This preview shows pages 1–8. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: December 2004 Math 100 Exam Solutions 1. Short  Answer Questions. (a) 2
. x—l , ($—1)(x+1) _ m+1 2
1 ————=1 ———:1 =—=~2.
zmm2~3as+2 xLHi(x—1)(x—2) gigs—2 ~1
(b)
1 1
1. 132—1 _1. $20?) WI. 1—? w 1—0 1
zir20$2_3x+2—x320 2 3 2 $320 3 2 ~1_0+0—~
m(1——+—2) ——+—2
J} (I? (I? {I} (c) Differentiate using the Chain Rule:
1 23: f'(’13)= mm? 12$) : *W (d) Differentiate: y' = 1  6—2“: + m  (—2641) = (1 — 2:06—21, and therefore
3/” = —2  6—22: + (1 — 2$)(—26"2$) = (4:1: — 4)e‘2$. Since 6—21” > 0 for any 3:, the sign of y” is the same as the Sign of 4x — 4. Note that
4x—4=0when$=1. If m < 1, then 42: — 4 is negative. Thus y” is negative on the interval (—00, 1). If a: > 1, then 4% — 4 is positive. Thus y” is positive on the interval (1, +00). We conclude that y changes the concavity at a: = 1, thus (I: = 1 is the inﬂection
point. e Divide 333 + x2 to m2 + 1 usin the lon division:
8 g x—l—l $2+1 3:34—12
933+J; (II—‘1' Therefore the slant asymptote is y = :1: + 1. (f) The equation of the tangent line is y — y(:r0) = m(a: — :50), where the slope
m = 9/070) Differentiate using the implicit differentiation: 31'2 + 3y2y’ = cosa: 1
In the above equation, let 1: = 0 and y = 1. Then 0 + 33/ = 1, so y’ = So the desired tangent line has equation 1 1
y — 1 = — 0)7 or equivalently y = 313 + 1. (g) Differentiate: y' = —— sinw. 1 1
Note that the slope of the line y = —; is m : —5 so we need — sin 3:0 2 —E, or 1
equivalently sin $0 = Thus 2:0 : (h) Take “In” on both sides of the given equation: lny :1n(a:1”) = (lnzr)(lnx) = (ln x)2 Differentiate both sides of the above equation with respect to m: y’ 1
_: 21 ._
y < m) 96
Therefore
y, : 21mm .y: 2lnac 'mlnz.
m a; (i) The linear approximation is :
m m f(4) + f’(4)(w — 4)
when (I: is close to 4. So for m = 4.1 we get:
f(4.1) m f(4) + f’(4)(4.1 — 4) = 3 + «m  0.1 = 3 + 50.1 = 3.5. (j) If f is differentiable for all x, then it is continuous for all 33. Thus f is
continuous at x = 1 i.e. 11m me) = $1_igl+f($) = N) 33—d— So lim = lim IE3 : 1 x—d— z—’1* lim f(x)= lim (AmPB) =A+B I—>l+ x—'1+ So we need A + B = 1. Next we differentiate f(3:): {33:2 ifac< 1 Mr): A ifar>1 The function has to be differentiable at a: z 17 i.e. f'_(1) : or equivalently
A z 3. Therefore A = 3 and B = —2. (k) Let = tan‘1 x — $2. If 3:0 2 1 is the initial guess, the second estimate m1
is given by the Newton Method formula: 1
a;1=.’E0 {(560) z 1 ~ f,()
f (330) f (1)
Differentiate f’(:r) = 1 — 2x so f’(1) : l — 2 2 —§
1 +x2 7 2 2'
Therefore 7T
tan‘1 1 — 1 Z _ 1 7r 1
xlzl— —§ —1+—§—‘é+§.
2 2
. . . f"(Z) 2
(l) The max1murn error 18 given by: R1(a:) = 2 (m — a) , where a = 0, x = 0.1 and0<z<0.1. The second derivative is f”(z) : (310‘:—1 < clown—1 = 60 = 1. 1 1
Therefore R1 < §(0.1)2 : §(0.01) = 0.005. (In) The Taylor series for cosx is: 1 $2 + $4
s = — — —— — co m 2 4
The Taylor series for sin(2w) is:
2 3
sin(2w) = 2x — ( m) 3!
Thus the Taylor series for f is:
x2 2:4 (2x)3 1 2 8 3
f(g;):(1_3+Z_...)+(2m_ 3! +... :1+2$_§x _6$ +.... (n)The Maclaurin series for f is Full  Solution Problems. 2. Let t be the time in minutes after turkey is put into the oven, any y = y(t) be the
thermometer’s reading 2 the temperature of (inside of) turkey.
By Newton’s Law of cooling: dy
— = k — 2
dt (2/ 00) dy _ du
_ __ 2 Z _ Z . T ‘ ' 1 _ . . 
Let u v y 00, so dt dt 0 dt hus u satlsfies dt — ku, so the solutlon of this differential equation is u = C  6]“.
It is given that 71(0) 2 20 — 200 and u(30) = 30 — 200. From here: u(0) = —180 2
C  e“ = C, C' = —180. From the second condition u(30) : —170 : —180e30k.. After solving the exponential nat'on we t k ~ 1 In 17
M _ . _.
eq 1 ge 30 18 Therfore turkey is done when y(t) = 80, or equivalently when u : 80 — 200 = ~120.
So, 1 17 2
—1—t=l —
<30 n18) n
_301n§
_ ln% 3. Differentiate using the Product Rule: (m2 — 1)(2:L') — : 2:33 — 2a; — 2$3 —2a: / _ _ ——
f (x) ‘ (x2 — 1)? (x2 — 1)? ($2 — 1)2
Since (a:2 — 1)2 is always strictly positive, the sign of f’ is the same as the Sign
of —2x. Note that —2ac : 0 if and only if m = 0. Note also that f’ do not exist when
a: = 1 and x = —1. Look ﬁrst at the interval (—00, ——1). There, f’ is positive (use for example the
‘test point’ a: = —2). It follows that f(:r) is increasing in the interval (—00, ~1).
