ABC2-Solutions - Solutions to the exercise sets in The ABC\'s of Calculus Vol 2 PDF

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Solutions to the exercise sets in The ABC's of Calculus Vol 2 ...

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The ABC’s of Calculus, Volume 2 Solutions to Exercise Sets Revised September 10, 2019

Exercise Set 1.

1. P~Q = (−2, −1), P~R = (1, −3). So P~Q + P~R = (−1, −4). ~ −P~ (with points (insert word) as vectors) and so P~ R = (2, −1) − 2. P~R = R ~ = S~ − R ~ = (0, 1) − (2, −1) = (−2, 2). ∴ 2 P~R − (1, 2) = (1, −3), while RS ~ RS = 2(1, −3) − (−2, 2) = (4, −8) ~ + RS ~ = P~S (addition rule for vectors) so P~S = S ~ − P~ = 3. P~Q + QR (0, 1) − (1, 2) = (−1, −1). ~ − 2P~S = 3(Q ~ − R) ~ − 2( S ~ −P ~ ) = (−7, 8). 4. 3 RQ ~ − 5P~Q + QR ~ =Q ~ −S ~ − 5( Q ~ − P~ ) + R ~ −Q ~ = (12, 3). 5. SQ ~ + SQ ~ − QP ~ = (−1, −3). 6. SR ~ + 2SQ ~ − 3P~S = (2, 3) 7. QS ~ − QS ~ = (3, −5). 8. P~R + QR ~ − RP ~ + 3SQ ~ =Q ~ −R ~ − (P ~ − R) ~ + 3(Q ~ − S) ~ = (−5, −1) 9. RQ ~ + RS ~ − SP ~ + P~R = (1, −4) 10. QR p √ √ ~ − R| ~ = |(−2, −1)| = (−2)2 + (−1)2 = 4 + 1 = 5 11. |P~Q| = | Q ~ = |P~S|, | P~S| = |S ~ − P~ | = √2 12. ∵ |P~R + RS| √ ~ − |P~S| = 2|(S ~ − Q| ~ − |S ~ − P~| = 2 − 2 13. 2| QS| ~ = |2( Q ~ − P~ ) + 3(R ~ − Q)| ~ = √89 14. |2 P~Q + 3QR| ~ , |P~R| = | RP ~ | ∴ |P~R| − |RP ~ | = 0. 15. ∵ P~R = − RP

2 ~ | = |2 · ~0| = 0, since P~Q = −QP ~ . 16. |2P~Q + 2QP √ √ ~ = |3( ~R − P~ )| − 3|R ~ − Q| ~ = 3 10 − 3 13. 17. |3P~R − 3|QR| ~ | = 2|P~S| − 2|SP ~ | = 0, (Since |SP ~ | = | − P~S| = |P~S|). 18. 2|P~S| − |2 SP √ √ ~ + 2|P~R| − |P~Q| = 1 + 2 10 − 5 19. | SQ| √ ~ + SQ ~ − RS| ~ = | − RS| ~ = | RS| ~ = 2 2 (Since QS ~ = 0). ~ + SQ 20. | QS 21. v = (2, 3) → tan θ = b/a and (2, 3) is in Q1, so θ = arctan(b/a) = arctan(1.5) = 0.9828 radians. 22. (−1, 2) is in Q2 → θ = arctan(−2) + π = − arctan(2) + π = −1.1071 + 3.1416 = 2.0344 rads. ~ −P ~ = (−1, 0) − (0, 1) = (−1, −1) is in Q3. Thus, θ = arctan(1) + 23. P~Q = Q π radians. π = 4 + π = 5π 4 24. (2, −5) is in Q4, hence θ = arctan(−2.5) + 2π = − arctan(2.5) + 2π = 5.0929 radians. 25. ~v lies along y-axis, hence θ = π/2 radians. 26. (−2, 4) is in Q2 so θ = arctan(−2)+ π = − arctan(2)+ π = 2.0344 radians. 27. (3, −4) is in Q4 so θ = arctan(−4/3) + 2π = − arctan(4/3) + 2π = 5.3559 radians 28. P~Q = (0, 1) − (2, −1) = (−2, 2) is in Q2, so θ = arctan(−1) + π = . − arctan(1) + π = −π/4 + π = 3π 4 29. The point is in Q3 so θ = arctan(1) + π =

π 4

+π =

5π 4

30. Here the point is in Q4 so θ = arctan(−1) + 2π (insert word)

Exercise Set 2.

