University of Toronto Scarborough Department of ...
University of Toronto Scarborough Department of Computer & Mathematical Sciences MAT B41H
2012/2013 Solutions #4
p 1. (a) If r = (x, y, z), r = r(x, y, z) = kr(x, y, z)k = x2 + y 2 + z 2 . Taking the gradient of r(x, y, z), we have ! p x y z ∇r = ∇ x2 + y 2 + z 2 = p = , p , p 2 2 2 2 2 2 2 x +y +z x +y +z x + y2 + z2 1 r p (x, y, z) = , r 6= 0. r x2 + y 2 + z 2 ! 1 y 1 x ,− 2 , = − 2 (b) Now ∇ =∇ p 2 2 3/2 2 2 2 2 r (x + y + z ) (x + y + z 2 )3/2 x +y +z z 1 r − 2 =− 2 (x, y, z) = − 3 , r 6= 0. 2 2 3/2 2 2 3/2 y + z ) (x + y + z ) r
(x +
1 r 1 1 1
∇
= − krk = = 2 = 2 .
r r3 r3 r x + y2 + z2 2. (a) A direction vector for the line is (2, −3, 1) − (−1, 1, 3) = (3, −4, −2), so a parametric description of the line is (−1, 1, 3) + t (3, −4, −2), t ∈ R. If (x, y, z) is a point on the line we have (x, y, z) = (−1, 1, 3) + t (3, −4, −2), for some t. This gives x = −1 + 3t, y = 1 − 4t and z = 3 − 2t, so a rectangular x+1 1−y 3−z description is (t =) = = . 3 4 2 (b) A pair of direction vectors in the plane are v = (2, −3, 1)−(−1, 1, 3) = (3, −4, −2) and w = (−2, 0, 3) − (−1, 1, 3) = (−1, −1, 0). A parametric description of π is (−1, 1, 3) + t (3, −4, −2) + s (−1, −1,0), s, t ∈ R. x = −1 + 3t − s y = 1 − 4t − s . The third equation If (x, y, z) is on the plane π, we have z = 3 − 2t 3−z gives t = . If we substitute this into the second equation we get y = 2 3−z 1−4 − s = 1 − 6 + 2z − s which gives s = −y − 5 + 2z. We substitute the 2 3−z −(−y−5+2z) = values for t and s into the first equation giving x = −1+3 2 17 7 + y − z. Rewriting we get 2x − 2y + 7z = 17 as an equation for π. 2 2 (An alternate approach would be to find n = v × w.)
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(c) A direction vector for this line is a normal vector for the plane π, n = (2, −2, 7). A parametric description of the line is (0, 1, 0) + t (2, −2, 7), t ∈ R. To see where the line and the plane meet we must find a t such that (2t, 1−2t, 7t) satisfies the equation of the plane. For this we need 2(2t) − 2(1 − 2t) + 7(7t) = 1 17 =⇒ 4t − 2 + 4t + 49t = 17 =⇒ 57t = 19 =⇒ t = . The point of 3 2 1 7 intersection is , , . 3 3 3 (d) The two lines will intersect if there is some t, say t = t0 , and some s, say s = s0 such that (1, 0, 1) + t0 (3, 3, 5) = (3, 6, 1) + s0 (4, −2, 7). Equating the components 1 + 3t0 = 3 + 4s0 3t0 = 6 − 2s0 . Subtracting the second from give a system of 3 equations: 1 + 5t0 = 1 + 7s0 14 2 . the first gives s0 = . Substituting this value into the second gives t0 = 3 9 14 2 Now substituting both values into the third we get 1 + 5 = 1+7 or 9 3 42 70 = . Clearly, this is not possible and we have an inconsistent system. Hence, 9 9 we conclude that the lines do not intersect. e1 e2 e3 3 5 = 31 e1 − e2 − 18 e3 = 31, −1, −18 . 3. (a) u × w = det 3 4 −2 7 e1 e2 e3 (b) Let x = (x, y, z). Now x × (1, 0, 1) = det x y z = y e1 − (x − z)e2 + 1 0 1
(−y)e3 = y, z − x, −y . Hence the equations become y, z − x, −y = p y 2 + (z − x)2 + y 2 = 1 and p x·(1, 0, −1) = x − z = 1. Substituting the second into the first we have 2y 2 + 1 = 1 =⇒ y = 0. Hence the solution is given by x = z + 1, y = 0, z ∈ R or we can write the solution set as x = (z + 1, 0, z) | z ∈ R . 4. The tangent plane to the graph of z = f (x, y) at the point (a, b, f (a, b)) is ∂f ∂f (a, b) (x − a) + (a, b) (y − b). z = f (a, b) + ∂x ∂y x 1 4y x2 + 4y 2 , fx = p , fx (2, 3) = √ ; fy = p , 2 2 2 2 10 x + 4y x + 4y r √ 2 fy (2, 3) = 3 and f (2, 3) = 2 10. 5 r √ 1 2 The equation of the tangent plane is z = 2 10 + √ (x − 2) + 3 (y − 3) which 5 10 √ can be rewritten as x + 6 y − 10 z = 0.
