of the object is :c’ (t) = 3t2 β€” 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)β€”WW_32β€˜12_ 5 - (22+2(2))2 - 64 β€”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 = β€”β€” ( ) wβ€œ) β€”40 β€”40w' I __ __ I __ h (t) - W2 (w (m β€” m? At w = 10, w' = +2. Therefore: β€”40(2) β€”β€”80 h, t = β€”β€”-β€” : β€”β€”-β€” :: β€”β€” . ( ) 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) β€” 2:1:(232 β€” 1)) 2 1 β€” ($371)? (β€œβ€™2 + W 1 + x2 First let’s simplify the numerator of the first term: 223(z2 + 1) - 221(172 -β€” 1): 22:3 + 2:1: -β€” 29:3 + 2x = 4:1: Next we simplify the denominator of the first term: ( 1_(22β€”1)?)($2+1)2=(z2+1)2 W 2:2 + 1 (a:2 + 1)2 _ (2:2+1)2\/z4+2:52+1β€”:r4+2z2β€” 1 β€˜ (22 + 1) = (x2 + mm = (9:2 + 1X22β€œ) Thus our derivative becomes: Therefore, f (as) = constant. We put in a: = 1 and get f(1) = sin"1 0 β€” 2tan"1 1 = β€”27r/4 = -7r/2. Thus we have f(x) = -β€”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) = Keβ€œ'2/8(1 β€” (2:2/4Β». We know that K > 0 so Keβ€˜zz/8 > 0 for all 1:. Thus the sign of f’(:c) depends 2 (17 only on the sign of 1 β€” I. 2 2 :1: 2: 1β€”β€” 0 1 β€” 4 2 4> <=> >4<2> >zc Β’>2>zor β€”2<:c So f(z) is INCREASING on β€”2 < a: < 2. For decreasing, we just reverse all the inequalities and get -β€”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: = β€”2 and increasing on the right of a; = β€”2, the point a; = β€”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 (β€”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 $2 2 _ W) = Kegβ€”Xe” β€œ5 (1 β€” I) + KW β€œ(5-) f"(2) > 0, and is concave DOWN if f”(:v) < 0. 3 ... As in part a, the sign of f"($) depends only on the sign of 17β€˜ 1612z' The function is concave UP if 2:3 β€”β€” 122: 16 <=>a:(a:2-β€”12)>0 <=> bothx>0ANDz2β€”12>00rbothx<0AND:c2β€”12<0 <=>m>\/fior0>:c>-\/T2_ >0<=>x3β€”12:z:>0 Thus the function is concave UP when 0 > a: > β€”\/1_2 or x > . To get concave DOWN, reverse the inequalities. So f (9:) is concave down on β€”-\/β€”1-2 > a: and 0 < a: < \fl-2,. The inflection points are when f”(a:) = 0 and that occurs at :1: = mix/1β€˜2. β€˜ e have the following picture: information from parts a and b, and the asymptotes f β€”+ 2 as a: β€”-> :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 = 4β€” 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 β€” 22:)(13) = (8 β€” 8x β€” 42: + 49:2)(93) = 8m β€” 121:2 + 42:3 f} V’(:c)=8β€”24w+12x2=1222β€”24a:+8=0 y 95 H’ g Dividing through by 4, we get: . 3x2β€”6z+2=0 be, x=6d=β€˜/36-4(3)(2)=1i_\/_Tβ€”3: β€˜ w:2Β»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:vβ€”24=24(zβ€”β€”1) / X;’_ 3, We see that we get a max if x < 1. So the max volume occurs when a: = 1 β€” J35 m 0.423. 3 So the maximum volume is 42:3 β€” 12:1:2 + 8a: N 1.54 cubic metres. Milena-E 4. The speed of the ant’s shadow is 2β€”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 β€” 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 β€” 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 β€” 0 _ 29:0 2 β€” 3 _ 2:0 β€” 3 $ β€” 2% (:1: 3) y β€” 2:0 β€” 3 Now, the y-intercept occurs at a: = 0, so we have: 21230 β€”6Β£Eo = β€”3 : g 11:0 β€” 3( ) 2:0 β€” 3 The rate of change, then is: ,(t) = β€”6$6(.'B0 β€” β€” (β€”6220X116) y (930 β€” 3)2 At 2:0 = 1,2:6 = β€”1, we have: 3,, _ β€”6(β€”1)(1~ 3) β€”(β€”6>(1)(β€”1) β€”12 β€” 6 _ \ (1 β€” 3)2 4 So the ant is moving downwards at 4.5 units per minute. _β€”9 β€˜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 β€” 45)e-'=/2 = 45 + 3555β€œ2 i = i = β€”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 10β€”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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