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Unformatted text preview: mama Math 100 Sample Exam Solutions
7 1. SHORT ANSWER QUESTIONS (a)
A . $2+4mz5 . (x+5)(z—1)_1+5_6
mz2+5x46‘l§'i(z+6)(z—1) ‘ 1+6_ 7 (b) 1 1— 1+1
lim(x—\/:c2+:v)=lim [ — x2(1+l>]=limz<1— 1+—)=lim————z z—NDO z—mo a: 2—H» a: :c—ioo l/cc We“ a “3;”; .‘mnaie‘rwowaewww mwu "  Now we make a substitution y = l/z. So a: —> 00 ¢ y —> 0. We have: lim ———l — ' + y —) 9
u—io y 0
Thus we can use L’Hépital’s rule:
_ 1
nm 14— VI” = lim Wu =_ 1 =1
y—w y y—w 1 2 1 + 0 2 So 1im(z — V22 +2) = —0.5. 2—)00 (c) At 3 =0, the limit takes the form 0/0 so we may apply L’Hépital’s rule: w W ‘.n,_wmw~mm_ a . l—cosz . sina: 0
11m .— = hm +—— —> —
z—ro msma: z—>0 s1nx+$cosz 0 So we apply L’Hépital again: _ sin a: _ cos a: 1
hm ———— = 11m 1
:—>0 sinac + zcosa: z—vo cosz — zsinx + cosa: _ 1 — 0+ 1 _ 2
(d) The function is differentiable so it must be continuous at :1: = 1, and its derivative must also be continuous at
a: = 1.
So we have: A(1)2=13+B =>A=1+B
2A(1) = 3(1)2 => 2A = 3
Therefore A = 3/2 and B = A — 1 = 1/2. (e)
9(01) = 933M932)
So we use the product and chain rules: g'(:1:) = 3m2h(z2) + msh’(a:2)(2m) = 39:2h(a:2) + 2z4h’(m2)
At x=2, we get: g'(2) = 3(22)h(22) + 2(24)h’(22) = 12h(4) + 32h’(4) = 12(2) + 32(—2) = 24 — 64 = —40
(f) We use the product rule to get f’ and f”: f’(m) = zcosz + sinx f”(9:) = x(— sinzv) + cosa: + cosx = 2cosa: — wsinm ) If f’ = —2 f , then f (9:) is an exponential function and we have: (s
f (z) = Ae‘“
Atz=1,f=2sowehave:
2 = As"2
A = 262 f(0) = Ae° = A = 2e2 (11) Differentiate implicitly to get:
’69“1 + yea”1 + e” + :ceyy’ = 0 y
Plugin 2: = 1,1; :0:
y’e°+Oe°+e°+le°y’=y’+1+y’=2y’+1=0
1
I— ——
y ’ 2
(i) To ﬁnd all inﬂection points, set y" = 0:
y=zv+sinx
y’ =1+cosx
y" =—sin:v er a: is a multiple of 7r. So the coordinates are (n1r,mr) where n is an integer. The sine function is 0 whenev (j) The next approximation is the a: intercept of the tangent line at the initial guess. Thus we set 3/ = 0 in the tangent ‘ line equation and get:
0 = 10x + 19
10:1: = —19
as = 1.9 (k) The speed of the object is :c’ (t) = 3t2 — 4. The rate of change of the speed is a)" (t) = 6t.
At t = 1, the rate of change of speed is 6(1) = 6 > 0 and so it is INCREASING at 6 metres per second per second. (1) Using the quotient rule, we have: 2 .. _
f,(x) : 2:1:(3: + 2:12)$2 im:m)22)(2x + 2) f’(2)—WW_32‘12_ 5
 (22+2(2))2  64 —16. =width, l=length. Then the area A = wh = 40. (m) Let 10
Since the width and length are changing with respect to time, they are functions of time t.
So we have: 40
l h t = ——
( ) w“)
—40 —40w'
I __ __ I __
h (t)  W2 (w (m — m?
At w = 10, w' = +2. Therefore:
—40(2) ——80
h, t = ——— : ——— :: —— .
( ) 102 100 0 8
Thus the height is decreasing at a rate of 0.8 cm/s.
(11) Using the chain rule, we get:
ftp) __ < 1 > (221272 + 1) — 2:1:(232 — 1)) 2
1 — ($371)? (“’2 + W 1 + x2 First let’s simplify the numerator of the ﬁrst term: 223(z2 + 1)  221(172 — 1): 22:3 + 2:1: — 29:3 + 2x = 4:1: Next we simplify the denominator of the ﬁrst term: ( 1_(22—1)?)($2+1)2=(z2+1)2 W 2:2 + 1 (a:2 + 1)2
_ (2:2+1)2\/z4+2:52+1—:r4+2z2— 1
‘ (22 + 1) = (x2 + mm = (9:2 + 1X22“) Thus our derivative becomes: Therefore, f (as) = constant.
We put in a: = 1 and get f(1) = sin"1 0 — 2tan"1 1 = —27r/4 = 7r/2.
Thus we have f(x) = —7r/2. (0) Since we will be multiplying the Taylor series of cosine by an 23" term, we only need to expand cosine to 9:6.
