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Unformatted text preview: MathematicanZZ Lecture 4 Wilson
Final Exam May 11, 2005 Answers Probleml
Let 6227+2f—kand172 —z+k. —» (a) (b) Calculate 17 ~ 17.
ANSWER: 1717: 2 x (—1) +2 >< 0+ (—1) x 1 = —3. Find the angle 6 between 17 and 17. ANSWER: cos6= "177'. 17: v4+4+1 = 3 and 17: \/1+1 = ﬂ, so cos6= % 2 —§ and6= 3—”. El E Find the vector projection of 17 on 17. ANSWER: The vector projection of 17 on 17 can be calculated in several ways. Thinking it
through, instead of just using a formula7 we are looking for a vector lying along 17 whose j_—g
u magnitude is the scalar projection of 17 onto 17. The scalar projection of 17 onto 17 is ITJIJ which is :72 or —%\/§. A unit vector in the 17 direction is $17 which is £17 or —§17+ gig. Putting
these together we get the vector projection to be :7; X (—gT—i— g) which simpliﬁes to gf— gig. Calculate 17 X 17. ANSWER: Set up and evaluate the determinant —’ 1’ 37 k a
17X17=det 2 2 —1 =2z7—j7—l—2k.
—l 0 1 BroblenLZ Evaluate the integrals: (a) V9 — x2
/ —$2 dx ANSWER: I would draw and label a triangle as above. Now we can set x = 3cos6 and
V9 — x2 = 3 sin 6, so dx = —3 sir16, and substitute into the integral to get /(3 s1r16)( 3sin6d6) : /s1n edg : /ta1126d9. 9 cos2 9 cos2 6 Next use tar129 = se029 — 1 to get — /(sec29 — 1) d6 = —tan§ + 6 + C'. Now we need to x/9—x2 convert back to 36 instead of 6: We can read tanQ from the triangle, and get —— + x
V9 — .762 arcsin T + C. (There are other, equivalent, ways of writing the answer.) /e_$ cos(2x) dx ANSWER: This begs for integration by parts. If we let u = cos(2x) and do = e’xdx,
du = —2sin(2x)dx and v = —e_$. Using the integration by parts formula we have M) — f v du = —e’33 cos(23c) — 2 / 6’3“" sin(2x) dx. We apply integration by parts again to this latter integral, with u = sin(2.7c) and do = 6—wdx so du = 2cos(2x)dx and v = —e_w. Now we
have —e’$ cos(23c) — 2(uv — /vdu) = —e’33 cos(2$) — 2(—e’$ sin(2.7c) + 2/6700 cos(2x)dx).
Expanding this and recalling the original integral we have f 6—9” cos(2$) dx = —e_w cos(2x) + 26"” sin(2x) —4 f 6’95 cos(2x) dx. Solving for the integral, and noticing that we need “+0” that 1
got omitted in going from do to v, we get /e_$ cos(23c) dx = 56—00 (— cos(23c) + 2 sin(2x)) +0. Problem 3
(a) For each series: Tell whether it converges absolutely, conditionally, or not at all, and justify your answer. (i) (ii) (iii) 00 Z(—1)"e<1/"> n:1
ANSWER: This is an alternating series and the terms are decreasing, but they are decreasing
with limit 60 = 1 and not 0. Hence this series does not converge. oo (_1)n
£11101 + 1) ANSWER: The denominators are getting arbitrarily large so the fractions decrease with limit
zero, so this alternating series converges. But if we make the signs positive we have a series
whose terms are larger than i and so diverges. Hence this series converges conditionally. ANSWER: If we take the absolute values of these terms, the numerator is never greater than
3
1% and the denominator is 715. That series converges by comparison to the p—series with terms
1 , which converges since p = g > 1. Hence this series converges absolutely. Eel (b) Find the radius and interval of convergence for the power series Z ( 2 23m .
