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Unformatted text preview: Mathematics 222, Spring 2008
First Midterm Exam Lecture 1 (Wilson) February 28, 2008 ANSWERS NOTE: Many of the integral answers that include trigonometric functions could be written in other
forms, using various trig identities. l have not tried to show all of the correct versions! Problem 1 (18 points) Evaluate the integ rals: (a) Problem 2 (a) e233 sin(x) dd. ANSWER: This looks like a candidate for integration by parts. You could either let u = 6203 and do =
sin(x)dx or let u sin(3c) and do = e2$do I’ll make the ﬁrst choice. If u = 62””, then
do = 262wdx. With do = sin(3c)dd, o = — cos(3c). So the original integral can be rewritten as
2st —e2$ cos(d) + 2/6200 cos(d)dd. This latter integral needs parts again. I will choose u = e again, with do = cos(d)dd, so do = 262xd5€ and o = sin(x). Using these on the above partial result, we get /€2w sin(d)dx = —e2$ cos(3c) + 26290 sin(x) — 4 / e2” sin(3c)dd. Now add 4/6203 sin(d)dx to both sides and get 5/6200 sin(x)dx 2 —e2"’3 cos(3c) + 262": sin(.7c), divide by 5, and remember this is an indeﬁnite integral so we need the constant7 to get
1 2 /e2$ sin(x)dx = —ge2x cos(3c) + 5620: sin(x) + C. /sec4(x)dx. ANSWER: I will separate sec4(d) as sec2(x) sec2(d) and use the identity sec2(x) = 1+tar12(x) on one of the
factors. We have /sec4(d)dd = /(l+tanw(x)) sec2(d)dd = /sec2(x)dx + /tan2(d) sec2(x)dx.
If we remember that the derivative of the tangent is the square of the secant the ﬁrst is trivial. The second is not far behind: We can let u = tan(d) and have do = sec2(d)dd. So we get tan(d) + 13 tan3(x) + C’. (18 points)
2 2x dd
Evaluate —.
0 v4 — $2
AN E :
The ﬁrst thing we must notice is that this is an improper integral. At x = 2 the denominator
fth 't d t s t 1' '1; /2 2mm 1' b 2mm
0 e in e ran goes 0 zero. 0 we mus use a 1m1 , e. . — = 1m —.
g 0 g 0 vii—$2 b—>2’ 0 vii—362 2% dx _ V4 — x2 —
b —/u_%du = —2u% + C'. So we must evaluate lim [—2v4 — ﬁlo. At x = O the quantity b—r2’
inside the brackets gives —2\/Z = —4. At 36 = b it gives 2V4 — b2 and as b —> 2— that has
limit 0. So the answer is O — (—4) = 4. 2 Now 2x is almost the derivative of 4 — x , so we can let u = 4 — x2 and have 96 f (96)
3 2
. . . . 3.5 1
(b) Five measured values of a function f are glven 1n the table at the 4— 3
5
right. Use the trapezoidal rule to approximate / f dx. You do not 4 5 4
3 _
need to calculate an error estimate. 5 6 AN E : We can do this straight from the formula for the trapezoidal rule. The interval [3, 5] is divided
into four intervals, each of length Ax = The values of f at the division points are the
nu5mbers in the second column of the table. So we have /3f(x)dxm %(f(3)+2f(3.5)+2f(4)+2f(4.5)+f(5)) = :(2 : 2 : 6 : 8 : 6): 244 =6. Problem 3 (18 points) L 2dx
a E aluate /N§ —.
( ) V 0 V1 — 4x2 ANSWER: This might look like an improper integral, but the quantity inside the square root sign will be   2 2 1   1  1 1 a
pos1t1ve so ong as 436 < 1, 36 < 1, or for pos1t1ve values of x, 96 < 5. S1nce m < 5 we aie safe!
I will use a trigonometric substitution to evaluate this integral7 starting by drawing a right
triangle: From the triangle we infer 2x = cos 6, so x : écosd and dx = —% sinddd, and 2:1: sinQdQ
\/ 1 — 4x2 = sin 6. Hence we can rewrite the integral as — / , 6 = — / d6, with the new
sm
limits of integration yet to be found. Looking at the triangle again if x —> 0, the triangle
tends toward being just two vertical lines7 so 6 —> At x 2 2—3/5, the two legs of the triangle each become g, so the triangle becomes an isoceles right triangle and 9 = %. So we now have (‘0) Evaluate / x2 111(96) dx. ANSWER: We use parts again: If we let u = ln(x) then du =
i 933
3 3 dx will cancel against part of 1). Since 30. l
1‘
2 f 0; dx = 9033 11m) 13
9 and the integral becomes . d1) = dex, v = Problem 4 (10 points)
A hyperbola in standard position, its center at the origin and its foci on one of the coordinate
axes, passes through the (rectangular coordinates) points (0, :12), and it is asymptotic to the lines
12
= ::—$.