1. Yes. ~v · w ~ = (0, −1, 1) · (1, 1, 1) = 0 · 1 + (−1) · 1 + (1)(1) = 0 2. ~v · ~z p = (0, −1, 1) · (0, 1, −1) √ = 0 − 1 − 1√= −2. On the other √ hand |~v| = 02 + (−1)2 + 12 = 2 while |w| ~ = 2 also ∴ |~v||w ~ | = ( 2)2 = 2 so that ~v · ~z = −|w||~ ~ v|. Hence ~v, ~z are antiparallel (even though they are actually negatives of each other here). 3. ~x =

1 w ~ 2

so ~x is a multiple of w ~ of ~x, w ~ must be parallel.

4. No. (~v + w) ~ · ~z = (1, 0, 2) · (0, 1, −1) = 1 · 0 + 0 · 1 + 2(−1) = −2 so they cannot be orthogonal. 5. We want (α~v + w) ~ · ~z = 0. So this means that (α(0, −1, 1) + (1, 1, 1)) · (0, 1, −1) = 0 or ((0, −α, α) + (1, 1, 1)) · (0, 1, −1) = 0, i.e., (1, 1 − α, 1+ α) · (0, 1, −1) = 0. Evaluating the dot product we get 1·0+(1 −α)− (1+ α) = 0 or −2α = 0 which forces α = 0. Hence w ~ ⊥ ~z, or α~v + w ~ ⊥ ~z for α = 0. 6. Yes. (~v + ~z )·~z = ((0, −1, 1)+(0, 1, −1)) · (0, 1, −1) = (0, 0, 0)· (0, 1, −1) = 0. (Note that ~0 · ~z = 0 for any vector ~z.)

3 7. Look at the vectors P~R and P~Q. Since R is to be the midpoint of the line segment from P to Q it’s clear that | P~R| = 21 | P~Q|. On the other hand (draw a picture) we see that P~R = 21 P~Q too! Hence, (x − a, y − b, z − c) =   0 0 z1 −z0 P~R = 21 P~Q = 21(x1 − x0 , y1 − y0 , z1 − z0 ) = x1 −x , y1 −y from 2 , 2 2 y −y which x − a = 12 (x1 − x0 ), y − b = 1 2 0 , z − c = z12−z0. Solving for a, b, c we get the result. 8. This is similar to #7 previously except that we replace ” 12 ” there by ”α”. 9. −2. w ~ + ~z − ~x = (1/2, 3/2, −1/2) so ~v · (w ~ + ~z − ~x) = (0, −1, 1) · (1/2, 3/2, −1/2) = −3/2 − 1/2 = −2. 10. −7/2. ~v − w ~ = (−1, −2, 0), ~x +~z = (1/2, 3/2, −1/2) and the result follows. 11. This question should have read: Evaluate (x − 12 u) · (w + x). The solution is the scalar, 9/4. This is because 21 ~u · (w ~ + ~x) = 0 and x · (w + x) = 9/4. 3 4

while w ~ ·w ~ =3 √ √ √ √ 13. −4. Observe that ~v · ~z = −2, ~v · ~v = 2 and w ~ ·w ~ = 2 Yes, this is always true and called the Schwarz... (insert word) 12. −9/4. Note that ~x · ~x =

14. −3/2. Here ~x · ~v = 0, ~x · ~x = 3/4, ~v · ~v = 2.   15. 2. Observe that ~x + ~v = 12 , − 12 , 32 and (~x + ~v) · (~x + ~v ) = 11/4 while ~x · ~x = 3/4 √ 16. 17 √ 17. 2 13 3 4 18. ( , ) 5 5 2 2 1 19. ( , , ) 3 3 3 20. −i − 3 j 21. −18 22. −

15 4

√ 23. |u| = 14; u + v = (3, 8, 6); 3u + 4v = (9, 30, 25).

u − v = (3, −4, −8); 2u = (6, 4, −2);

Exercise Set 3.