71:1 71 ANSWER: Taking the absolute value of the ratio of succesive terms a”“, and simplifying alge— an
braically, we get §$— "—H .The limit of this 1atio as n —> 00 is —x so the se11es converges absolutely
n3 3 3 when gm < 1, i.e. —§ < x < 5. Hence the radius of conve1gence is 5. Now we need to check
whether the series converges or diverges at the ends of the interval. At x— — —— the series amounts
to Z(—1)”n—12, and at 36 = g the series amounts to Z %. Since the second of those is a 19— series with p = 2 > 1 it converges, and the second series is just the absolute values of the ﬁrst so both 3 3 converge. Hence the interval of convergence is [—5, 5]. BroblemA
Label three points in space as P = (2,1,3), Q = (2, 2, 5), and R = (1,1,6). (a) Find an equation for the plane passing through points P, Q, and R. ANSWER: First we ﬁnd a vector 71’ perpendicular to the plane. We do this by ﬁnding two
vectors in the plane and constructing their cross product. For vectors in the plane I will use
the vector from P to Q and the vecto1 f1orn P to R, but other pairs would wo1k also. If
we let u be the vector from P to Q to =01 + j + 2k: Similarly the vector from P to R is
v: —z+0j+3k: Nown=u><27 231—2]+k Next we ﬁnd the plane perpendicular to 71 passing through any one of the points P Q or R:
I will use P. The equation for this plane is 3(36 — 2) — 2(y — 1) + 1(2 — 3) = 0 which simpliﬁes
to 3x—2y—1—227. (b) Find equations in both parametric and symmetric forms for the line that goes through Q and
is perpendicular to the plane you found in (a). ANSWER: We already know a vector 7? in the correct direction for the line, so we can im— mediately w1ite down the pa1arnetric equations x— — 3t + 2, y— — —2t + 2, and 2— — t + 5. In symmetric form we have 9%2 = y—__22 : 2—15. Problem 5 Consider the two circles 7’ = 3 sin 6 and 7“ = — 3 cos 9: Find the
area of the region which is inside both of these circles.
ANSWER: The shaded region at the right has the area we
need to calculate. It may look symmetric about the tilted line
connecting its “corners”, but note that one circle curves rnore
sharply than the other: we really need to calculate the whole
area as the sum of two integrals. The two circles intersect where 3 sinQ = —f = g, tanQ— — %, so 6: 7r. The “upper”
part of the shaded region, bounded by the smaller circle, starts with 6— — g and continues to 5%, while the remaining part goes from 5% to 7? bounded by the larger circle. The area we need can be computed then as 21% 3cos6)2 d6+— :/_: (3si116)2 d6: g/T?COS26d6l—:W 5lsin 2.6d6 2 6 Using the half—angle formulas for sin2 9 and cos2 6 this can be evaluated as f — i. Pr 1m (a) Find the solution of the differential equation y” — 24y + 169 = 0 that satisﬁes 34(0) 2 —1 and
y’(0) = 17. ANSWER: The auxiliary equation r2 — 247” + 169 = 0 has roots T = 24:— V2’100 2 12 :: 52'. Hence the general solution is y(x) = 612$(01 cos 5x+ 02 sin 5x). Taking the derivative, y’(x) =
6123(1201 cos 5x + 1202 sin 5x — 5C1 sin 5x + 502 cos 5x). Putting in the initial conditions we have Cl 2 —1 and then using that fact 17— — —12 + 502 so 02: . Thus the desired solution is
y(x )2 612x(— cos 5x +23 9sin 5x). or e l eren 1a equa ion — — = , e roo s o e aux1 iar equa ion are n = —
b F th d'ff t' l t' y” 3/ 6y 0 th t fth 'l' y t' 2
and 7’2 2 3. Find all solutions of the non—homogeneous equation /I_ [—6 __1063x
LL; ANSWER: From the roots of the auxiliary equation we know the homogeneous solution is
yh : 016’233—1—0263‘73. We need to ﬁnd a particular solution 3),, to the nonhomogeneous solution.
Since 3 is a single root of the auxiliary equation we try yp = 0366393 for some constant C'. Then 206390 + 30966390 and y”— — 60639” + 90966390 Putting those back into y” — y’ — 6y we get
51287639”, so C' must be —2 in order to give —10€3’”. Hence yp = —2xe3w and the complete
solution is yp + yh = —2xe3x + 015233 + 026393. Problem 7
Evaluate; /2 dt
,2 344 — t2
ANSWER: This is an improper integral: The integrand “blows up” with denominator zero at each 0 dt ’7 dt
end of the interval of integration. We rewrite the integral as filing+ / m + bliI2Il_ /0 m. (You could break at some point other than 0.) The integral f \/— gives a1csin —, eithe1 f1om a b
formula or by drawing a triangle. Thus we have lim (arcsin(0) — arcsin(g)) + lim (arcsin(§) — a—>—2+ 2 b—>2*
arcsin(0))= (0 — (—5)) + (g _ 0) : Problem 8
The position vector of a particle is given at time t by F(t) 2 (2t — t2)?+ tj—l— etk‘. (a) (b) (C) Find the velocity vector 17(25) as a function of t. ANSWER: 50:) = F’(t) = (2 — 2107+ j+ 6t]; Find the acceleration vector 5(t) as a function of t.
