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Unformatted text preview: MATH 313%} .131] {lewmﬁer 331m
Sniutiuns
{311335333113 3. £54 13133333333 marks 333%}
a) The limit has an indeterminate form so we can use L’Hospital’s rule to evaluate it as follows
32333333 “3n! 71}me ﬁmW. 3:: 13:13 3:; _ L
333233: +343 313231"+121“}+E 5 b) The ﬁrst step is to calculate the derivative
3313:} = 2.3 ~14 (1 . 1)
and then use Eq 1.1 to ﬁnd that at (1,2)
NEE”211:5 «rm'33 (1.2) Now we a slope (Eq 1. 2) and a point so we can detemine the tangent line to be
33. 11133321: :3 21 31:11+213L3f1 4 3:: __1 3323+4 c) Use the product rule to get
1 1 3: 111—3 15113 E31:11—32:11:E—31;}_§+(——2.:3)5£13‘13:14—321~23731.13"'].3 (1) Use the quotient rule to get £1313: —23+..1 3 _ (2.3: — 29(1)
1132 — 23+ 21
2 (.31)
1.32  3.3 + 312
e) Use the chain rule and the given initial conditions to get the required derivative as follows 11 1+(f(x))21=é11+(f(X))21_%(2f(x)f'(x)) dx 1 =é11+(5)21'5 12(5)(%)1 5
31% 1) We want f (0) & f '(0) to be deﬁned and continuous. For f (O) ,
lim f(x x)=A — lim_f(x)= x—>0* x—>0 soA=1. For f’(0),
lim f’ (x):— =—1= 1im_f’(x)=—B x>0 x—>0 soB=1. h) j) k) 1) 2 3 f(x) = sin ax <:> f’(x) = acos ax <:> f"(x) = —a sin ax <:> f”'(x) = —a cos ax Find the derivative by implicit differentiation as follows dy
d ( 1 ( ) 2 2 ) 1+ 3 dy
—tan_ x+y +x —y =0 <:>———+2x—2y~—=0 (1.3)
dx 1 + (x + y)2 dx
Instead of getting an explicit formula just plug in the given coordinates and solve for 2—)}. The
x slope is —1 so the equation of the tangent line is simply y = —x . Note the following general relationship between instantaneous and average speed between
times a and [3 X017) _ x00)
b — a
Now use Eq 1.4 to ﬁnd the required time to make the statement true 1212—1=&Q_ﬁt§:0)c>1217'=4<::>t=i\/E=iL
12 tl t0 J3 Since time is a positive quantity, we reject the negative result.
The acceleration is the second time derivative of position so x"(t)212t2—2=O<::>t=_1/——2—=i\/I
12 6 but time is a positive quantity so we reject the negative result.
The linear approximation for a function at a is f(X)zf(a)+f'(a)(xa) (15)
Now use Eq 1.5 with the given values to get f(28) e f(3) + f'(3)(28 — 3) instantaneous = average c> x’(t) = (1.4) m 5 + 2 (—0.2)
m 4.6
Recall that the volume of a sphere of radius r is g—nrz’ . Now we take the time derivative to get
the ﬁnal answer as follows
7/ ’7/
V 237”} <:> ii: =47rrg: C> 6 = 47r(10)ﬂ <:> ﬂ=—3—cm/s
3 dt dt dt dt 2071’
Recall that exponential growth is 200’?
A(t)=A0e”,r >0 (1.6)
Using Eq 1.6, at t=1
A(1) = Aoer = 10,000 (1.7)
and at t=2
A(2) = Aoezr = 20,000 (1.8) Since Eq 1.7 and E 1.8 form a system of equations with 2 unknowns we can easily ﬁnd the
required unknowns. One way is to divide Eq 1.8 by Eq 1.7 to ﬁnd r and then substitute into
either equation to ﬁnd A0. A(2) (3 Age” _ 20,000
A(1) Aoe’ 10,000
Now substitute Eq 1.9 into either Eq 1.7 or Eq 1.8 to get _ 1,,2_ _10,000
A(1)—A0e 4000042140 em <:>e'=2<:>r=ln2 (1.9) 25,000 (1.10) Combine the previous results to get the ﬁnal answer
A(5) = 5,000e”“2 = 5, 000e5an = 160, 000 n) The differential equation for Newton’s Law of Cooling is d—T = k(T — To). dt 0) Here is the graph: p) Start out by expressing the Taylor series for the exponential about x=0 to get e1: — (1.11) Now “substitute” x = x2 into Eq 1.11 to get the ﬁnal answer
°° 2n 2 4
ex :Zx :1+_)f__+_x__+...
