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Unformatted text preview: Mathematics 222, Spring 2007 Lecture 3 (Wilson)
First Midterm Exam March 1, 2007 ANSWERS Notice that some problems can be done in various ways, and I have given only one possible answer.
Also remember that answers could be correct despite looking quite different from those given: In
particular this applies to indeﬁnite integrals, where the “——C” can take many forms. For example
sin‘1 6 = — cos’1 6— a constant that could be included in ——C. Broblemi
Evaluate the integrals: 2
(a) / ac ln(m) dw
M1—
AN E :
We can do this using integration by parts. If we let u = lnac, then du = idm. Now do has to
be whatever is left to make up the integrand, so do 2 ac dw, hence 1) = Using the integration
2 m2 2 2 m2 1
by parts formula fudv 2 1w — fvdu, we have /1 a; 111(00) d1; = —/1 1 = [anacxgﬂf— = (21112— am) — (2— i) = 21112— g. (b) cos3 dw ANSWER: This odd power of the cosine can be integrated by keeping one copy of cosine to go with due and
replacing cos2ac by 1 — sin2 1;, then (if you want to be formal) substituting u = sinac so du = cos(ac)d1;. We have /cos3(1;) dac = /(1 —sin2 1;) cos(ac) d1; = /cos(ac) dm—/sin2(ac) cos(ac) d1; = sin(ac) — g si113(ac) + 0. (Note that this answer can be written in other ways that don’t look at all the same, using trig identities to produce an answer that differs by a constant which is swallowed in C . ) BroblennZ
, 5w — 3
Evaluate the integral / — d$.
(m + l)(w — 3)
ANSWER:
This seems to call for a partial fraction rewriting of the integrand. We know we can write % due 2
ﬂ + 00—133, for some constants A and B. Multiplying both sides by (m + l)(w — 3) we have 535 — 3 = A(1; — 3) + B(ac + 1). Expanding we get 51; — 3 = (A + B)ar + (—3A + B). Equating the 1; terms we
have A + B = 5, and from the constants we have —3A + B = —3. You can solve these for A and B
in several ways. One way: Multiply A + B = 5 by 3 to get 3A + 3B 2 15. Add that equation to
—3A + B = —3 and you get 4B 2 12, so B = 3. Putting that in A + B = 5 gives A = 2. — 2
So now we know / L due 2 / dm+ / 3 dw. Each of those integrals is of the form
(ac+1)(ac—3) JI—l—l 93—3 fidu, so the answer is 2lnlw+ll +3lnlm—3] +0. Problem 3 A parabola centered at (0,0) and opening upwards goes through the point (—4, 1). Find an equation for this curve. What are the coordinates of its focus? (Write out the equation and
the coordinates explicitly!) ANSWER: We know we can write the equation for a parabola that is in standard position and opens upward as y 2 41—10002, for some number 19. But since (—4, 1) is on the curve, 1 = ﬁ(—4)2 = %. Hence p = 4, and the equation for the parabola is y = 1—16352. We also know that a parabola in this position has its focus at (0,1)) on the y—axis, i.e. at (0,4). Problem 4
i f (w)
1 3
A function f obtained from real—world measurements takes on these values: 7 2
T 3
T 1 4
Estimate / f d$ using one of our numerical integration techniques, Simpson’s Rule
1 or the trapezoidal Rule: Be sure to specify which you are using! ANS W E B: We have the interval [1,4] divided into n = 3 subintervals. Simpson’s rule only works for an even
number of subintervals so we have to use the trapzoidal rule. (Since most people think that is easier
to use, I suspect most people taking the test would have gone this way even without that argument!)
Our subinterval end points are $0 = 1, m1 = 2, $2 = 3, and $3 = 4. The corresponding function values
are f(1) = 3, f(2) = 2, f(3) = 3, and f(4) = 1. The length of each subinterval is Ax = 1. Using the
trapezoidal rule we want to calculate A290 I 2f(2) : 2f(3) : = %(3+4+6+ 1) = %>< 14 = 7. Problem 5 2
Find parametric equations ac = f (t) and y 2 g(t) describing motion along the hyperbola —% + if = 1,
such that the point (1;, y) is at (0, —2) when t = 0 and it moves to the right as 75 increases.
ANS W E B: We recall that sect and tant satisfy the identity sec2 t = 1 + tan2 t, or equivalently sec2 t — tan2 t = 1.
2
If we let sec2 75 = y? and tan2 7: = % then the equation for the hyperbola will be satisﬁed. That means we can use in = :4 tant and y = :2 sect, where we still have to Choose the signs.
Putting in t = 0 has to give us in = 0 and y = —2: Since sec 0 = 1, this forces us to pick the — sign for
y y = —2 sec 75. Now as 7: increases from 0, tant also increases. We want 1; to be increasing, so that the point moves to the right, so we have to choose the + sign for m. Hence the parametric equations
are 00 = 4tant and y = —2sect. lm For the curve 7“ = 4sin 36, ﬁnd the slope where 6 = g. The
plot to the right shows roughly what this curve looks like:
You must calculate the slope using derivatives to get credit. AN E
We use the formula 3—: = W with f(6) = 7" = 4sin 36. Then f’(6) = 12 cos 36. At the point where 6 = %: sin6 = cos6 = g, f(6) = 4sin36 = 4sin§T7r = 2\/§, and f’(6) =
12 cos 36 = 12 cos :37” = —6\/§.