5 (a) Problem 5 Find an equation for this hyperbola. ANSWER:
Since its foci are on the y—axis, the equation will have the form —‘:—: + 2—: = 1. The points where it crosses the y—aXis will be (0, ::a), so a = 12. The asymptotes will have slope : a so IE,
g i _
5 144 — CL b = : Since we already know a = 127 I) must be 5. So the equation is —% + Where are the foci of his hyperbola? (Give speciﬁc coordinates.) ANSWER:
The foci will be at (07 :0) where c2 = a2 + b2 = 144 + 25 = 169. So 0 = x/ 169 = 13 and the
foci are (0, :13). What is the eccentricity of this hyperbola? AN E The eccentricity is i = E 12. Note this is greater than 1, appropriate for a hyperbola. (10 points) A curve is described by the equation (a) (b) 00—2 2
y__
100+36—1' What kind of conic section is this curve? AN E This is an ellipse we can tell from the form of the equation. Graph the curve: You should label and give speciﬁc coordinates for any places where it crosses
the x— and y—axes, foci7 etc. Your graph must not have sharp corners where it should really
be smooth7 but otherwise you won’t be graded on artistic ability. ANSWER: Here is a graph as produced by
Maple. The foci are shown at
(:8, 0), the x—intercept points (for— mally these would be called the
ends of the major axis) are (:10, 0),
and the y—intercept points (the ends
of the minor axis) are (07 :6). Um: (c) What is the eccentricity of this curve? ANSWER:
In the usual notation, c = 8 (from the foci) and a = 10 (from the ends of the major axis), so
the eccentricity e = g = % = g. Problem 6 (12 points)
An object is intended to move so that its position at time t, for 0 S t g 27T, will be given by x = 3cost and y = 5sint. (a) Describe this motion using words and/ or an equation in rectangular coordinates. Be sure to
tell how far up/ down and how far left / right the object would move and to tell what kind of
conic section it follows. Tell where it starts7 what direction it moves7 and where it ﬁnishes. ANSWER:
Since g 2 cost and g = sin t, and cos2t + sin2t = 17 we have 389—2 + g—: = 1 for any value of t.
Hence the motion follows along that ellipse. It extends left and right between (:37 O) and up
and down between (0, :5). When 75 = 0, i.e. at the start, the position is 96 = 3cos0 = 3 and
y = 5sinO = 0, so the object starts at the farthest rightmost point (3,0) where the ellipse
crosses the x—axis. As 13 increases from 0, cost decreases so x decreases, and sint increases so
y increases. Hence the object is moving upward and to the left along the ellipse from that
7 starting point, and it continues on around the ellipse going counter—clockwise. When 23 reaches 27r the object stops, having made exactly one trip around the ellipse, where it began, at (3, 0). (b) At the instant when t = E, the forces holding the object on that curve fail and the object ﬂies
away along the tangent line to the path it had been on. Find an equation for that tangent line. ANSWER:
We need to ﬁnd two things in order to construct the equation for the tangent line: Where was 4 the object at that time and what is the slope of the tangent line. “Where” comes from the values of 96 and y at the time t = E: 36 = 3cos§ and y = 5 sin g, so the object is at the point 4 d_y (3%, The slope at any value of t is given by 2: = (3:; = fg‘s’isntt. When t = g both cost and sint have the value g, which will cancel out in the derivative, so $1 2 —3. So now we need an equation for the line through (¥, with slope —§. We can write that as y — 5—‘5 2 —§ (x — 3—‘/§) That can also be simpliﬁed as y 2 —§ x + 5\/§ Below is a plot
2 3 2 ‘ 3 ' of part of the ellipse with that line superimposed, as a check. Note that l have plotted this
with different scales on vertical and horizontal axes in order to reduce the size of the picture! Problem 7 (14 points) Consider two curves given in polar coordinates:
(curve r = 1.
(curve (ii)) T = 2 cos 6. (a) Find the point(s) where these two curves intersect. Give both the polar and the rectangular
coordinates of each intersection point7 being sure to label which coordinates are polar and
which are rectangular. ANSWER: Here is a graph with both curves plotted.
There are two intersection points. If we set
T = 1 and T = 2 cos 6 equal and divide by 2
we have cos 6 = %, so 6 = ::E or some angle
differing from those by a multiple of 27T. At either of those values for 6 we have T = 1 \mmm 1
\ \\\\\«\\\\\ 1'5 since the point is on curve You can
check that at those values of 6 the other
curve has T = 2cos9 = 2 X % = 17 con—
ﬁrming that the points really are on both
curves. So in polar coordinates the inter—
section points are (1, and (1, To convert to rectangular coordinates we use sin g — 7, 96 =
_ 1 _ 1 _ M3 _ \/§ 1  i 1 __\/E
x — 1 X 5 — 5, and y — 1 X 7 — 7, so the coordinates are (§,__7). (b) At the right are two
graphs. Which one
(A or B) is curve
(i), and which one is
curve (ii)? ANSWER:
Curve (i), T = 1, is centered at the origin, so that is A, and curve (ii) is B. (c) One of the intersection points (and only one) is in the ﬁrst quadrant, where 36 2 0 and y 2 0.
At that point7 ﬁnd the slope and the tangent line (equation in rectangular coordinates) to curve (ii).
ANSWER:
The intersection point in the ﬁrst quadrant is7 in rectangular coordinates, ($7 We need to
ﬁnd the slope there, for the tangent to curve (ii) T = 2 cos 6. If we call that T = f(6) = 2 cos 6
we can use the formula derived in class for the slope7 3—: = W. At the intersection
ipoint: = T = 1. f’(6) = —2sin6 = 2singr = 2 X ‘ég = sinQ = cosQ = So the slope is dy_ ﬁx? I 1x:_ 1 _ 3N05774 dw —\/§><%—1><§ —\/§ 3 i . \/§ _ x/3 f The tangent line is y — 7 — 3 (x — which can be simpliﬁed as y = 73x + Below is a
plot of that line superimposed on the two circles7 as a check: \\\\\\ z\«\\«\«:\\\\\\\\\\\\\\\\\\\\\\\‘x\\\“ ...
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 Spring '08
 Wilson

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