# Math 100 Dec 97 - Math 100 Final Exam December 1997 Solutions Problem 1a differentiating using the quotient rule 3(1 e3'xi-3x2 — x3-3e3x

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Unformatted text preview: Math 100 Final Exam December 1997 Solutions Problem 1a: differentiating using the quotient rule 3 (1 + e3'xi-3x2 — x3-3e3x Substituting 1 for x gives us... ,———— (1 + e 3x 2 {I + e > 2 3 ) Problem 1b: we need to use the law of logs to simplify the problem so we get... l 2 i 2 l [(2x—1) (2x+l) log(2x - 1) — log(2x + 1) differentiating, we get substituting 1 for x gives us (2 — ;= :i 3 Problem 2a: The linear approximation method uses Taylor’s polynomial... f(x) .: f-(a) + ﬁ-ﬁaﬂx — a) now we are given values of 3 so f(x) is approximately... {(3) + 3—f(3)-(x — 3) now plugging in our desired value of 2.98 we get... x . Problem 2b: L% the quadratic approximation simply adds the next term to Taylor's polynomial ’2 + 4-(2.98 —V3)m-: 1.92 L—\____ - 2 2 f(x) .2 f-(a) + i-f(a)~(x * a) + game-(x ‘28) since we have the value of the second derivative, once again with a=3... 2 _ . _ . _ 2 # plugging in 2.98 gives us f(a) :2 2 + 4-(2.98 — 3) ~ adding up the factors we get 1.918 f(x) .= 2 + 4-(x — 3) + Problem 3a: the problem requires implicit differentiation to solve properly so thus ﬂ'XB + y-(3x)2 + g-Syz .: 10 solan for dy/dx will is simple dx dx . 2 dy 10 — 3x ‘y dy r 3 2 ,_ 2. —:=—— a-\x +3y ) .- 10— (3x) y dx )3 _ (3y)2 10—3-12-2_10—6 10—6‘_4 13+3.22 ‘ 1+12 1+12‘ 13 now we plug in the x=1 and y=2 and we get Problem 3b: as the function's second derivative is negative at (1,2) this tells us that this point lies at 1 the top of the curve and thus the curve is concave and opens downward ’7’< 1 Problem 4a: this is a tricky question, so we use L'Hospitals rule twice (f'/g') to reduce the question to a determinant form 2 . . 2 —X '2 1 we differentiate again to the correct answer. —— ‘: 0 ex ex Problem 4b: once again we need L'Hospitals rule (f'/g') applied twice to get the correct answer... 1— cos(x) + x-sin(x) __ 2 differentiating we end up with 1 _ cos“) ' 1 Problem 5: we need to now the differential of arcsin d . g I solve for x, with d/dx=2 gives us d—arcs1n(x) I x 2 1 :2 is 2 “7 ._-ﬁ plugging in this we can get the points on the curve it; El and (£4) 2 3 2 3/ Problem 6: 2 d—f(x)15x4 — 20x3 s 13 11— f(x) 60x3 — 60x2 dx dx2 there are two roots here, x=0 and x=1 so there are two inflection points (0,0) and (1,11) Problem 7: here we needed to use the standard growth and decay function Y(t) tam-e"t j—Ym := kY I let Y(0)=1, now we are given that Y(9) is 3, so we can easily get the value of k ,: MO) 9 3-y y-ek'9 3 := e log(3) := 9-k k now we can easily calculate the doubling time tl_ 9-log(2) 1 (3) evaluating the logs we end up with approximately 5.678 hours 0g 2 e‘“ log(2) := kt Problem 8: this problem can be looked at as being a triangle, x=300, y=400, and thus I = 500 by Pythagoras we are given 37 y I: 20 3x ‘: ’ 15 now since i2 : x2 + y2 differentiating both sides l t d d w d 2~x-—x + 2-y~—.y _ 2.1._] I ' I dt dt at now we can easnly solve fordth once again usnng Pythagoras to get us | 3500 300-015) + 40020 50011 d. _ d_l 7 d1 dl 500 dt Problem 9 differentiating We get using the product rule 1R -: 1Q? + d_p.Q dt dt dt now we know dP=.08 and d0 is -.02 so is we divide both sides by P0 (R) and plug in the given values 3R z—R t t ‘ L z t + _ g: 202 + 0.08 so then then the answer IS .06 R PQ PQ R Problem 10a: 1 Newton's method is v_ f<xn> using x1 as 7, f(7)=1, f'(7)=3 we get x2 ‘2 7 — — xn+ 1 _ x11 _ 3 d—w dx Problem 10b: critical points are roots to the derivative, i.e. f'(x) =0, so let g(x) = f'(x) so using Newton again Xn+l : xn ‘ —<d plugging in values we get x2 '= 7 —34 x2 ‘= é ‘ Problem 11: given that the volume is fixed at 4500, V=lwh so here V :iw11 V'=3wzh we need to compute the area so we add up the area of the 6 sides and thus we get 4500 . . . . . A ‘= 8~w~h + 6-w2 but remember h -= ——2 now to find the minimax we differentiate 3-w 12000 . d—A ‘= 0 9—A -: +12-w w '= 10 now knowrng w, we can solve for h h ‘: 15 dw dw WZ so the dimensions of the box is 10x30x15 ’7 ~ 24000 d A‘: 3 +12 r) dw~ w this is positive for all w>0 so the curve is concave upwards and thus we have a minimum Problem 12: solving for x=0 and y=0 will give us two important points if x=0 then y=1, while if y=0 then x=—1/2 d x2 x2 if(x) : Zve' — 2v(2-x +1)-x-e‘ dx solving for zero we get 1/2, and -1, when x=1/2, y=1.558, and when x=-1, y=-.0379 now we can find asymptotes, first noting the function is continuos for all x however computing the limits in both direction both are zero (exponentials overpower polynomials) so we see they y=0 is a horizontal asymptote Now lets look at the plot Problem 13: 2 d Ay = 6x dx the standard equation for a line is y=mx+b in slope intercept form, now if x=a then m=6a, the slope and so y=6ax+b but since (a,3x2) lies on the graph y:= 3x 2 3-a2 = 6-a2 + b b = —3-a so the tangent y=mx+b, fits nicely y ‘= 6-a-x - 3.212 I now to pass through (2,9) we simply plug in the x and y values, and we get two points: a? and a=3 so there are two points so the curve that are tangent to the parabola (1,3) and (3,27). 2 730 aa on P 1: t “‘1' _ Problem 14a: (1A ‘= .05-A + 2000 we can solve by substitution u ‘= .05-A + 2000 :—u '= .OS-S—A dt 1 t d—u ': 05‘u u '= u -e'05‘I substituting back we get d1 ‘ 0 now our original substitution comes back, and we know A0 is 50,000 so solving for A .05.: 4500. A 2000 . . . . . A '= —~e—— t=40 as that IS how long he has until age 65 so plugging In t and solvmg .05 A : 90000-60540 — 40000 A 625015.04 Problem 14b: This requires a slightly different differential equation d r- .d_ :—A .05~A — w -: .05-A — w (Tyy " '05 (“A y: yo.e'°5" t we know A(0) trom above, and we knowt is 20, so therefore y0 + w : 31250.75 yoe + w 0 —y0-e w yO l w 49437.98 6 ...
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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 The University of British Columbia.

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Math 100 Dec 97 - Math 100 Final Exam December 1997 Solutions Problem 1a differentiating using the quotient rule 3(1 e3'xi-3x2 — x3-3e3x

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