This preview shows pages 1–7. 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 Document
Unformatted text preview: Math 100/120/153 Solutions/Dec. 1994 Math 100/120/153 Solutions Dec. 1994 1. This question tests your ability to take derivatives using the Product Rule and Chain Rule. (11) f (2:) = (21: + 3)e' (b) y=ln(a:+\/:c§‘—l =ln(z+(z2—1)§) iv
dz dy dz f'(z) = 2e’+(2z+3)e'
= (2.'t:+5)ez f’(1) = (2+5)e1=7e
1 ;+— ,/:2:i(1+%(2'1)_§(2$) l a: x+Vz§—l(1+ 1:21) 1 z+\/:z:2—l :c+\/12—1. «22—1
1 22—1 (c) Slope of y = ‘/arctan(z2) at 2: = l Marctan (2:2) 1 + $4 1
2 (arctan (2:2)) 'é .‘B l Atx=l,y=‘/arctan =f§=azﬁ Ly
dz 1:! l Varctanl . 1 [Oh— 1 2:1: ) Page 1 (d) To find the points where y = 22' — 2’ has a horizontal asymptote, solve dy _  21 _. 2 1
d1: — 0. Notice that 2 — (2 ) 
y = (2’)’ — 2'
3% = (22V ln(2’) 2’ 1:12 22‘2ln22’ln2
(22“! — 2') ln2 = 0 Thus, since In 2 ;£ 0, 22241 _22 = 0
22141 _ 2:: To solve the equation, we equate the exponents. 22: + 1 = z
z = —1
Hence y has a horizontal tangent at :1: = —1. 2. The definition of the derivative of f (x) at a: = a is, I _  f(z)f(a)
f(a)— Inn—r; 2H! Now apply it to f(1:) = Vx’ + 5. I _ . f(r)f(2)
“2)  lama?
_ 1i ¢x2+5—\/22+5
— 1:312 3—2
¢z2+5—3 ¢z2+5+3 lm——— 'Jz2+5+3 2—»2 :z—2 (, Math 100/120/153 Solutions/Dec. 1994 Page 2 _ 2:2 + 5 — 9 3 dzy
= lim——————— —+3+9———+23—3 = 0
=42 (:r—2)(\/z2+5+3) 2 ( )dri:2
— lim ___i'—4——— 12% — g = o
2—»2 (12)(‘/32+5+3) :5 (ﬁg 5
= “m ___(3=+_2)(£i 2:2 = ﬂ
1H2 (a: — 2) (V32 + 5 + 3)
_ =42 2:2 + 5 + 3 (c) The piece of information f (3) = 1 tells you that the curve passes through
_ 4 _ 2 the point (3,1). Part a) tells you that the slope of the tangent line to the
 é " 5 curve at that point is —%. So the tangent line has a slope of —%. Additionally,
. 1 _ . I I _ part b) indicates that the curve is concaveup at (3.1), so our tangent line
'Il'y checking the answer by ﬁnding f'(z) usmg the differentiation rules and lies b ﬂow the curve_
substituting in a: = 2.
Y
3. (a) Usmg impliCit differentiation y 1”,” m
dy dy
2_ 2___ =
33] dz + 213g + 3 dz 0 I my"! “1‘: concave upward
slopez —L
Now when x = 3, y is equal to 1, so ’
dy dy
2_ 2_ =
3(1) dz: 3:3 + 2(3)(1)+(3) dz 2:3 0
Rewrite th's as
‘ :12 _ :1 _ _1
d: 3:3 ‘ 3 + 9 ” 2 3 ‘ (b) To ﬁnd f”(2:) we implicitly differentiate again the ﬁrst line of the pre vious calculation: 4. (a)
dy 2 ,d’y d9 ‘13! 2J2}!
6y(—) +3y—+2y+2z—+2z—+z——=0 t c
dz dz” d3 dz d“r’ lim M(t) = lim 4 .6— = lim 4" = 9 = o
2—4—00 tboo 1 + e“ et l—i—oo e' + 3 3
When 1: = 3, we know that y = f(3)=1 and £ = f’(3) = mg.
lim M(t)= lim 4 =—4—=4 2 t—boo t—roo l + '1 1+ 0
6(1) “mg +2(1)+ 2(3) +2(3) + (202% = o e Math 100/ 120/153 Solutions/Dec. 1994 (b) Now w d 1 _ __ —t 
dt _ dt4(l+3e)
= 4(—1) (1 + 3e")_2 (—3e')
__ 128“
(1+3e")2
14 1 4 4
MMTM) ‘ Z'1+3e¢(4_1+3et) 1 4 + 12¢" — 4
1+3e“ ( 1+3e“ )
12e“
(1 + :se—t)2 Since expressions (1) and (2) are identical, we have proven that 1
ZM(4 — M). (c) The maximal rate of growth of M means the maximum of ﬁnd this by solving dM
F. We J‘M dM
dtz — 0. We already know —dt from part b). 2’14. 1
dt’ "dt l2e“
(1+ 33")2 Use the quotient. rule to take the derivative. (PM
dt2 —l2e"(1 + 3e")2 — 12e“(2)  (l + 3e“)(—3e") (1 + 3e")4 —12e"(l + 3c" — 6e") (1+ 3e“)3 —12e"(1  3e") (1 + 3e“)3 dM (1) (2) Page 3 d2
Now we set (it)? = 0 and solve for t. We know that —12e" ¢ 0 so we must
have
1 — 3e"‘ = 0
1
1 _ _
e ' 3
—t — ln(3‘l)
t — ln3
. . . dM
To ﬁnd the maxxmum value, we substitute t = ln3 Into our formula for 71?.
