Calculus- Early Transcendentals, 2021a

592 Selected Exercise Answers
13.2.9 No limit; use x = 0andy = 0
13.2.10 Limit is zero
13.2.11 Limit is −1
13.2.12 Limit is zero
13.3.1 −2xysin(x 2 y), −x 2 sin(x 2 y)+3y 2
13.3.2 (y 2 − x 2 y)/(x 2 + y) 2 , x 3 /(x 2 + y) 2
13.3.3 2xe x2 +y 2 ,2ye x2 +y 2
13.3.4 yln(xy)+y, xln(xy)+x
13.3.5 −x/ √ 1 − x 2 − y 2 , −y/ √ 1 − x 2 − y 2
13.3.6 tany, xsec 2 y
13.3.7 −1/(x 2 y), −1/(xy 2 )
13.3.8 z = −2(x − 1) − 3(y − 1) − 1
13.3.9 z = 1
13.3.10 z = 6(x − 3)+3(y − 1)+10
13.3.11 z =(x − 2)+4(y − 1/2)
13.3.12 r(t)=〈2,1,4〉 + t〈2,4,−1〉
13.3.16 height
13.4.1 4xt cos(x 2 + y 2 )+6yt 2 cos(x 2 + y 2 )
13.4.2 2xycost + 2x 2 t
13.4.3 2xyt cos(st)+2x 2 s,2xyscos(st)+2x 2 t
13.4.4 2xy 2 t − 4yx 2 s,2xy 2 s + 4yx 2 t
13.4.5 x/z,3y/(2z)
13.4.6 −2x/z, −y/z
13.4.7 1. V ′ =(nR − 0.2V)/P
2. P ′ =(nR + 0.6P)/2V
3. T ′ =(3P − 0.4V)/(nR)
13.5.1 9 √ 5/5
13.5.2 √ 2cos3
13.5.3 e √ 2( √ 3 − 1)/4
13.5.4 √ 3 + 5
13.5.5 − √ 6(2 + √ 3)/72
13.5.6 −1/5, 0
13.5.7 4(x − 2)+8(y − 1)=0
13.5.8 2(x − 3)+3(y − 2)=0
13.5.9 〈−1,−1 − cos1,−cos1〉, − √ 2 + 2cos1+ 2cos 2 1
13.5.10 Any direction perpendicular to ∇T = 〈1,1,1〉,for
example, 〈−1,1,0〉
13.5.11 2(x − 1) − 6(y − 1)+6(z − 3)=0
13.5.12 6(x − 1)+3(y − 2)+2(z − 3)=0
13.5.13 〈2 + 4t,−3 − 12t,−1 − 8t〉
13.5.14 〈4 + 8t,2+ 4t,−2 − 36t〉
13.5.15 〈4 + 8t,2+ 20t,6− 12t〉
13.5.16 〈0,1〉, 〈4/5,−3/5〉
13.5.18 1. 〈4,9〉
2. 〈−81,2〉 or 〈81,−2〉
13.5.19 in the direction of 〈8,1〉
13.5.20 ∇g(−1,3)=〈2,1〉
13.6.1 f xx =(2x 3 y − 6xy 3 )/(x 2 + y 2 ) 3 ,
f yy =(2xy 3 − 6x 3 y)/(x 2 + y 2 ) 3
13.6.2 f x = 3x 2 y 2 , f y = 2x 3 y + 5y 4 , f xx = 6xy 2 ,
f yy = 2x 3 + 20y 3 , f xy = 6x 2 y
13.6.3 f x = 12x 2 + y 2 , f y = 2xy, f xx = 24x, f yy = 2x, f xy = 2y
13.6.4 f x = siny, f y = xcosy, f xx = 0, f yy = −xsiny,
f xy = cosy