(a) f (x, y) =
p
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x+y x + 2y 1 1 . fx = − , fx (2, 3) = −1; fy = 2 , fy (2, 3) = and 2 3 x x x 4 5 f (2, 3) = . 4 1 5 The equation of the tangent plane is z = − 1 x − 2 + y − 3 which can be 4 4 rewritten as 4x − y + 4z = 10. x+y ∂f −x2 − 2xy + y 2 − 1 1 ∂f (c) f (x, y) = . = 2, 3 = − ; 2 , 2 2 x +y −1 ∂x ∂x 18 x2 + y 2 − 1 2 2 ∂f x − 2xy − y − 1 ∂f 1 5 = 2, 3 = − and f (2, 3) = . 2 , ∂y ∂y 8 12 x2 + y 2 − 1 1 5 1 5 x − x−2 − y−3 = − + The equation of the tangent plane is z = 12 18 8 12 18 1 y 3 1 1 65 1 1 65 − + =− x− y+ which we can rewrite as x+ y+z = . 9 8 8 18 8 72 18 8 72 1 1 1 1 x+y . fx = − 2 , fx (2, 3) = − ; fy = − 2 , fy (2, 3) = − and (d) f (x, y) = xy x 4 y 9 5 f (2, 3) = . 6 5 1 1 The equation of the tangent plane is z = − (x − 2) − (y − 3) which we can 6 4 9 rewrite as 9x + 4y + 36z = 60. 2x 4 2(x2 − 1 − y) 1 − x2 + 2y . f = − , f (2, 3) = − , fy (2, 3) = ; f = (e) f (x, y) = x x y y2 y2 9 y3 1 0 and f (2, 3) = . 3 4 1 The equation of the tangent plane is z = − (x − 2) which we can rewrite 3 9 as 4x + 9z = 11.
(b) f (x, y) =
5. (This question is a slight generalization of the situation discussed in class.) The linear approximation to f (x) at a point a is given by T1 (x) =f (a)+Df (a)(x−a). 1 1 ez z 2 Here we have f (x, y, z) = (x + e + y, yx ) so Df (x, y, z) = and 2xy x2 0 1 1 1 Df (1, 1, 0) = . The linear approximation to f at (1, 1, 0) is 2 1 0 T1 (x, y, z) = f (1, 1, 0) + Df (1, 1, 0) (x − 1, y −1, z) x−1 3 1 1 1 y−1 = + 1 2 1 0 z x+y+z+1 = . 2x + y − 2
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6. (a) (i) We want the tangent plane to the level surface g(x, y, z) = x2 + 2y 2 − 3z 2 − 3 = 0 at (2, −1, 1). The tangent plane is given by ∇g (2, −1, 1) · (x, y, z) − (2, −1, 1) = 0. Now ∇g = (2x, 4y, −6z) so ∇g (2, −1, 1) = (4, −4, −6). Thus the tangent plane is (4, −4, −6)·(x−2, y +1, z −1) = 4x−8−4y −4−6z +6 = 4x − 4y − 6z − 6 = 0 or 2x − 2y − 3z = 3. The normal line is (2, −1, 1) + t (2, −2, −3), t ∈ R. (ii) We want the tangent plane to the level surface g(x, y, z) = sin(xyz)−x−2y − 3z = 0 at(2, −1, 0). The tangent plane is given by ∇g (2, −1, 0) · (x, y, z) − (2, −1, 0) = 0. Now ∇g = (yz cos(xyz)−1, xz cos(xyz)−2, xy cos(xyz)−3) so ∇g (2, −1, 0) = (−1, −2, −5). Thus the tangent plane is (−1, −2, −5)·(x− 2, y + 1, z) = −x + 2 − 2y − 2 − 5z = −x − 2y − 5z = 0 or x + 2y + 5z = 0. The normal line is (2, −1, 0) + t (1, 2, 5), t ∈ R. (b) The tangent plane to the graph of z = f (x, y) defined implicitly by xy+yz+zx = 3 at (1, 1, 1) is also the tangent plane to the level set of g(x, y, z) = xy+yz+zx−3 = 0 at (1, 1, 1). This tangent plane is given by ∇g (1, 1, 1) · (x, y, z) − (1, 1, 1) = 0. Now ∇g = (y + z, x + z, y + x) so ∇g (1, 1, 1) = (2, 2, 2). The tangent plane is (2, 2, 2) · (x − 1, y − 1, z − 1) = 2x − 2 + 2y − 2 + 2z − 2 = 2x + 2y + 2z − 6 = 0 or x + y + z = 3. 