ANSWER: 50:) = 17’(t) = —2r+ 6%. Find equations for the tangent line to the path of this particle at the instant when t = 0. ANSWER: When 75 = 0, 27(0) gives 27+ f+ I; as a vector in the direction of the tangent line.
(Note: When I ﬁrst posted these I had an error here, I had 2]" rather than f.) At that instant
the position is given by 77(0) : I: i.e. the particle is at the point (0,07 1). In parametric form
the tangent line is x 2 225 + 0 2 225, y = t + O 2 2t, and 2 = t + 1. In symmetric form we have
Héd. 2—1—L Problem 9
A conic section has the following properties: 1. It is symmetric about the y—axis and the x—axis.
2. It crosses the y—axis at y = ::3.
3. It has foci at (O, :5). (a) (b) What kind of curve is it? (Circle, Ellipse, Parabola, Hyperbola?) How do you know? ANSWER: This must be a hyperbola since its foci are further from the center ((0, ::5)) than
the points where it crosses the axis they line on ((0, ::3)). Find an equation for this conic section. Your equation should contain only x’s and y’s and
numbers i.e. you should ﬁnd values for all parameters. ANSWER: The equation will be of the form —9;—: + if: = 1, since this is a hyperbola opening
up and down. Putting in the fact that y = ::3 when .76 = 0 we get a = 3. We know that the
parameters a b, and c satisfy a2 + b2 = C2 where c is the distance to the foci i.e. c = 5. Hence
we have 9 + b2 = 25 so I) = 4. Thus the equation is —% + y; = 1. Tell what other facts you can about this curve: If it is a circle be sure to include center and
radius. If it is an ellipse be sure to include how long the axes are and which is horizontal, and
where it crosses the coordinate axes. If it is a parabola be sure to describe its orientation, e.g.
opening to the left. If it is a hyperbola be sure to include equations for the asymptotes and
tell whether it opens vertically or horizontally. In any case give the eccentricity. As a hyperbola opening up and down the main things we can give about it are its eccentricity 3 and asymptotes. The eccentricity is g = g. The asymptotes are the lines y = :130. Problem 10 The terms of the Maclaurin series for sin(x) through
for 36 6 (—12,12).
How big might the error be in this approximation? Be sure to justify your answer using theorems ] l . ] . ﬂ '
ANSWER: You can answer this using either the remainder term from Taylor’s theorem or the
alternating series error estimation process. Since for any x the odd powers of x that appear in the
series will either all be positive or all negative, the alternating signs in the Maclaurin series for
sin(x) give an alternating series. The terms we are to use actually go through 3377 since the term in the sin(x) Maclaurin series with x8 has coefﬁcient zero: We can take advantage of this to sneak
some extra accuracy. ﬂ 8, are going to be used to approximate sin(x) _ f(”+1)(0)93“+1
_ (n+1)! f(”+1)(c) means the (n + 1)“ derivative of ﬁx) 2 sin(x) evaluated at some number 0 between 0
and x. In the case we care about, 71 seems to be 7: We can actually do better since the 81th degree
term is 0 and so the polynomial we are to use is actually the same as the one with n = 8. I will use n = 8 and hence n + 1 = 9. Since all derivatives of sin(x) are :: sin(x) or :: cos(x), the largest 1 1 9
that f9(c) can be is 1. For 95 E (—5, 5), the largest that x9 can be in absolute value is (g) = 5%.
1 Hence the size (absolute value) of R8(3c) is at most 512><(9I)‘ That is a satisfactory answer, or you I will use the Taylor’s theorem remainder term. That has the form Rn(x) where can approximate it as 5.3823 x 10—9. ...
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 Fall '08
 Wilson
 Calculus, Geometry

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