dM 12e'n3 12(1) 1 71? mm, = (1+ 3eln3)2 6. Start off by (implicitly) differentiating the given equation with respect
to t: _L§ _ _iﬂ _ LE
C2 dt R2 dt 52 dt
2?.  52 1.19 _ i151
dt _ 02 dt R2 dt
dS  dR .
We want to know Et— when R = 250 Q, S = 1000 Q, I = 100 Q/minute and
% = 10 Q/minute. We have everything we need to fill in all the unknowns in the above equation, except C. We get this value by plugging our R and S
values into the original equation. 1 _ 1+1 0 “ R s .1. _ L+__L
C ” 250 1000
l _ 4+1 C ‘ 1000 C = ESQ—0:200!) Math 100/120/153 Solutions/Dec. 1994 Now put it all into the. differential equation. d5 2 1 1
Et— _ 1000 (—2002(10)———2502(100))
200252(10) _ 250242000) 2002 2502
2510—16100 250  1600 = 1350 Q/minute
Thus 5 is decreasing by 1350 ﬂ/minute. 6. Let a: be the height of the box, y be the depth of the box, and 2 be the
width of the box (see the following diagram). We can write down a formula
for the volume of the box: V = zyz 5ft 8ft However, we have the following constraints. 22+2y=8 y=4—a:
2z+z=5 2:5—22: We can make our formula for volume into a function of a: only by substituting in for y and z.
V(a:) = 2(4 — z)(5  22:) Page 4 I This is the function that we want to maximize (with respect to 1:). (‘11—: = (4 — x)(5 — 21:)  1(5  2x)  2:0(4 — 2:)
= 62:2 — 262 + 20
= (6x — 20)(:r: — 1) = 0
Thus 0;: = 0 for z = (3—0 = g or a: = 1. When 2: = 13—0, our other values are
y = g aid 2 = —§. But 2 cannot be negative! So instead try :2: = 1. Then y = 3 and z = 3. The volume in this case is 9 ft’. This is the maximum
volume. 7. (a) Use f’(a:) to ﬁnd intervals over which f(z) increases, decreases, or
has local maxima or local minima. , __ 2(1 — 1:2) __ 2(1— z)(1 +2)
“3) ‘ (1+z2)2 ‘ (1+z2)’ Since (1 + :22)2 > 0, we need to consider only the sign of (1 — 1:)(1 + z). l — a: 1 + 2: f’(z)
— negative
—1 < a: < 1 + positive
1 < a: — + negative So, f has a local minimum at (—1,0), and a local maximum at (1,2). f(2:)
is increasing on (—1,1) and decreasing on (oo, —1) and (1,00). (b) We do the same thing we did for part a), except this time we use f”(2:)
instead of f'(x). 2(—2:c)(1 + 1:2)2 — 2(1— 2:2)(2)(2:c)(1 + 2:2)
(1 + zz)‘
2(—21:)(1 + z2)2 — 2(1— $2)(2)(2x)(1 + x2)
(1 + 1:”)4
2(—21:)(1 + 1:2) — 2(1 — $2)(2)(2:c)
(1 + :52)3 f"(z) Math 100/ 120/ 153 Solutions/Dec. 1994 —4a:(1 + 2:2)  83(1 — 2:2) (c) See the following diagram.
(1 + z’)3
4a:(l + 2:2 + 2  22:2)
(1 + 11:2)3
—4a:(3 — 1:2)
(1 + 22)3 Now we know that f”(2:) = 0 when a: = 0 or a: = is/ﬁ. The denominator
of f ” (x) is always positive, so we only need to worry about the sign of the
numerator. z :c—s/ﬁ a:+\/§ f"(:c) negative —\/5 < a: < 0   + positive 8. (a)
0 < 1: < x/i +  + negative “m “13 z 9
J5 < z + + + positive 2—H sin(1r:1:) 0
Using L’Hépital’s theorem,
.1.
lim .ln1 = lim —5———
2H sm(1rz) z—u 1rcos(7r:1:)
_ 1
' ’— “(1)
f (:c) is concave up on (—\/§,0) and (Jim). 1 f (z) is concave down on (—00, —\/§) and (0, J5). 7r
Hence f (2:) has points of inﬂection at a: = 0,:l:\/§. (these are the points
where the concavity changes sign.) ('3) Page 5 Math 100/ 120/ 153 Solutions/Dec. 1994 Again, use L’Hopital's theorem. 1. e2! _ 1 — 2e2t
e336 te2t _ 151(1) 6"" + 2te”
2
= —— = 2
1 + 0 (c) The In function is continuous, so lnlimz = limln 2:. 1“ [Hm (“exp/2,] = lim [ln(seczrl’z] 2—90 1—)!)
