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Unformatted text preview: Math 100/120/153 Solutions/Dec. 1995 Math 100/120/153 Solutions Dec. 1995 1. The derivative of f at the point a: is deﬁned to be
m + h)  my I __ .
““l‘i'lft h Applying this, we get f'(r) “m m: + h)  n2)
h—DO h
l 1 lim(ac+h)2+1 —1:2+l
hso h
l l . (z’+2hz+h2+l)_z2+l
11m —————
hH) h
lim (19+ II)  (/5?2 + 2’13 + h2+ l)
1.40 h(a:2 + 1)(:¢:2 + 2ha: + h2 + 1)
I'm  li(2::+h) __ 2:1:
Ill+0 11(32 +1)(:r:2 + 2hx + h2 + 1) ‘ (1:2 +1)2 2. We compute the slope of the tangent line to each of the given functions;
that is, their derivatives: —22 22_ 22_ —22 . .
mm) = ———————(l+e £18355 2e ’ f’(o)=————2 4:2 2:3
, _ 1 1(3) , _
9")  Wmsx) 99"
h'(:c) = Iccosk:i:+‘/—1Tk——W h’(0)=2k (a) The tangent lines to f and g are parallel at 0 since the derivatives are
equal there. Page 1 (b) The tangent lines will be perpendicular if h’ (0) = ———!— which happens f’(0) if and only if 2!: = —§, that is, if k = %
3. (a) Using implicit differentiation
5y‘y’ + y2 + 2zyy' + 3x2 = 4
201f’(y')2 + 5y‘y” + 2yy' + 2yy' + 2::(y')2 + 2zyy" + 61: = 0 (a) At 2 = 2, y = 1, so the ﬁrst expression above becomes 5y' +4y’ +1+12 = 4
which implies that y' = —l; similarly, substituting y = 1 and y’ = —1 in the
second expression gives 20 + 5y" — 4 + 4 + 4y” + 12 = 0 which implies that u____§_2_ 31—9 (b) The tangent line approximation is y = yo +y(’,(1: — :50), which in this case
is y = 1 — 1(2: — 2). When 2: = 1.97, y z 1  (1.97  2): 1.03. (C) Math 100/120/ 153 Solutions/Dec. 1995 4. (a) Taking w to be the width of the box (side length of the square base)
and h to be the height, we can write the volume and cost as follows: volume V = wzh = 320 => h = 37229
cost C = 10w + 4wh = 10w + valid for 0 < w < 00 w Since C —) 00 as w ) 0 or as w —> oo, 0 has a minimum which must occur
at a critical point. d0 1280 a = 20"”?
d0 71; o if and only if 20:»3 = 1280 if and only if w = 4. Hence the minimum occurs when w = 4 and h = 20 (units are centimeters). 5. Let S be the amount of undissolved sugar. The rate at whichthe sugar dissolves is then dif = kS. This implies that S = Soc“. We are also told that 0.755.; = Soe", so that k = ln(.75). (a) .2530 = Soc” => ln(.25) = ln(.75)t so _ ln.25 _
 In .75 _ t 4.8min (b) We have 10 = Soe2‘5" = Soe2‘5‘“ '75. Thus so = 10e‘2'7'“ '75 = 20.5gm. Page 2 6. Referring to the sketch of the triangle, and using the law of cosines, we
have 32 =1+$2 — 2zcosl20° = 1+12 +a:. When 3 = 2, 4 = 1 + :c + 12, which implies that X a: + 2:2 — 3 = 0
a: _ —1 d: \/'1+12
I " 2
1 s a: > 0 a 2: = :1+2_\/ﬁ
So
ds dz:
283t — (21: +13%,
d3 dy , d: 28
when 3 — 2, (—1? — 750 2.2.7 _ (—1+x/i§+l)(—E which means that a? — E km/min, that is (in rounded ﬁgures) approximately 7.77 km/min. 7. We begin by taking the derivative of f and noting where it is zero: f'(:1:) —(z3 « 5)e" + 32“" = —e"’(:c3 — 312 — 5)
f’(2:) = 0 if and only if g(1:) = 9:3 — 31:2 — 5 = 0 We’ll work with g(:z:) although it would be okay to work with f’(:1:) directly.
We compute a few values of 9(1): 9(0) — 5;
9(1) = 8;
9(2) — ~9;
9(3) = 5;
9(4) = 11 By the mean value theorem, the unique critical point is in the interval (3, 4).