7. (a) f (x, y) = x2 sin πy; p = (2, 1), v = (4, 3). We first note that ∇f = (2x sin(π y), π x2 cos(π y)) and ∇f (2, 1) = (2(2) sin π, (2)2 π cos π) = (0, −4 π). Now Dv f (p) v 12 π (0, −4 π) · (4, 3) √ = ∇f (p) · =− = . 2 2 kvk 5 4 +3 (b) f (x, y, z) = xy 3 z 2 ; p = (3, 1, 2), v = n = (2, 1, −4). Now ∇f = (y 3 z 2 , 3xy 2 z 2 , v 2xy 3 z) and ∇f (3, 1, 2) = (4, 36, 12). Hence Dv f (p) = ∇f (p) · = kvk −4 (4, 36, 12) · (2, 1, −4) p =√ . 21 22 + 1 + (−4)2 (c) f (x, y, z) = x2 + y 2 − z 2 . We want Dv f (3, 4, 5) where v is a tangent vector to the curve of intersection of the surfaces 2x2 + 2y 2 − z 2 = 25 and x2 + y 2 = z 2 which has positive x–direction. To find v we put g(x, y, z) = 2x2 + 2y 2 − z 2 − 25 and h(x, y, z) = x2 + y 2 − z 2 . Since v will be in the tangent planes of both g and h at (3, 4, 5), it must be orthogonal to both normal vectors. Now ∇g (x, y, z) = (4x, 4y, −2z), ∇g (3, 4, 5) = (12, 16, −10); ∇h (x, y, z) = (2x, 2y, −2z) and ∇h (3, 4, 5) = (6, 8, −10). Hence v will be in the direction of ∇g × ∇h = e1 e2 e3 det 12 16 −10 = −80 e1 + 60 e2 + 0 e3 . We can choose v = (4, −3, 0). 6 8 −10 ∇f (3, 4, 5)·(4, −3, 0) (6, 8, −10)·(4, −3, 0) 2 √ Hence Dv f (3, 4, 5) = = = 12 − k(4, −3, 0)k 5 16 + 9 12 + 0 = 0.
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yz xz xy 0 2x2 z . 8. f : R3 → R3 is given by f (x, y, z) = (xyz, x2 z 2 , y 3 x), so D f = 2xz 2 3 2 y 3y x 0 1 1 0 0 z y g : R3 → R4 is given by g(x, y, z) = (x + y, yz, z − 2x, x2 y), so D g = −2 0 1 2xy x2 0 1 1 0 3 2 2 0 xy xz . Now the chain rule gives and D g f (x, y, z) = −2 0 1 2x3 yz3 x2 y 2 z 2 0 D(g ◦ f )(x, y, z) = D g(f (x,y, z)) D f (x, y, z) 1 1 0 yz xz xy 3 2 2 0 xy xz 2xz 2 0 2x2 z = −2 0 1 y 3 3y 2 x 0 2x3 yz 3 x2 y 2 z 2 0 yz + 2xz 2 xz xy + 2x2 z 2x2 y 3 z 2 + x2 y 3 z 2 3x3 y 2 z 2 2x3 y 3 z = −2yz + y 3 −2xz + 3y 2 x −2xy 3 2 4 3 2 4 4 4 4 2 3 4 2 3 2x y z + 2x y z 2x yz 2x y z + 2x y z 2 yz + 2xz xz xy + 2x2 z 3x2 y 3 z 2 3x3 y 2 z 2 2x3 y 3 z . = 3 2 −2yz + y −2xz + 3y x −2xy 4x3 y 2 z 4 2x4 yz 4 4x4 y 2 z 3
To compute directly, we have g ◦ f : R3 → R4 and g ◦ f (x, y, z) = g(f (x, y, z)) = g(xyz, x2 z 2 , y 3 x) = xyz + x2 z 2 , x3 y 3 z 2 , y 3 x − 2xyz, x4 y 2 z 4 Now D(g ◦ f ) 2 2 yz + 2xz xz xy + 2x z 3 2 2 3x2 y 3 z 2 3x y z 2x3 y 3 z which is the same as we had when we used = y 3 − 2yz 3y 2 x − 2xz −2xy 4x3 y 2 z 4 2x4 yz 4 4x4 y 2 z 3 the chain rule.
