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Unformatted text preview: Math 180/ 100
Final Exam
December 2002
Solutions 1.Short Answer Questions.
(3.) Evaluate the following limit: 1” $2+3x+2
I—l’n—ll 32—:5—2 Solution: Call this limit L. Then . (z+1)(:1:+2) .
2 l —— I ll .
1.1331 (:1:+1)(:r—2) 13195—2 3 a2+2 _ —1 (b) Determine an equation of the tangent line at (1, 5) to the curve y = 3:3 + 4.
Solution: dy/da: : 3:122, so the slope at .7: = 1 is 3 ~ 12 = 3. Thus the equation of
the line is y— 5 = 3(2: — 1). (c) Calculate the derivative of 9(3) : arctan{ln(s2 + 1)} Solution: We use the chain rule: g'(5) = 1 1 1+ln2(s2+1).32+1  (23) ((1) Find the second derivative of Solution: We use the chain rule:
f'(m) = 2cos(2m)esjn(2z) f"($) = (4 cos2(2:c) — 4sin(2x))65in(2z) (e) Evaluate the following limit, if it exists _ sing:
11m
r—~0 tan 3m
Solution: Call this limit L. Then
3
L: lim c?” z 'sina:
1—*0 Sll'l 3m . 3:1: sina: 1
: hmcos3x~—.————
:—’0 sm 3x a: 3 DJIH :1.1.1. (f) A bead is moving on a straight wire. It’s position a: is given as a function
of time t below:
$(t) = t3 + t + 1 Find the value a > 0 so that the average velocity of the bead over the interval
[0,a] is equal to 5.
Solution: The average velocity is :1:(a) — 22(0) # a.3 + a a (l Therefore a : 2. (g)Suppose the function f is deﬁned as f(a:) : 0 if a: < 0 f(z)=Azz+Bif:1:_>_0 for constants A and B. For what value or values of A and B is f differentiable
for all 2:? Solution: For f to be differentiable it must be continuous. Thus 0 = limxnm f(2:) = f(0) = B, so B = 0. For f to be differentiable we must
have 0 : lim,,.0+ f’(a:) = f’(0) = A  0, or A 0 = 0. This puts no restriction
on A. Thus B : 0 and A can be any value. (h) Find the tangent line at (1,1) to the curve msin(zy  yz) : {1:2 — 1. Solution: We differentiate implicitly.
dy dy  _ 2 _ 2 _ __ _ __ :
sm($y y )+a:cos($y y ) (11+ xdz Zydx) 2:5
Therefore at (1, 1) we have
dy dy
1 — — — : 2
+ dx da: and thus dy/da: = —1 when :1: 2 1. Therefore the equation of the tangent line
isy:1—(a:—1). (i) Indicate on the graph below where the estimates 2:1 and $2 to the root of
f (x) will be located if calculated using Newton’s method starting with initial
estimate (guess) $0. y! y: ICU) (j) Determine the slant asymptote of 31:3 +2.7:2 — 1 f(.'1:)= $2~4$+3' Solution: We use polynomial long division to determine that 472: — 43 =3 ——.
f($) 1+14+$2_4$+3 Therefore the slant asymptote is given by
y = 33: + 14. (k) If f(l) : 2 and f’(1) = 3 use a linear approximation to estimate the value
of f( 1.15). Solution: f(1.15) 2 2 + 3(0.15) : 2.45. (1) In the previous question, if it is known that If”(:1:) < 2 for all :12, what is
the maximum error in your approximation? 1
Solution: Error 3 5  2 (0.15)2 = (0.15)2. (m) Determine the ﬁrst three nonzero terms in the Taylor series based at
:1: = 0 for
cos(2:l:). Solution: We ﬁrst compute the series for cost: :1:2 3:4
:1— — — —
cosm 2 + 4!
Therefore 2 2 4
2
cos(2a:)=1(:) +(Z) ... and the ﬁrst three nonzero terms are thus
1 — 2x3 + (2/3)::;4 — . .. (n) Evaluate . cos(2:1:) — 1 + 211:2
km _————.
1—»0 (1:4 Hint: The previous problem will be helpful here.
Solution: Call this limit L. First we use from above that cos(2:1:)=1— 23:2 + (2/3):1:4 + . . ..
Therefore cos(2n:) — 1 + 2m2 2 (2/3)a:4+ higher order terms. Therefore 4 5...
L: m (2/3»: t“ )
1—)0 :1: = 2/3+0 = 2/3. 2. Theory Questions.
(a) Evaluate the derivative of f (9:) : x/a: + 1 using the deﬁnition of the derivative,
_ f (a? + h)  f(m)
Solution: We will multiply by the conjugate. vm+h+ —\/m+1v—_____a:+h+1+\/m+1 :17 =11m——h—
ﬂ) heo ¢m+h+1 +¢aT
— —1im ($+h+1)_ (14—1)) =lim h h—v0h(\/$+h+1+\/IE+1hﬂoh(\/$+h+1+\/$T) And evaluating we have
1 I f (m) ‘ war
(b) Consider functions f(z) deﬁned on the interval [0, 1] that have the
following properties:
(i)f is continuous in the interval [0, 1].
