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Unformatted text preview: Math 100 Final Exam December 1997 Solutions Problem 1a:
differentiating using the quotient rule
3 (1 + e3'xi3x2 — x33e3x 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) + ﬁﬁaﬂx — 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) + if(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 + gSyz .: 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—3122_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) + xsin(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) tame"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 3y yek'9 3 := e log(3) := 9k k now we can easily calculate the doubling time tl_ 9log(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 + 2y~—.y _ 2.1._] I ' I dt dt at now we can easnly solve fordth once again usnng Pythagoras to get us  3500
300015) + 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 + 6w2 but remember h = ——2 now to find the minimax we differentiate
3w
12000 .
d—A ‘= 0 9—A : +12w 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(2x +1)xe‘
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 3a2 = 6a2 + b b = —3a so the tangent y=mx+b, fits nicely y ‘= 6ax  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 ‘= .05A + 2000 we can solve by substitution u ‘= .05A + 2000 :—u '= .OSS—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 : 9000060540 — 40000 A 625015.04
Problem 14b: This requires a slightly different differential equation d r .d_
:—A .05~A — w : .05A — 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 —y0e 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.
 Fall '08
 LAMB
 Math

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