1. We need to find a vector ~u in the direction of ~v. So let ~u = ~v/|~v| where  p √ |~v| = (−1)2 + 22 + 02 = 5. Then ~u = √15 (−1, 2, 0) = − √1 5 , √25 , 0 ∴ cos α = − √1 , cos β = 5

√2 , cos γ 5

= 0.

2. ~u is already a unit vector so its coordinates are its direction cosines. cos α = √1 , cos β = 0, cos γ = − √12 . 2

4 p √ ~ v 3. As before we let ~u = |~ where |~v| = 22 + (−1)2 + 12 = 6 so that ~u = v |   2 2 1 (2, −1, 1) = √ √1 , √1 . So, cos α = √ √ , − , cos β = − √1 , cos γ = √16 . 6 6 6 6 6

6

√ ~ v where |~v| = 02 + (−2)2 + 12 = 5. So, ~u, the unit vector 4. Here ~u = |~ v| in the direction of ~v, has direction cosines given by cos α = 0, cos β = − √25 , cos γ = √15 . p

p √ 5. Here |~v = (−2)2 + 32 + 42 = 29 so that ~u, given by ~u = √1 (−2, 3, 4), 29 is a unit vector in the same direction as ~v. It follows that cos α = − √2 , cos β = √3 , cos γ = √429 . 29 29 6. Since ~u is already a unit vector we know that cos α = √13 , cos β = − √13 and √ √ cos γ = √13 . So α = arccos(1/ 3) = 0.9553, β = arccos(−1/ 3) = 2.1863 √ and γ = arccos(1/ 3) = 0.9553 rads. Remember that α, β, γ must all belong to the interval from 0 to π 7. ~v = (−3, 0, 3) means that ~u = |~~vv | is a unit vector in the same direction √ as ~v where |~v| = 18 (why?) So, ~u = √1 (−3, 0, 3) means that cos α = 18 − √3 , cos β = 0, and cos γ = √318 . Using our calculator we get α = 18     arccos − √318 = 3π/4 = 2.3562 rads; β = π/2, and γ = arccos √318 = π/4 = 0.7854 rads.

√ √ Note: We could reduce 3/ 18 to 1/ 2 using the simplifications √318 = √3 = √93√2 = 3√3 2 = √1 but since we need to use our calculator anyhow, 9·2 2 this isn’t necessary. √ 8. ~v = (1, −1, 2) so |~v| = 6 and ~u = √1 (1, −1, 2) is in the same direction 6 √ √ as ~v. It follows that cos α = 1/ 6, cos β = −1/ 6, cos γ = √2 6 or α =       arccos √16 = 1.1503; β arccos − √1 6 = 1.9913; γ = arccos √26 = 0.6155 rads. √ √ 9. ~v = (−3, −2, −1) has |~v| = 9 + 4 + 1 = 14 so cos α = − √314 , cos β =     √2 − √2 , cos γ = − √1 and α = arccos − √3 = = 2.5011; β arccos − 14 14 14 14   2.1347; γ = arccos − √114 = 1.8413 rads.

