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Unformatted text preview: Math 100 December 2000 Solutions x3—1_131 0 la. lim — x—H xl l—i 0
, ‘34 _ (x—Ixx’+x+)
' xl _ x—l um2 +x+1=(l)2 +(1)+1=3
x—51 1b. =x2+x+l. = — Thus we must factor either using the factoring rules or long division. lim («Ix2 +5x—Vx2 —x)= lim (\lx2 +5x—\/x2 — 1:)(x2 +5x+.x2 —x)
"H” "W” (1sz +5x+slx2 —x) x2 +5x— (x2 ~x) :2 lim = iim———————  lim————————— “’"(sz +5x+1lx2 —x) “’“(s/x: +5x6+xlx2 —x) G’Hm‘lxz (1+5)6<1~‘/x2 (1—L . 6x .
= hm ———————= lim "**‘°x( l+§+ l—ﬁ) "’“°( 1+§)6+‘/1_§:—0'__(J1+ )6+Ji0 =1—+1 1c. lg{xsin2(—)} We know I s sine? S i , therefore 0 s sin2 49 51
x x ' =3 1 Since we know that sin 219 cannot be bigger than 1, and IS positive, then 0 < sinz < 1. Then lim(x 0)< lim{xsin2 (—)}< lim(x 1). And we know lim(x l): 0. x—+0 Then by squeeze theorem lin(1){x sin2(—)} = 0 .
1—? x 2a.Usethequotientrule:1(3x+4)=W= 2 2
x+2 (J:+2)2 (x+2) dx _ 2
2!). Use the product rule: {ii—[(1 — x2 ) arcsin(x)] = (—2x) arcsin(x) + (1 x )
l— x2 2c. Use chain rule: %(e*z’3x) = e""‘(2x  3)dx 3. f (x) = x2g(sin x) f(x) = 2xg(sin x) + x2 g' (sin x) cos x
f' (g) = 2(§)g(sin12’—) Hg)? g'(sin gym; =nig(1)1+(§)2[g'(1)1(0) =7r
4. f(x)=xln(l+x2) .‘.f(x)=ln(1+x2)+ 2x2 2 1+x
2 _ 3
.‘.f"(x)= 2x2+4x(1+x 2) 24x : 2x + 4x
l+x (1+x) 1+1:2 (1+1?)2
..f"(2)=~+—3—=3§=1.12 525 25 x = (——2x arcsin(x) + ill  x 2 )dx 5. For x < 0,f'(x) = Acos(x) — Bsin(x).
To be differentiable at 0, it must be continuous at O and contain no kinks. This means that for the split function x2 +1 = Asin(x) + Bcos(x) , and 7202 +1): i(Asina) + Bcos(x)) at 0. Thus 02 +1 = Asin(0) + Bcos(0) = A(0)+B(l), B =1
And 2): = A cos(x) — B sin(x) at 0, so 2(0) = Acos(0) — (l)sin(0) = A(l) — 1(0) , A = 0
So, A=0, B=l makes the function differentiable at 0. 6. Use implicit differentiation on x2 + xy + 2y3 = 4 —2(—2>—(1) = 3 ' —2,1 =
y( ) (~2)+6(1)2 4 7. For a linear approximation, we create a secant line. y =mx+b wherewearegiventheslope=m= l atthe(2,4) 4 = (—l)(2) + b , therefore, b = 6. Now we have the equation: y = x + 6 to use as our linear approximation
So, to approximateﬂl), y = —(1)+ 6 = S.j(1)= 5. 8. y=(x2 +4)"',wherewe are given dx= 3 _ 2xdx
y'= —(x2 +4) 2(2x)oix = W . Now, let’s ﬁndy’ atx = 2.
y'(2)= 2(2X3) ___12___12_=_.3_=—0.1875 (22 +4)2 _ —(s)2 ‘ 64 1_6
9. To start this question, we are told that 4 grams decays to 3 grams in 10 seconds.
We start with the equation Ce” .  Then 4ek(10) =3 .ek(10)____3_'
9 9  4 1n(e"“°’) = ln(%) . So, k(10) = 1n(i—)t 3
111(3)
Fromthatwecanﬁnd k=——10—.
Now to ﬁnd how long it takes to go from 4 grams to 1 gram, we set:
w 3 / 4 ) r 111(3/4)‘ l
4 e ‘0 = l , .'.ln(e ‘0 )=ln(Z) 111(3/4)
10 ln(l / 4)
ln(3 / 4) ) E 48.2 seconds. t=ln(%),sot:10( 10. Newton’s Method: xl =xO + y, —yo .
f(xo) We will use y = 3J— :— to approximate y = f (x) . We can do this because they share a common tangent line at (l, 0.5)
And because we are using x = l as our initial guess. Let’s use Newton’s Method to ﬁnd xi: Xi = 1 +2 T.‘ 05
f0)
1 l
3 '5 3 _, 3
' :— a '1 :—l 2 :—
f(x) 2(x) f() 2() 2
—0.5 —1 2
x; =1+——=1+—=‘
1.5 3 3 11a. Write the Taylor series for ln(l+xz). Let us make a substitution, u = x2. Taylor series nearx=0 is: f(x)=f(0)+fl1('0)x+f"2('0) x2 +f"3:'(0)+x3 +... _ _ u (1)142 (2)143 (6)144 (24)u5
Thenf(u)—ln(l+u)—0+—ﬁ— 2! + 3! _T+T+m =u—— —————+—— ———.—+
2 3 4 5 6 7 s
4 6 8 10 l2 l4 l6
Now, replaceuwithxzz ln(l+x2)=x2 —£—+L~i—+L_L+L_L+m
2 3 4 5 6 7 8
3 5 7
11b. sinu=u—u——+5——f——+...