(ii) f is differentiable in the interval (0, 1).
(iii) f(0) = 0 and f(1) = 1.
Find the function deﬁned on the interval [0, 1] that has these properties that
makes the maximum of ]f’(:z)] on the interval (0, 1) as small as possible. Hint:
begin by sketching some graphs that satisfy the above properties.
Solution: f(z) : x. Here’s the proof. If f(a) < a for any a then f’(c) > 1 for
some c between a and 1, by the mean value theorem. If f (a) > a for any a
then f'(c) > 1 for some c between 0 and a, also by the mean value theorem.
Therefore only f (as) 2 x will minimize the absolute value of the derivative. Word Problems 3. A bead is constrained to move along a wire bent into the shape of the graph
y = sin :12. At a certain instant in time the bead is at :1: = 1r/4 and dar/dt = 2.
Find the rate of change of the distance between the origin and the bead at this
instant in time. Solution: The distance to the origin of the point (z,y) is «11:2 + yz. So if the
point is on the curve y = sinx then the distance is given by f(m) : V152 + sin2 2:. We are asked to ﬁnd df/dt and we use the chain rule.
df df dx _:1:+sin:ccosm dz E 2 E 25 _ W i 3'
At the instant cc 2 7r/4, dm/dt = 2, and thus
df _ 1r/4+1/2 — _ —————— 2.
dt 7r2/1S+1/2 4. Find the point or points on the graph y 2 x2 closest to (0, 3/2). Show that
you have carefully checked all possibilities. Solution: Note that minimizing the square of the distance is equivalent. Thus
we minimize f(a:) 2 2:2 + (:1:2 — 3/2)2. We compute f’(x) = 21: + 2(12 — 3/2)  21: = 43¢»? — 1). Thus the critical points are :1: 2 0, a: 2 1 and a: 2 —1. We evaluate and ﬁnd
that f(0) 2 9/4 and f(1) 2 f(—1) 2 5/4. Furthermore we note that lim f (:5) 2 lim f (as) 2 00. Therefore the minimum is obtained at the two
I—‘m IH—m points (1, 1) and (—1, 1). 5. A 1 kg model boat is put into water at rest and its engines are turned on.
Let v(t) denote the velocity of the boat in m/s, with t in seconds after the
engines are turned on. It is known that v(t) solves the differential equation d1) 1) d_t_ _E‘ (this equation represents mass times acceleration equal to net force of the
engines and friction). ' (a) Solve the differential equation for t. (b) Evaluate 31.120120). Solution: We note that Q 2 —1/10(v ~ 10). dt
Let u 2 v — 10. Then we note that
du dz)
— 2 — 2 —1 10 .
dt dt / "
This is an exponential growth differential equation and has solution
u(t) 2 nos—”10. We also know that 110 2 0 and so no 2 —10. Thus v(t) 2 10 — 1064/10.
Therefore tlim v(t) = 10.
6. Sketch the graph of
Hi) = we‘r’ showing all of the following if they are present: (i) yintercept (ii) 1 intercept (iii) critical points, local maxima and minima, intervals where f is increasing
or decreasing (iv) inﬂection points and intervals where f is concave upward or downward (v) asymptotes (horizontal, vertical and slant) Give the (x,y) coordinates for all of the points listed above. Solution: f (0) = 0 is the only root and the only place where a: = 0. To ﬁnd the
critical points we compute f’(:z:) = (1 — 2m2)e—I2. Therefore the critical points are where (1 / ﬂ, 6'1/2 / ﬂ) and (—1/\/§, —6_1/2/\/§). When a: < —1/\/§, f is decreasing. When —1/\/§ < :r < l/ﬁ, f is increasing. Lastly we note that when :r > l/x/i, f is
decreasing. Therefore (1 / ﬂ, 6—1/2 / ﬂ) is a local max and (—~1/\/§,—e”1/2/x/§) is a local min. Now we inspect the second derivative,
f”(a:) : e"Iz(—6m + 413) Therefore the possible inflection points are :1: = 0, a: 2 J37? and a: = ~m.
We note that for at < —\/§—/_2, f is concave down. For —\/IV§ < a: < 0, f is
concave up. For 0 < a: < \/3/—2, f is concave down. And for :1: > m, f is
concave up. Therefore the inflection points are (M,°\/3/—26~3/2) and (—\/3’/—,—\/2We—3/2). Finally we note that
lim f($) : lim f(:1:) = 0.
(ta—00 1—900 Therefore there is a horizontal asymptote of y = 0. Thus the graph should
look like this: —‘/2
(—L)"L'e > {OCctl WMX The End ...
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