√ √ 10. ~v = (−1, 2, −2) has |~v| = 1 + 4 + 4 = 9 = 3. Hence, cos α = −1/3, cos β = 2/3, cos γ = −2/3 which imply that α = arccos(−1/3) = 1.9106; β = arccos(2/3) = 0.8411; γ = arccos(−2/3) = 2.3005 radians.   11. ~u = √12 , 0, c means the cos α = √12 , cos β = 0 and cos γ = c. But |~u| = 1

also (insert word) that cos2 α + cos2 β + cos2 γ = 1 or 12 + 0 + cos2 γ = 1 1 . However c < 0 (given), or cos2 γ = 1/2 which implies that cos γ = ± √ 2 1 √ or γ = 3π/4 (or 2.3562 radians (insert word)) so cos γ = − 2

12. ~u =





1 , b, − √1 √ 3 3 cos2 β + 13 =

means that cos β = b. But |~u| = 1 also implies that √ 1 or cos2 β = 1/3, i.e., cos β = ±1/ 3. Since b > 0 is + √ given we get β = arccos(1/ 3) = 0.9553 radians. 1 3

5 13. ~u = (a, −1/2, 1/2) and |~u| = 1 implies that cos2 α + 14 + 41 = 1 or cos2 α = 1/2, i.e., cos α = ± √1 . Since a < 0 we get cos α = − √1 or alpha = 3π/4. 2 2  1  1 + cos2 γ = 1 or cos2 γ = 14. ~u = 0, 4 , c and |~u| = 1 both imply that 16 √ 1 15/16,i.e., cos γ = 4 15 (since c = cos γ > 0 is given). Hence γ = √ arccos 415 = 0.2527 radians.

  15. ~u = 12 , b, − 21 , |~u| = 1 gives us 14 + cos2 β + 41 = 1 or cos2 β = 1/2, i.e., b = cos β = ± √12 , i.e., cos β = √1 since b > 0. Hence, β = π/4 radians. 2

16. w ~ = (a, b, c) and we know that √b10 = cos β, √c10 = cos γ. But cos β = 0 =⇒ b = 0 while cos γ = − √1 =⇒ c = −1. But w ~ ⊥ (1, −1, 3) means 10 that (a, b, c) · (1, −1, 3) = 0, i.e., a − b + 3c = 0. Combining our results we get a = b − 3c = 0 − (−3) = 3, ∴ w ~ = (3, 0, −1).

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Exercise Set 4.

1. ~v × w ~ = (1, 2, 3)

(Use the expansion of the determinant)

2. w ~ × ~v = (−3, 1, 2) 3. (2~v) × w ~ = (2, 8, 4) 4. ~v × (3w) ~ = (−24, −3, 6) 5. ~v × w ~ = (2, 1, 3) 6. n1 = ~v × w ~ = (0, 3, 0) and n2 = w ~ × ~v = (0, −3, 0) 7. n1 = ~v × w ~ = (3, −1, 2) and n2 = w ~ × ~v = (−3, 1, −2) 1 1 1 1 8. n1 = ~v × w ~ = ( 12 , −21, − 18 ) and n2 = w ~ × ~v = (− 12 , 2, 8)

~ × ~v = (−1, − 12 , − 41 ) 9. n1 = ~v × w ~ = (1, 12 , 41 ) and n2 = w 10. n1 = ~v×w ~ = (−0.32, 0.2032, 0.48) and n2 = w×~ ~ v = (0.32, −0.2032, −0.48) 11. Since w = −2v the vectors are antiparallel. 12. Since v = 2w the vectors are parallel. 13. Since v ×w = (1, 2, 1) 6= 0 the vectors are neither parallel nor antiparallel. 14. Since v = 2w/3, the vectors are parallel. 15. Since w = −2v the vectors are antiparallel. 16. 0, the zero vector. 17. 1 18. −1 19. 0, the zero vector. 20. −1 21. θ = Arcsin (1/2) = π/6. √ 22. θ = Arcsin ( 3/2) = π/3 √ √ 23. θ = Arcsin ( 3/ 5) = 0.8861 rads. √ √ √ 24. Find c√first: c = ± 8 = ±2 2. It follows √ that√v · w = ± 8. Since |v| = 5 and |w| = 3 we get θ = Arccos (± 8/(3 5)) = 1.1355 or 2.006 rads. 25. v×w = (b+a, −b+a, −2) and v×w = (2, 2, −2) both imply that b+a = 2 and −b + a = 2. So, a = 2, b = 0. Thus, θ = Arcsin (1) = π/2. 26. Since √ the area is equal to (1/2)|v × w| we get Area = (1/2)|(2, −1, 4)| = 21/2. √ 27. Area = 54/2. Same idea as the previous one. 28. We don’t really need S since the area of the parallelogram is simply twice √ the area of ∆P QR. So, Area = 2(1/2)|(6, 2, −8)| = 104.