The Taylor Series for cosx is: So the series for cos(:z:/3) is: 2. (a) The function f is increasing when f’ > 0 and decreasing when f’ < 0. We have f’(:r) = Ke“'2/8(1 — (2:2/4». We know that K > 0 so Ke‘zz/8 > 0 for all 1:. Thus the sign of f’(:c) depends
2 (17 only on the sign of 1 — I. 2 2 :1: 2:
1—— 0 1 — 4 2
4> <=> >4<2> >zc ¢>2>zor —2<:c So f(z) is INCREASING on —2 < a: < 2. For decreasing, we just reverse all the inequalities and get —2 > a: and a; > 2. For critical points, we change all the inequalities to equal signs and get a: = :l:2. Since f is decreasing on the left of a: = —2 and increasing on the right of a; = —2, the point a; = —2 is a local minimum.
Similarly f is increasing on the left of a: = 2 and decreasing on the right of a: = 2 so this point is a local maximum.
Since the function approaches 2 as a: approaches 00, f (—2) = 0 < 2 is an absolute minimum. Similarly f (2) = 4 > 2 is
an absolute maximum. (b) For concavity we must ﬁnd the second derivative f”(2:). Using the product rule we get: ._ x 2 $2 2 _
W) = Keg—Xe” “5 (1 — I) + KW “(5) f"(2) > 0, and is concave DOWN if f”(:v) < 0.
3 ...
As in part a, the sign of f"($) depends only on the sign of 17‘ 1612z' The function is concave UP if 2:3 —— 122:
16 <=>a:(a:2—12)>0
<=> bothx>0ANDz2—12>00rbothx<0AND:c2—12<0
<=>m>\/ﬁor0>:c>\/T2_ >0<=>x3—12:z:>0 Thus the function is concave UP when 0 > a: > —\/1_2 or x > .
To get concave DOWN, reverse the inequalities. So f (9:) is concave down on —\/—12 > a: and 0 < a: < \fl2,. The inﬂection points are when f”(a:) = 0 and that occurs at :1: = mix/1‘2. ‘ e have the following picture: information from parts a and b, and the asymptotes f —+ 2 as a: —> :too, w U6 .(c) Given the 3. Let a: be the length of a side of a corner that is
removed. Since we are removing a corner from
each side, the length of the box will be L = 4— 22:,
and the width will be W = 2  22;. The height of the box will be equal to 1.
To maximize volume we take the derivative and set it to zero: V = Lch = (4  21:)(2 — 22:)(13) = (8 — 8x — 42: + 49:2)(93) = 8m — 121:2 + 42:3 f}
V’(:c)=8—24w+12x2=1222—24a:+8=0 y 95 H’ g
Dividing through by 4, we get: .
3x2—6z+2=0 be,
x=6d=‘/364(3)(2)=1i_\/_T—3: ‘ w:2»gx
(2X3) 6 X < y 0423,1577 Now we check second derivativeTo/determine whether the critical Thus we have two critical points a; n
points are max or min: f7 V"($)=24:v—24=24(z——1) / X;’_ 3,
We see that we get a max if x < 1. So the max volume occurs when a: = 1 — J35 m 0.423. 3 So the maximum volume is 42:3 — 12:1:2 + 8a: N 1.54 cubic metres. MilenaE 4. The speed of the ant’s shadow is 2—3:, where y(t) is the
coordinate of the ant’s shadow on the y axis. Because light travels in a straight line, the position of the shadow
on the yaxis will be the y — intercept of the line joining
the ant and the light source. Given an arbitrary point along the ant’s path, the slope of the line will be:
1/0 ~ 0 _
x0 — 3 (gab) Mm?
Note that $0 is the wcoordinate of the position of the ant and is a function of time since it is moving.
The equation of the line is
y — 0 _ 29:0
2 — 3 _ 2:0 — 3
$ — 2% (:1: 3)
y — 2:0 — 3
Now, the yintercept occurs at a: = 0, so we have:
21230 —6£Eo
= —3 :
g 11:0 — 3( ) 2:0 — 3 The rate of change, then is:
,(t) = —6$6(.'B0 — — (—6220X116)
y (930 — 3)2
At 2:0 = 1,2:6 = —1, we have:
3,, _ —6(—1)(1~ 3) —(—6>(1)(—1) —12 — 6
_ \ (1 — 3)2 4 So the ant is moving downwards at 4.5 units per minute. _—9
‘2 5. According to Newton’s Law of Cooling, if T(t)=object temp. at time t (t in minutes), Ta=ambient temp., T0=initial
temp., then we have: for some constant k.
We are given: Solving for k, we have:
50 = 45 + (400 — 45)e'=/2 = 45 + 3555“2 i = i = —k/2 355 71
7l=ek/2
k: 21n71 Now we plug this in and solve for T at t = 5. T(5) = 45 + 3555mm)5 = 43 + 355(e’n71r10 = 45 + 355(7110) = 45 + 1.09 x 10—16
So T is VERY close to 45. End of Solutions. ...
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 Fall '08
 LAMB
 Math, Derivative, Inflection Points, Product Rule, SAMPLE EXAM SOLUTIONS, L’Hépital’s rule

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