2012/2013 Solutions #4
p 1. (a) If r = (x, y, z), r = r(x, y, z) = kr(x, y, z)k = x2 + y 2 + z 2 . Taking the gradient of r(x, y, z), we have ! p x y z ∇r = ∇ x2 + y 2 + z 2 = p = , p , p 2 2 2 2 2 2 2 x +y +z x +y +z x + y2 + z2 1 r p (x, y, z) = , r 6= 0. r x2 + y 2 + z 2 ! 1 y 1 x ,− 2 , = − 2 (b) Now ∇ =∇ p 2 2 3/2 2 2 2 2 r (x + y + z ) (x + y + z 2 )3/2 x +y +z z 1 r − 2 =− 2 (x, y, z) = − 3 , r 6= 0. 2 2 3/2 2 2 3/2 y + z ) (x + y + z ) r
(x +
1 r 1 1 1
∇
= − krk = = 2 = 2 .
r r3 r3 r x + y2 + z2 2. (a) A direction vector for the line is (2, −3, 1) − (−1, 1, 3) = (3, −4, −2), so a parametric description of the line is (−1, 1, 3) + t (3, −4, −2), t ∈ R. If (x, y, z) is a point on the line we have (x, y, z) = (−1, 1, 3) + t (3, −4, −2), for some t. This gives x = −1 + 3t, y = 1 − 4t and z = 3 − 2t, so a rectangular x+1 1−y 3−z description is (t =) = = . 3 4 2 (b) A pair of direction vectors in the plane are v = (2, −3, 1)−(−1, 1, 3) = (3, −4, −2) and w = (−2, 0, 3) − (−1, 1, 3) = (−1, −1, 0). A parametric description of π is (−1, 1, 3) + t (3, −4, −2) + s (−1, −1,0), s, t ∈ R. x = −1 + 3t − s y = 1 − 4t − s . The third equation If (x, y, z) is on the plane π, we have z = 3 − 2t 3−z gives t = . If we substitute this into the second equation we get y = 2 3−z 1−4 − s = 1 − 6 + 2z − s which gives s = −y − 5 + 2z. We substitute the 2 3−z −(−y−5+2z) = values for t and s into the first equation giving x = −1+3 2 17 7 + y − z. Rewriting we get 2x − 2y + 7z = 17 as an equation for π. 2 2 (An alternate approach would be to find n = v × w.)
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(c) A direction vector for this line is a normal vector for the plane π, n = (2, −2, 7). A parametric description of the line is (0, 1, 0) + t (2, −2, 7), t ∈ R. To see where the line and the plane meet we must find a t such that (2t, 1−2t, 7t) satisfies the equation of the plane. For this we need 2(2t) − 2(1 − 2t) + 7(7t) = 1 17 =⇒ 4t − 2 + 4t + 49t = 17 =⇒ 57t = 19 =⇒ t = . The point of 3 2 1 7 intersection is , , . 3 3 3 (d) The two lines will intersect if there is some t, say t = t0 , and some s, say s = s0 such that (1, 0, 1) + t0 (3, 3, 5) = (3, 6, 1) + s0 (4, −2, 7). Equating the components 1 + 3t0 = 3 + 4s0 3t0 = 6 − 2s0 . Subtracting the second from give a system of 3 equations: 1 + 5t0 = 1 + 7s0 14 2 . the first gives s0 = . Substituting this value into the second gives t0 = 3 9 14 2 Now substituting both values into the third we get 1 + 5 = 1+7 or 9 3 42 70 = . Clearly, this is not possible and we have an inconsistent system. Hence, 9 9 we conclude that the lines do not intersect. e1 e2 e3 3 5 = 31 e1 − e2 − 18 e3 = 31, −1, −18 . 3. (a) u × w = det 3 4 −2 7 e1 e2 e3 (b) Let x = (x, y, z). Now x × (1, 0, 1) = det x y z = y e1 − (x − z)e2 + 1 0 1
(−y)e3 = y, z − x, −y . Hence the equations become y, z − x, −y = p y 2 + (z − x)2 + y 2 = 1 and p x·(1, 0, −1) = x − z = 1. Substituting the second into the first we have 2y 2 + 1 = 1 =⇒ y = 0. Hence the solution is given by x = z + 1, y = 0, z ∈ R or we can write the solution set as x = (z + 1, 0, z) | z ∈ R . 4. The tangent plane to the graph of z = f (x, y) at the point (a, b, f (a, b)) is ∂f ∂f (a, b) (x − a) + (a, b) (y − b). z = f (a, b) + ∂x ∂y x 1 4y x2 + 4y 2 , fx = p , fx (2, 3) = √ ; fy = p , 2 2 2 2 10 x + 4y x + 4y r √ 2 fy (2, 3) = 3 and f (2, 3) = 2 10. 5 r √ 1 2 The equation of the tangent plane is z = 2 10 + √ (x − 2) + 3 (y − 3) which 5 10 √ can be rewritten as x + 6 y − 10 z = 0.