7 √ √ 29. Area = (1/2)( 150 + 72) ~ × OQ ~ = (2, 2, 2) and the required volume is (1, 1, 1) · (2, 2, 2) = 6. 30. OP 31. 10. Same idea as the preceding one. 32. Expand both sides and simplify. 33. Expand both sides and simplify. 34. Expand both sides and simplify. 35. Expand both sides and simplify. 36. Expand both sides and simplify. 37. Expand both sides and simplify. 38. Expand both sides and simplify. 39. |v × w|2 = sin2 θ|v|2 |w|2 and |v · w|2 = cos2 θ|v|2 |w|2 . ∴, |v × w|2 + |v · w|2 = |v|2 |w|2 (sin2 θ + cos2 θ) = |v|2 |w|2 . 40. Expand both sides and simplify.

Exercise Set 5.

1. 8x + 13y + z = 32 2. 2x + 2y + z = 2 3. z = 0 4. z − y = 3 5. y = 2 6. x = −2 7. 2x − y + 3z = 4

x y + −z =2 2 2 9. x + y − z = 0 8. −

10. x + 2y + z = 0 11. Parallel 12. Not parallel 13. Not parallel 14. Parallel 15. 0.9553 rads (or 54.7o ) 16. 1.3806 rads (or 79.1o ) 17. 1.3508 rads (or 77.4o )

8 18. 0.5148 rads (or 29.5o ) 19. 20.

y − 1/2 z + 1/2 x = . = 1 −2 3

x−1 z−0 y−1 . = = 2 1 −1

21. Let v = (0, 1, 0) and P (0, 0, 1). Then we must have Ax+By +C(z −1) = 0 and B − C = 0. Combining these two equations we get that the plane Π : Ax + By + B(z − 1) = 0 contains both v and P . Since A, B can be any two numbers, this gives an infinite number of planes.

Exercise Set 6.

1. 2. 3.

z−1 y−0 x+1 = = , or x = −1 + 2t, y = −t, z = 1 + t. −1 2 1 x−2 y−3 z−0 = = , or x = 2 − 2t, y = 3 + t, z = 3t. 1 −2 3

y+1 z−5 x−2 = = , or x = 2 + 6t, y = −1 − 2t, z = 5 + 9t. 6 −2 9

x+3 z+9 y−4 = = , or x = −3 + 9t, y = 4 + 11t, z = −9 + t. 11 9 1 z+3 y−5 = , or x = 1, y = 5 + t, z = −3 − t. 5. x = 1, 1 −1

4.

x−0 z−1 , y = 0, or x = 2t, y = 0, z = 1 + 2t. = 2 2 x+2 7. = t, y = 1, z = 0, or x = −2 − t, y = 0, z = 0. −1

6.

8. 9.

y−0 x−1 = , z = 1, or x = 1 − 2t, y = −t, z = 1, −2 −1

x−0 y−0 z−1 = = , or x = −t, y = 2t, z = 1 + t, 0 ≤ t ≤ 1. 2 −1 1

10. x = −1, 11. 12.

z−2 y−0 = , or x = −1, y = 2t, z = 2 + t, 0 ≤ t ≤ 1. 2 1

x−1 = t, y = 1, z = −1, or x = 1 − t, y = 1, z = −1, 0 ≤ t ≤ 1. −1

x+2 1 3 5 y − 1/2 , z = 0, or x = −2 + t, y = − t, z = 0, 0 ≤ t ≤ 1. = 2 2 2 −3/2 5/2

13. x = −1 + t, y = 1, z = 0. 14. x = t, y = −t, z = 1. 15. x = 1 + t, y = 3 + 2t, z = −3 − t. 16. x = 1, y = 1 − t, z = 1 + t. 17.