So let the initial approximation be 2:1 = 3.5. Math 100/ 120/ 153 Solutions/Dec. 1995 Newton’s method gives us the iteration 23,—3zﬁ—5 :1: 1:3 —
n+ " 32%6zn or, after simplifying
x _ 2x3, — 3x3, + 5
"+‘ ' 3x2, — 62,, Then, starting with x1 = 3.5, we obtain the sequence of approximations
x1 = 3.5,x2 = 3.4285,:3 = 3.4259, . . .. Since 12 and 1:3 differ by the required
accuracy, the last approximation is sufﬁciently good: r = 3.4259. 8. (a) f(a:) = 33(12  10km + 251:2) = 23(2: — 5k)2. So f(:r) = 0 if :1: = 0 or
a: = 51:. Hence we can represent the values of f, roughly, as follows:  + +
ﬁx) ———i———————i———
0 5k Computing the derivative, we ﬁnd f '(z) = 5:1:4 — 401:9:3 + 751332 = 5:1:2 (:1:2 —
8kx+ 15k”) = 522(35k)(z 3):). So f’(a:) = 0 if and only if a: = 0, a: = 3k,
or :r = 5k. We can therefore see that f (1:) is increasing on (—oo,0), (0, 3k)
and on (51:, oo). [(23) is decreasing on (3k,5k). Hence f has a local maximum
at 3]: and a local minimum at 5k. + +  +
f'(x) ———I————l———i——
0 , 3k 5k (b) f"(a:) = 202:3 ~120k22+150k2x = 10:2:(23:2 — 12kz+ 151:2). So f”(:i:) = 0
if and only if a: = 0 or a: = {3:i: a?” z {4.2k, 1.71:}. These are the inﬂection
points because f” alternates sign between these points as shown:  +  +
f '(x) ———!————l'———*—
conc. down 0 cane. up 1.7k down 4. 2k cone. up Page 3 (c) With the above information, we can sketch the graph of f : 9. (a) We begin by sketching situation. a picture of the A formula for the parabola is y =
kzz; since we are given 100 =
Ic(100)2 we have k = 1/100. Thus
the highway is described by the
parabola y = £3512. To an
swer the question, find where the
slope of the tangent line is equal
to the slope of the line from the
parabola to the statue: the slope
of the tangent line to this curve is given by the derivative 1211.
‘dz‘loo' Math 100/ 120/ 153 Solutions/Dec. 1995 2 2
The slope of the line from (9:, £6) is 50 _ a:— :00 . Equating these slopes
we solve to ﬁnd 2 .21 = iii => 32—200z+5000=0
100 100 — :1:
SO
2 = 200 a: «(2920?  20000_ Since a: < 100, the correct choice here is a: = 100(1 — 3?). So the askedfor
point is a: = 29.3m (east on the road) and y = 8.6m (north on road). 2
(b) Let f (1:) = (1: —100)2 +(—1%—0  50)2. We want to minimize f (2:). Clearly
this has a minimum which occurs at a critical point where 2
, _ _ 35. _ .2: =
f (z) — 2(2: 100) + 2(100 50) 100 0,
that is, if and only if
22:3
1—100+I66§—z—0. . . . 3 1003 . _1 4 .
This happens if and only if z = —2, that is, a: = 100/ (2 / ). In this case
y = 100/ (2‘2/ 3), in other words: ' = 79.4m
y = 65.0m. Page 4 10. Consider the ﬁgure at the right and compute: tan(0 + 01) = 3423, for a: > 0.
tan(91) = 1%, for z > 0. 0 = arctan(3izi) — arctan(l=zﬂ), for 0 < :1: 5 10. c
Since a: = 0 implies that 0 = 0 and 0 is 1
continuous on (0, 1], the maximum oc &
curs at a critical point in (0, 10), or when r
2: = 10.
D E
2'3 _ _1_ :9. + _;_ i
dz _ 1+(3_ﬂ)2 2:2 1+(Ltz)2 $2 3
_‘3__+__1_
2z2+63+9 2z2+2z+1 So E‘Zzouandonly if6$2+6z+3=212+6m+9ifand only “422:6, d3:
in other words a: = 3?. Since 9(10) < 9025), the maximum occurs when 1:393. Note: :—:>0forz<3§§andgg<0forz>[email protected] Hence the maximum 9 is 0 = arcane—‘5?) — arctan(2—f7‘g§), for 0 < a: _<_ 10, or
approximately 0 = .221 radians or 12.7 degrees. Math 100/120/153 Solutions/Dec. 1995 11. (3) Compute f’(0) from the deﬁnition:
1m f(0+h) — rm) h—bO
7r
h2 cos— . 2h
1'31}: h f'(0) 1r
— 112.11%: = 0 since [cos 5H 5 1 using the squeeze lemma. (b) Let g(:1:) — :1: 22lc03( ) for :1: ¢ 0. Then 9 ’—(:1:) — 2:: cos(—) + —12rsin(—). Now
“m f(1/3+ h) — 10/3) = “m g(1/3+ h) +g(1/3)
540+ h 1140+ ’1
7r
— —g<1/3) — 5
and
f(1/3+ 121 mm = lim g<1/3+ 1:) —g(1/3)
hl—rO‘ h—ro— h
= (1/3) = l
9 2
Therefore Ilia}, W does not exist hence f’(1/3) does not
—’
exist.
(c) A similar situation occurs at all points where :1: ¢ 0 but cos(27r—I) = 0.
1r 1 1
That 13, whenxaéOand 5;_ n1r+ 2,i.e., E —2n+1, son: — 2n+1' 12. (a) The mean value theorem implies that there exists a with 1 < a < 2 and “2 f1
f’=<a) —’.‘—’= 21 =1 Page 5 (b) f(2:) = c, for :1: < 0, implies that f’(:c) = f”(a:) = 0 for :1: < 0. (By
continuity, f’(0)= f"(0)= 0 also.) Choose b such thata —— = k where a is the number 1n (1,2) with f’ (a) = 1. Thatis, F =a—boa.rb=a—— Since0<lc<l ,b<a—2. Sincea<2,
b < 0. Therefore 1" (b): 0. Now apply the Mean value theorem to f’ (1:) on
the interval [b,a]. I _ I __
It implies that there exists c with f”(c) = M = 1 0 = k. Thus a—b a—b
f"(C) = ...
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