Look next at (—1,0]. In this interval f’ is positive (use for example the ‘test
point’ :c = —0.5). It follows that is increasing in the interval (—1,0]. Next look at the interval (0, 1). In this interval f’ is negative (use for example
the ‘test point’ x = 0.5). It follows that is decreasing in the interval (0,1).
Finally, look at (1,00). In this interval, f’(a:) is negative. (To show this, we can use
a ‘test point’ like at z 2) . It follows that f is decreasing in the interval (1, We conclude that the point (0,0) is a localﬁiahﬁm. m whim/m (ii) Next we ﬁnd the second derivative; differentiate using the Product Rule and the
Chain Rule: ,, m2—122—2x2232—1 29: 2952—2 —82
f($):_( )()($éa)1)4( )( )2 ( ($2_)1)3w Since 2+6m2 is always positive the Sign of f’(x) is the same as the sign of (a:2 — 1)3.
Look ﬁrst at the interval (—00, —1). There, ($2 —— 1)3 is positive, so f”(:r) is positive
(use for example the ‘test point’ x : —2). . Next look at the interval (—1, 1). In this interval (x2 — 1)3 is negative , thus f”(m)
is negative (use for example the ‘test point’ a: = 0). Finally, look at (1, 00). In this interval, f”(x) is positive. (To show this, we can use
a ‘test point’ like :3 = 2) . Therefore function f (ac) changes its concavity in m = —1 and 1' = 1 but these points
don’t belong to the domain. Thus f does not have any inﬂection point. (iii) Study the existence of horizontal asymptotes : 1' —1' x2 —1' $2 —1' 1 —1
1—13—1100 ﬂ (1:2 — 1 — z—ir—noo 12(1 _ — z—{r—Iloo 1 — 512— —
Similarly
$3351me) = 1
So y = 1 is horizontal asymptote.
Study the existence of vertical asymptotes :
lim f(:1:): —oo; lim f(a:) = +00;
z—v—IJF m—b—l—
so as = —1 is vertical asymptote liIn = +00; lim 2 ——oo :c—>1+ z—al‘ so ac : 1 is vertical asymptote yA .V Lighl Let y = y(t) :the height of ball, and m = m(t)= the distance from the shadow to
the ball’s landing spot.
By similar triangles a:_x+3 ac
y 8 8
Differentiate both sides with respect to t +§
8.
yi—“Eﬁt :ldg
y2 8dt' d
Note that when y = 4, 1% = #20. Substitute these values and :1: = 3 into the above equation to get:
dz w 2 11m
16 8 dt
d
So, d—C: : —30. Thus the shadow is moving with 30 m/sec. (b) Let s = 5(t) be distance between the ball and its shadow. Since 82 = 3:2 + 312,
we can differentiate both sides with respect to t: ds d2: dy
2—:2— 2—
s a: +ydt dt dt
d d '
Substitute IL‘ 2 3, y = 4 , d—g: = —20, E:— = —30:
d3
32 + 423% = 3(—30) + 4(—20) : —170. d
Therefore (1—: = —34. Thus, the distance between the ball and its shadow is changing at a rate of 34 m/sec. 5. Let a: be the width of pool and y be the length of the deep end, such that shallow
end has length 2y. The volume of the shallow end is = 2113;. The volume of the deep end is y  m . 2 : 217;, so the total volume is 4mg 2 800, so my 2 200, or 2
yz—Oq,forx>0. :v
The total surface area in contact with water is : S = my + 23:31 (bottom)
+ as + 211: (ends)
+ 2(2y + 23!) (sides)
+ :1: (middle) 1600 1600
:3$y+4x+8y:3200+4a:+—x—=600+4m+ w. Differentiate both sides with respect to w. 1600 _ 42:2 — 1600 _ 4(332 — 400)
2 _ 2 2 Sl(1L')——— 4 I} (I) (E Find the critical points from S’ : 0. Since at > 0, the only critical point is
m = 20. Let us look ﬁrst at the interval (0,20). On this interval .5” < 0. It follows that
S is decreasing in the interval (0,20). On the interval (20, +00), where S’ > 0, S(:r) is increasing. Therefore S takes
its minimum value when a: = 20. 1
6. (a) Let f(:z:) = 1 2. By the deﬁnition of the derivative:
— x
1 1
d 1 *, ~. f(m+h)—f(x) _. ﬁrm—ﬁe
3(1—332) £136 —}lL1—I'I%) h a:
_ 2 _ 2
thﬂ :3) (1 (w+h))
h—iO (1 — x2)(1 — (x + h)2)h
1—;r2—1+a:2+2$h—h2 : hall) (1 — m2)(1— x2 — 2xh+ h2)h
_1im (2x+h)h
_ h—»0 (1 — 9:2)(1 — m2 — 2Ih + h2)h _ 2z+0 2:1;
(1—x2)(1—$2—0—0) _ (1—3132)2 (b) Since f is differentiable for all :r, so is g(x), hence g(x) is continuous for all
$. Note that 9(0) = f(0)+f(1 — 0) = f(0) +f(1) and 9(1) = f(1) +f(1 — 1) = f(1) +f(0) = 9(0) Applying the Mean Value Theorem to g(m) on [0, 1], we conclude that there exists 1 — O 0
ac,0<c<1,suchthatg’(c)=gL£=I 1 _ 0 = 0, as required. ...
View
Full
Document
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, The Land

Click to edit the document details