0 ≤ t ≤ 1.

x−0 y−2 z+1 = = 5 1 −4

9 18.

x−1 y+1 z−1 = = 2 −5 3

19.

y+1 z−0 x−0 = = −1 10 7

20.

x − 1/2 z−2 y−0 = = −4 −1 3

21. x = −1, y = 2, z = 3 + t. 22. x = 1 + t, y = −3 + t, z = 2 + t. 23. x = 1, y = −1 − t, z = 0 24. x = −t, y = 2 − t, z = −1. 25. cos θ = 0 so θ = π/2 or θ = 90o . 3 26. cos θ = − √ √ < 0. So, θ = 2.618 and the angle between the planes is 6 2 π − 2.618 = 0.5236 rads, or 30o . 1 27. cos θ = √ > 0. So, θ = π/4 rads, or 45o . 2

1 28. cos θ = √ > 0. So, θ = 1.4615 rads, or 83.74o . 84 12 29. cos θ = − < 0. So, θ = 2.0714 and the required angle is π − 2.0714 = 25 1.0701 rads, or 61.31o . 7 30. cos θ = − √ < 0. So, θ = 3 and the required angle is π − 3 = 0.141 5 2 rads, or 8.13o . 31. x = −1 + t, y = 1 − 2t, z = 3t. 32. x = 1 − y/2, z = y/2 =⇒ x + y − z = (1 − y/2) + y − y/2 = 1 − y + y = 1 for any value of y. So, every point P (x, y, z) on the line also lies on the plane and the result is clear. 33. L1 has direction vector√v1 = (1, 2, 3). L2 has direction vector v2 = (2, −1, 1). So, cos θ = 3/ 84 > 0, i.e., θ = 1.237 rads or 70.89o . 34. The direction v of the line L of intersection of the planes is given by finding the cross product of their normal vectors. So, v = n1 × n2 = (3, −1, 3). Next, we need a point on L to completely determine L . It suffices to choose a point that lies on both the planes. For example, setting x = 0 and solving for y, z in the equations for Π1 , Π2 we get, y = 3, z = 3. Thus (0, 3, 3) in on L . The parametric equation of the line is then x = 0 + 3t, y = 3 − t, z = 3 + 3t. Incidentally, we now have three points namely, (0, 2, 6), (0, 3, 3) and (3, 2, 6) (set t = 1), that can be used to determine the desired plane. Its normal is then given by (0, −9, −3) or, more simply by (0, 3, 1). Its equation is then 0(x−0)+3(y−2)+1(z−6) = 0 or 3y + z = 12. 35. This is because we have seen that every unit vector v = (cos α, cos β, cos γ ) where the cosines are the direction cosines of v. Since the line has direction v, these terms must appear in the denominator of the expression for the symmetric equations.

10

Exercise Set 7.

√ 1. ( 2, 0) 2. (1, −1) 3. (1, 0) √ √ 4. ( 2/2, 2/2) 5. (3, 2) 6. (3.268, 1.522) √ √ 7. ( 2/2, −3 2/2) 8. (1, 0) √ 9. (1/2, 2/2) 10. (−3.933, −3.087) 11. (4.843, 1.241) 12. (−6, 3)

Exercise Set 8.

√ √ 1. ( 2/2, −3 2/2) √ 2. (0, 2/2) √ √ 3. (− 3 − 1/2, 1 − 3/2) 4. (−3.933, −3.087) √ 5. (3 3, 3) √ √ 6. (2.6 2, −1.4 2) 7. (1, 0) √ √ 8. (− 2/8, 3 2/8) √ √ 9. ( 3 + 3/2, 1 − 3 3/2) 10. (−2.455, 3.313) 11. (3.236, 2.351) 12. (5, −2) √ 13. ( 2, 0) √ √ 14. ( 3 + 1/2, 1 − 3/2) 15. (3, 1)

11 √ √ 16. ( 2/2, 2/2)

Exercise Set 9.