3! 5! 7!
6 10 14
sinx2 =x2 —x——+i£—
3! 5! 7!
2 _ 2
11¢. Find lim__—————————1n(1+x )4 5m“ )
x—eo x
4 6 6 8 [0 10 12 l4 l4
[JR—(5:2)} +" +(x )—" +x —(" )" +x +(" )+...]
=1,m_____2_3__}'___4__§__5!__6_.7___7'___
x—>0 x4
2 2 4 6 6 8 IO 10
=iim[l+x—+(x—)x—+i‘——(f—)"—+x——+(i—)+...] x40 2 3 3! 4 5 5! 6 7 7! =~l+0+0—0+0+...
2 = —0.5 12. The war went 6 km in 1/ 12 of an hour. This suggests and average speed of 72 km/h. But the speed limit
is 60 km/h. By the mean value theorem, the car must have at least reached 72 km/h at some time during the
travel. Thus, he must have sped. dv kv+g 133. —=— —kv, V:
dt g k
V = v+§. Differentiating this gives ﬂ =§1+0.
k d! dt
The previous line also tells us, v=V——g
dv g dV
.‘.———=— —kV—— =— kV+ =~kV=——— fromabove
d, g ( k) g g d1 ( )
V=ce""
Then we know v = V —% = c6,”r —%, where we‘re given v0 = v(0) = ce’klo) ~§
Solving forcgives us: c=v0 +%.
g —kI g
S , = v +—e ——
0 V ( o k) k
g g g
l3b.Ast—)oo, v: v +— 0 —=——
(o k)() k k
14a. Lethdenote the height of the wateratthe deep end.
At hsz, V=8'1°h'h =40h2
At h22, V=160+8.20(h~2)=160h—160
ﬂ=—1m3/min.Wemustﬁndﬂ.
dt dt
Weknow ﬂﬂgﬂ
dV d1
So, for h = 2.5, then V =160hl60
g1=160,..,_c&=;
dh dV 160
iiiﬂ =(—1)(—l—) = ——l m/min. Surface of water is falling at — —l—m/min
dt dV I60 I60 160
14b.h= l,then V=40h2
011.4%, 3:12;
dh dV 80h
dV dh ——— = (—1)(—!—) = — ——1— m/min. Surface of water is falling at — —1— m/min
dt dV 80(1) 80 80 15a. To determine intervals of increase, decrease, and maximum and minimums, we should ﬁnd the critical
points (f’(x) = 0). f' (x) = 0 only when (1 —4x2 ) = 0 or e'2‘1 = o. 211 . 1—4;;2 = For e“
l=4x2 e‘” :0
:+00
l=x2 X _.
4
il:x Butx=ioois notalocalpoint,
2 so we won't check it. f (— %) = —2 and f (%) = 2 . Since we only have two critical points, it’s pretty obvious that x =—;~ is a local maximum, and x = —% ’s a local minimum. And simply using the fact that [(0) = 0, the interval (2,2) must be increasing. That information can also be determined by plugging in a number between 2 and 2
into the f"(x) function. The intervals (oo,2) and (2, 00) are decreasing. 15b. For this question, we must determine the second derivative. Let u = —2x2 , then du = —4xdx _
f'(u) = K(l +2u)e" and we must ﬁnd where f"(x) = 0. f“(u) = K(2e" +2ue" )du = 2Ke" (l + u)du f' '(x) = 2K6”! (1 + (—2x2 ))(—4xdx)
= —3Ke‘2"[x(1~2x2)]dx 1 We only need the check the stuff in the square brackets because e'z" is always positive, and only
equal to 0 at inﬁnity. So it can, for the most part, be regarded as a constant. x(l—2x2)=0
Lx=0 ﬁ—
2 So the points of inﬂection are — J; ,0, J; . To check for concavity, we can check the points between the inﬂection points and end points. («xx—g) and (mg) are concave down. (—J;,0) and (g, 00) are concave up. 16. Once this question is understood, one will realize that a = % of circle removed. The original circle had
a circumference of 271R , whereas the reduced circle has a circumference of 27rR(l —a) . Let u = l—a . This substitution isn’t necessary, but makes the equation look simpler. We will need to maximize the volume of the cone: é—nrz h . We’re given r = (l —a)R = Ru.
hz‘le—RZGaf =‘/R2 —R2u2 =R l—u2 So we need to maximize %ﬂ'(Ru)2 (Rx/l — u2 ) , where R is a constant. f(u)=% R3ulel—u2 Let K:1 R3.
3 l
Then f(u)=Ku2(1—u2)5
l l
f'(u) = K[2u(1—u2)5 +u2(%)(l—u2)_i(—2u)]du Now, we have to set f (u) = 0 to ﬁnd maximum points.
1 x 1 K[2u(l — u2 )5 M2 (%)(r — u2 )7 (—2u)]du = 0 Now, multiply both sides by (1 — u2 )5
K[2u(1 — u2 )+u2 (%)(—2u)] = 0 =K[2u(l—u2)+(—u)(u2)] =1<[2u—2u3 —u3]=K(2u—3u3) We only nwd to check when 2143u3 =0.
2u—3u3 =0 2u=3u3 u=0
2=3u2 iJz=u
3
u=—‘/§,0,JZ. Wehad set u=l—a,then a=l+J§,l,l—JZ.
3 3 3 3
2 a 2 1+ J; is impossible. (I = 1 means that the whole circle is removed, which will give a minimum. So the answer must be a = l—Jg : 0.1835 that will give the maximum. ...
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 Fall '08
 LAMB
 Math

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