Putting those numbers into the formula we note that the sin 6 and cos 6 factors in the numerator f’(9)+f(9) :
f’(9)f(9) and denominator are all g and so they cancel out. That leaves us with 3—5 = 76312772312 _ i4 _ 1
4512—2312 —8 2‘ Find the points where 7“ = 1 + cos 6 and 7’ = 1 — cos 6
intersect. You can use the plot at the right as a check of
your work but you must show how you calculate the spe—
ciﬁc coordinates of each intersection point: Just reading the intersection points from the plot will not receive any credit. lb
9 ANSWER: First we try setting the two 7’ values equal. We get the equation 1+cos 6 = 1 —cos 6 or 2 cos 6 = 0,
so cos 6 = 0. That occurs for 6 = :5 We put that into each function and ﬁnd they both give
7’ = 1 — 0 = 1 so each curve passes through the point (1, and through the point (1, —g),
conﬁrming the two points of intersection the picture shows on the upper and lower y—axis. Now it appears both curves go through the origin which we think of as (0, 0), but that does not
work in either equation! But 7“ = 1 + cos6 takes on the value 0 when cos6 = —1, e.g. when
6 = 7r, and 7" = 1 — cos6 gives 0 when cos 6 = 1 e.g. when 6 = 0, so the origin i_s on each curve
just for different 6 values. So the origin, whether labelled (0, 0) or (0, 7?), is an intersection point forjhemmrlesr Problem 7 (a) Sketch the curve % + = 1. Be sure to show where it crosses the w—axis and/or the y—axis (write out the coordinates!), and where its foci are (write out the coordinates!). Your sketch will
not be graded for drawing ability but should resemble the correct curve. ANSWER: Here is Maple’s plot of this ellipse. The curve crosses the ac—axis where y = 0, so 1'2 = 9 and
as = :3. Similarly the y—intercept is where y = :5. In our usual notation a is the larger of those
sizes, 5, and b is the smaller, 3, so the distance from the center to a focus is v25 — = 4. Hence the intercepts are (::3,0) and (0, :5), and the foci are at (0, :4). (b) The equation 4.252 + 2\/§my + 2312 + 10/335 + 10y = 5 describes a conic section. (An ellipse,
parabola, or hyperbola, not a degenerate case such as a line or point.) (i) Find an angle 6 such that rotation of the coordinate system by 6 would eliminate the any
term. (You do not need to carry out the rotation!) ANSWER:
This equation has the form 14352 : Bmy : Cy2 : Dan : Ey : F = 0 where A = 4, B = 2V3,
C = 2, and the rest don’t matter for this problem. We can rotate by any angle 6 that ' — A—C — i — L — L — ' ' — ﬂ
satisﬁes cot 26 — 7P — 2% — ﬂ. If cot26 — V3, tan26 — So one ch01ce 1s 26 — 3, in which case 6 = 6 (ii) Which kind of curve (ellipse, parabola, or hyperbola) is this conic section? ANSWER:
We can use the discriminant B2 — 4A0 = (265V — 4 x 4 X 2 = 12 — 32 = —20. Since that
is negative, the curve is an ellipse. Problem 8 m3 d$ a Evaluate the inte ral / —.
( ) g #1 6 _ m2 ANSWER: It is possible to do this with a “u substitution” that does not involve trigonometry, but the
difference of squares suggests a trig substitution so I will do it that way. If we draw a triangle
and label it as 4 1/16—m2 we are led to 1; = 4cos6 so d1; = —4 sin 67 and 1/ 16 — x2 = 4sin 6. Thus the integral becomes (4 COS 9)3  3 3 2
/W(—4sin6)d6 = —/64cos 6d6. We separate cos 6d6 as (cos 6) cos6d6 and use
in cos2 6 = 1—sin2 6, and have —64/(1—sin2 6) cos6d6 = —64/cos6d6—1—64/sin2 6cos6d6. We use the substitution u = sin6 on the second integral and get —64 sin 6 + % sin3 6 + C . From the
triangle we read sin6 : V 1647332 and substituting that in gives —16\/ 16 — x2 + %(16 — x2)% + C. , 00 due 00 d$
(b) One of the integrals / T and / —3 converges, and the other does not. Evaluate the one
1 SB 1 an that converges. ANSWER:
00 due We could try evaluating each to see which one converges. But each is of the form / —p, which
1 m we know converges only if p > 1, whereas we have to choose between 1) = % and p = 3. So we go 0° dac
with p = 3 and evaluate / —3.
1 1;
. . . . . . . b dw . .
To evaluate this improper integral we need to set it up as a limit, liinlH00 / —3. The integration
1 1;
b
 i 72 _ i i _ i i  . . 1
(power rule) yields 7235 L — 72w — f2 — 5 <1 — b—g). Taking the limit as b —> 00 we get 5 as the answer. ...
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This note was uploaded on 05/15/2008 for the course MATH 222 taught by Professor Wilson during the Spring '08 term at University of Wisconsin.
 Spring '08
 Wilson

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