√ √ 1. (3 2/2, −3 2/2) √ 2. (1/2, 3/2) 3. (−1, 0) 4. (−1.763, 2.427) or (−3 sin(π/5), 3 cos(π/5)). √ √ 5. ( 2/2, 2/2) √ √ 6. (−3.098, −0.634) or (−3 3/2 − 1/2, −3/2 + 3/2) 7. (1, −5) 8. (7, −2) 9. O = (3, 2), x′ = x + 3, y′ = y + 2, x′ − y′ = 0, and x′ + y′ = 0. 10. O = (1, 0), and use (2.62)-(2.63), with θ = π/2. Then y′ = −x′

Exercise Set 10.

1. x = t, y = (t − 1)/2, 0 ≤ t ≤ 3. 2. x = t, y = t2 − 1, −1 ≤ t ≤ 4 3. x = t3 + t + 1, y = t, |t| ≤ 2. 4. x = t, y = t2 − 6, −6 ≤ t < ∞ 5. x = −t/2, y = t, 1 ≤ t ≤ 2 √ √ 6. x = t, y = 3 2 − t2 , |t| ≤ 2 7. x = t, y = (2t2 + 1)/3, −3 ≤ t ≤ 5 8. x = (t2 − 3)/2, y = t, 0 ≤ t ≤ 1 √ 9. x = 5 1 − 2t2 , y = t, 0 ≤ t ≤ 2 p 10. x = t, y = (1 − t2 )/2, −1 ≤ t ≤ 1 Exercise Set 11.

1. x = 2 cos t, y = 2 sin t, t ∈ [0, 2π ] 2. x = 1 + cos t, y = − sin t, t ∈ [0, 2π ]

2

12 3. x =

√ √ 2 cos t, y = −1 + 2 sin t, t ∈ [0, 2π ]

4. x = 1 + 3 cos t, y = 1 + 3 sin t, t ∈ [0, 2π ] √ √ 5. x = −1 + 5 cos t, y = 2 − 5 sin t, t ∈ [0, 2π ] √ √ 6. x = 7 cos t, y = 3 − 7 sin t, t ∈ [0, 2π ] √ 7. x = 3 cos t, y = −2 sin t, t ∈ [0, 2π ] 8. x = 2 cos t, y = 3 sin t, t ∈ [0, 2π ] √ √ 9. x = 3 cos t, y = 1 + 5 sin t, t ∈ [0, 2π ] √ 10. x = −2 + 2 cos t, y = −2 sin t, t ∈ [0, 2π ] √ 11. x = 1 + 12 cos t, y = −1 + 4 sin t, t ∈ [0, 2π ] p 12. x = −2 + 1/2 cos t, y = 1 + (1/2) sin t, t ∈ [0, 2π ] Exercise Set 12.

1. x =

√ √ 2 cos t, y = 3 sin t,

t ∈ [0, 2π ].

2. x = 3 cos t, y = 1 − 2 sin t, t ∈ [0, 2π ]. √ √ 3. x = −1 + 2 cos t, y = 2 + 3 sin t, t ∈ [0, 2π ]. √ √ 4. x = 3 + 5 cos t, y = −1 + 2 sin t, t ∈ [0, 2π] . 5. x = −3 + 2 cos t, y = −4 − 3 sin t, t ∈ [0, 2π] . √ √ 6. x = 12 (2 cos t − 6 sin t), y = 21 (2 cos t + 6 sin t), t ∈ [0, 2π]. c.c. orientation √ √ 7. x = 32 3 cos(t) − sin(t), y = 3 sin(t) + 23 cos(t) + 1, t ∈ [0, 2π]. c.c orientation √

√

√

8. x = 32 sin(t) + 2 2 cos(t) − 1, y = − 26 cos(t) + 23 sin(t) + 2, t ∈ [0, 2π ]. c.c. orientation √ √ √ √ √ √ √ 9. x = 22 ( 5 cos(t) − 2 sin(t) + 3 2), y = 22 ( 5 cos(t) + 2 sin(t) − √ 2), t ∈ [0, 2π]. c.c. orientation 10. x = −3 + 3 sin(t), y = −4 − 2 cos(t),

Exercise Set 13.