n! 1! 2!
n=0
q) The best way to solve this question is to build up the required series expansion by
using known ones.  00 n x2n+l
Slnx=Z(;(—l) (2H1)! (112) Use Eq 1.12 to build the required series as follows 0° 2n+1 (x+l)sinx=(x+1)Z(—l)n(:n+l)! = x + x ——+
3!
r) Use L’Hospital’s rule to get
. cosx—e"2 . —sinx—2xe"2
hm ——T—— = l1m—————
x>0 x x—>0 2x
. —cos x — 4x2e"2 — 26“2
= 11m —————————
x—>O 2
_ —1—0— 2
2
3 Question 2 [10 marks] The question is not clear about whether the kite is moving vertically and horizontally or
only horizontally. I will assume the latter. r m., ?? m/min 30 m. I Vii a" A _._____, 40 m, 10 m/min a) It helps to look at the geometry of the problem. From the above graph we obtain the relation x2+y2=r2 (1.13)
where y is a constant (an assumption). Taking the time derivative of Eq 1.13 gives
dx dy dr dx dr
2x—+2y—=2r—<:>x—+0=r—— (1.14)
dt dt dt dt dl Solving for i gives the ﬁnal answer dt {1.3.5} hi] Leaking a: the graph 13.61335 13;: 1:113:71 the 333131331313 233333 = m 113‘ :53
3
As heicrre take {be time derim 33.3133 hf Eq 3.13.1 ta gei
aft dr
3;! i m x . 3:33? 1' £3 .— 3; E"
_{ {IR33%;“ 2': _} E5 “friﬁﬂ—— E ., m} ....... ‘1 ................. I
:33“ 1.,~ 3" if! r"
r if: 1' {173*
.1333 333" 1%? (“131191 33311 3 3 . :
m—=— .3, """ ————————————,—' ~——wa’3'man €3,113}
d: 3““ $1118 I ‘33} " 35 " ' (53)} 5311111333
{j'uestiﬂn 3 £12 merits} 3113: mam3313;; 3.3 :13 1113311333: 3f the 313333431? frame: “a: From the picture we can obtain an expression for the perimeter (P) as follows P:l(2ﬂ[£])+x+2y
2 2 =[l+%ﬂ]x+2y (1.18)
and also the area (A)
2
l x
A=— _
2[7r(2) J+xy
1 2
=§7rx +xy (1.19) It is necessary to express Eq 1.19 in terms of a single variable, say x. Use Eq 1.18 to express y in terms
ofx as follows FEEﬂﬁzr 1 Substitute Eq 1.20 into Eq 1.19 to get 4 1" 1 2 EMF}: +3r—mr —m:r:: 3
' 1 E I+m 31r+ﬁvmlﬁtﬁ 13" .r’ 11‘” 1 2!
+1”!§m—i[l+ 4113:}
. "1’ " 1 L,_ f.
.l .2 all 121 .7 4 =1——:r——1—1x2+5x 1. 8 21’ Now ﬁnd the maximum value of Eq 1.21 as follows Now substitute Eq 1.22 into Eq 1.21 to get the maximum area of Questing 4 E12 marimi 54F d1”. 1 1‘
:m. [WWW dx 8 1
4 Recall that radioactive decay is given by
xiii} ; 3106’? E}
First ﬁnd the constant r by using the deﬁntion of half life as follows Start the timing at 1980 In 2000, In 2020, In 2030, A(t)= Aoe"<:>— —=A0 Aoertc>r= Now the amount of radioactive material at any time t is given by (ie 1:0), A 11 A1r)= 1e _l_n2!
_[1—20 ln21 A( )=100e 120 (t) = [100e120 _ln2
100e 120 ln220 —20
+100] {1 [1+1ﬂ)x
. 2 g 2 13:2 4 5x1
2 _ J. :[———7z— 111+5Hr 20
7r+4 5'2} n+4 1n1
__2 t _1n2t
+100 6 170
£20
6 120 +100] ‘— _ln2
120 ln21
e 120 m2 4 111131112 . (1.20) (1.21) (1.22) (1.23) (1.24) (1.25) (1.26) (1.27) (1.28) _E20 Jﬂgzo —E10
AU): 100e120 +100 e120 +100e120 ~253.3924kg (1.29) Question 5 [12 marks] 5 /
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 Fall '08
 LAMB
 Math

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