ln(sec z) 0 z lim
2—90 272 0 1
—— secz tan 2; lim ﬂit—— using L’Hopital’s theorem.
zw 2a: ll
.5: = % using L’Hopital’s theorem. ll
'5: 9. (a) The given limit is ﬁnite. Notice that the denominator goes to zero. lim 2:2 — 4 = 0
242 In order for the limit to be ﬁnite, the numerator also must go to zero. limf(:c)—5=0=>li_.n12f(2)=5 2—52 (h) The deﬁnition of the derivative of f (3) at a: = 2 is 1—92 1—2 We have to massage this to get it into a form where we can use the limit that
we were already given. Note that since f(a:) is continuous at a: = 2, we can Page 6 « say that f(2) = 5.
nn—quz+2 I _ .
f (2) _ :1: — 2 :1: + 2
= mgﬂnzwu+m
2—02 2: —4
= limf(:)5limz+2
:42 z — 4 :42
== 3  4 = 12
. dy dv . . .
10. First, we note that d— = v and E = a. Now 1t 15 Simply a matter of
integrating and calculating constants of integration.
(1—!) _ 6t if 0 < t 5 10
dt _ —9.8 if10<t<T
Um _ 3t2+C if0<tglo
“ —9.8t+D if10<t<T To calculate the constants C and D we are told that the rocket starts at rest,
so 11(0) = 0. Thus C = 0. When the fuel runs out (after 10 seconds) the
velocity function switches from one formula to the other. The velocity must
be continuous, so we can stipulate, [33”!=10 = [—9.8t+D]t=m
3(10)2 = —9.8(10)+D
D = 300+98 D = 398 So now we have the velocity function. v(t)_ 3:2 if0<t510
‘ 9.8t+398 if10<t<T Now we integrate v(t) to get y(t). g __ 3t2 if0<t310
dt ‘ —9.8t+398 if10<t<T
(t) __ t3+E if0<t510
3’ " —4.9t2+398t+F if10<t<T Math 100/120/ 153 Solutions/Dec. 1994 Again, we use the information we are given to calculate the constants of
integration, E and F. The rocket starts from the ground at t = 0, so y(0) =
0 => E = 0. Also, the height of the rocket is continuous, so, [t3]t=10 = [—4.9t2+398t+F]l=10
103 = 4.9(10)2+398(10)+F
F = 1000+490—3980
F = —2490 Thus, the height function for the rocket is (t); t3 if0<t$10
y ' 4.9t’+398t2490 if 10<t<T 11. (a) We can centre the tangent line approximation at a: = 2 since we
know f(2) and f’(2). The linear approximation function is ﬁt) z 14(3) = f(2) + f'(2)(17  2)
= 41(2:—2)
= 6—2: L(3) = 6—3=3 (b) To ﬁnd the smallest interval that we are sure contains f (3), we must
calculate a bound on the error of our estimate of f (3). We know that Im) — Len = gm“: — 2)’ for some c between a: and 2. We don’t know what that c is, so instead we
choose a c that makes our error estimate as large as possible (so that we don’t underestimate the error). man—Len s §crgllzagllf”(c)l(x2)2
lf(3)L(3)l s gﬂgllmcmszf Page 7 It is given to us that i S f "(3) 5 1 for all a: > 0. Thus, we are guaranteed 2a: a:
that I f”(:c) will be no bigger than % on the interval [2,3] (try substituting
1 l :1: = 2 and :1: = 3 into —z and E to try to get an idea of what :cvalue will
give you a maximum bound for f"(:r)). f(3) — Lam < 1 max moms — 2)” = g (1)2 = § — 2 «[2,3] The smallest interval that we can be sure contains the value for f (3) is
1 l 11 13
[3‘2'3+zi  [7’ Ii 12. (a) We are looking for solutions to the equation f (z) = a: for ze[a, b].
Deﬁne the function g(z) = f (z) — a: and try to show that g(c) = 0 for some
ce[a, b]. Notice that, since a S f (1:) 5 b Vze[a, b] (“for all a: in [a,b]”), we can say
9(a) = f(a) a 20
and
9(5) = f(b) —bS 0
Thus, since 9 is continuous, by the Intermediate Value Theorem, we
know that there exists some c in [a, b] such that g(c) = 0. Therefore, g(c)=0=f(c)c = f(c)=c as required. (b) We use the method of contradiction for this proof. Assume that there
are two :r—values that make I (1:) = 1:, and call them an and 2:2. Then by the
Mean Value Theorem, there is a c between 2:; and :2 such that no): “Ill“332) ___ 251552 :1 12112 $1—12 But we were given that f'(z) < 1 for a S :1: 5 b. The above result contra
dicts this constraint, so we know that there cannot be two (or more) roots in
[a, b]. This fact, together with part a) proves that there is exactly one root
to the equation f (1:) = 2:. ...
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

Click to edit the document details