1. x = t, y = 12 (t + 1)2 − 1 2. x = t2 − t, y = t 3. x = 3 − 4(t − 2)2 , y = t 4. x = t, y = 3 − 4(t − 2)2

t ∈ [0, 2π]. c.c. orientation

13 √ 5. x = −19t2 + (19/2) 2t − 229/152, 6. x = −6t2 + 6t − 9/8,

y = 19t2 − 113/152.

y = 6t2 + 5/8.

Exercise Set 14.

1. Hyperbola 2. Hyperbola 3. Ellipse 4. Parabola 5. Hyperbola 6. Parabola 7. Hyperbola 8. Ellipse √ 9. x = ±2 cosh t, y = ± 2 sinh t. √ √ 10. x = 1 ± 2 cosh t, y = −1 ± 2 sinh t. 11. x = ± sinh t, y = ± cosh t. 12. x = (1/2)(cosh t + sinh t),

y = (1/2)(cosh t − sinh t), t ∈ R

Exercise Set 15.

1. y =

p 3

x/2, x ∈ R.

2. y = x2 , x ≥ 0. 3. x2 + y2 = 9, x, y ∈ [−3, 3]. 4. y = sin(cos x), x ∈ [0, π/2]. 5. x2 + (y/2)2 = 1, |x| ≤ 1, |y| ≤ 2. √ 6. y = (x − 1)3/2 + x − 1 − 1, 1 ≤ x ≤ 2. 7. (x − 1)2 − y2 /9 = 1, x ≤ 0, or x ≥ 2. 8. y = cos(2 ln(x − 1)), x > 1.

Exercise Set 16.

1. y′ (x) = 1/(4t3/2 ), y′′(x) = −3/(16t7/2 )

14 √ 2. y′ (x) = 2 t + 1, y′′ (x) = 2 3. y′ (x) = − tan t, y′′(x) = − sec3 t 4. y′ (x) = −(2/3) tan t, y′′(x) = −(2/9) sec3 t 5. y′ (x) = 2 tan t, y′′(x) = 2 sec3 t 6. y′ (x) = (1/2)tanh t, y′′(x) = (1/4)sech3 t 7. y′ (x) = 3tanh t, y′′(x) = 3sech3 t 8. y′ (x) = 6t/(2t − 1), y′′(x) = −6/(2t − 1)3 9. y′ (x) = −2/(t − 1)2 , y′′(x) = 4/(t − 1)3 10. y′ (x) = −e−2t , y′′(x) = 2e−3t

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

15

Exercise Set 17.

1. 4π 2. 9π 3. 4π 4. 2π 5. 1/4 √ 6. 2 2/15 7. π 2 /4 8. 81/20 9. 1096/105 10. 1/60

Exercise Set 18.

1. 52/3 √ 2. 2 2 √ 3. 4 2 √ 4. 2 5 √ 5. 2 ln(2) 5 6. 2π √ √ √ √ 7. −(1/2) 2 − (1/2) ln(1 + 2) + 5 − (1/2) ln(−2 + 5) √ √ 8. (1/2) 5 − (1/4) ln(−2 + 5) R π/2 q 1 − (3/4) sin2 t dt = 19.38 9. 16 0

10. 3/2

Exercise Set 19.

1. (−3, 0) √ 2. ( 3, −1) √ √ 3. (−4 2, 4 2) √ 4. (2, −2 3)

16 √ 5. (1, 3) 6. (0, 4) 7. (0.97, 0.7) 8. (−0.9, −0.43) 9. (−2.5, 4.3) 10. (−1.73, 1) √ √ 11. ( 2, π/4) or (− 2, 5π/4) √ √ 12. (3 2, 5π/4) or (−3 2, π/4) √ √ 13. ( 5, −0.46) or (− 5, 2.68) 14. (4, π/2) or (−4, −π/2) 15. (4, π) or (−4, 0) √ √ 1...