(a) f (x, y) =
p
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x+y x + 2y 1 1 . fx = − , fx (2, 3) = −1; fy = 2 , fy (2, 3) = and 2 3 x x x 4 5 f (2, 3) = . 4 1 5 The equation of the tangent plane is z = − 1 x − 2 + y − 3 which can be 4 4 rewritten as 4x − y + 4z = 10. x+y ∂f −x2 − 2xy + y 2 − 1 1 ∂f (c) f (x, y) = . = 2, 3 = − ; 2 , 2 2 x +y −1 ∂x ∂x 18 x2 + y 2 − 1 2 2 ∂f x − 2xy − y − 1 ∂f 1 5 = 2, 3 = − and f (2, 3) = . 2 , ∂y ∂y 8 12 x2 + y 2 − 1 1 5 1 5 x − x−2 − y−3 = − + The equation of the tangent plane is z = 12 18 8 12 18 1 y 3 1 1 65 1 1 65 − + =− x− y+ which we can rewrite as x+ y+z = . 9 8 8 18 8 72 18 8 72 1 1 1 1 x+y . fx = − 2 , fx (2, 3) = − ; fy = − 2 , fy (2, 3) = − and (d) f (x, y) = xy x 4 y 9 5 f (2, 3) = . 6 5 1 1 The equation of the tangent plane is z = − (x − 2) − (y − 3) which we can 6 4 9 rewrite as 9x + 4y + 36z = 60. 2x 4 2(x2 − 1 − y) 1 − x2 + 2y . f = − , f (2, 3) = − , fy (2, 3) = ; f = (e) f (x, y) = x x y y2 y2 9 y3 1 0 and f (2, 3) = . 3 4 1 The equation of the tangent plane is z = − (x − 2) which we can rewrite 3 9 as 4x + 9z = 11.
(b) f (x, y) =
5. (This question is a slight generalization of the situation discussed in class.) The linear approximation to f (x) at a point a is given by T1 (x) =f (a)+Df (a)(x−a). 1 1 ez z 2 Here we have f (x, y, z) = (x + e + y, yx ) so Df (x, y, z) = and 2xy x2 0 1 1 1 Df (1, 1, 0) = . The linear approximation to f at (1, 1, 0) is 2 1 0 T1 (x, y, z) = f (1, 1, 0) + Df (1, 1, 0) (x − 1, y −1, z) x−1 3 1 1 1 y−1 = + 1 2 1 0 z x+y+z+1 = . 2x + y − 2
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6. (a) (i) We want the tangent plane to the level surface g(x, y, z) = x2 + 2y 2 − 3z 2 − 3 = 0 at (2, −1, 1). The tangent plane is given by ∇g (2, −1, 1) · (x, y, z) − (2, −1, 1) = 0. Now ∇g = (2x, 4y, −6z) so ∇g (2, −1, 1) = (4, −4, −6). Thus the tangent plane is (4, −4, −6)·(x−2, y +1, z −1) = 4x−8−4y −4−6z +6 = 4x − 4y − 6z − 6 = 0 or 2x − 2y − 3z = 3. The normal line is (2, −1, 1) + t (2, −2, −3), t ∈ R. (ii) We want the tangent plane to the level surface g(x, y, z) = sin(xyz)−x−2y − 3z = 0 at(2, −1, 0). The tangent plane is given by ∇g (2, −1, 0) · (x, y, z) − (2, −1, 0) = 0. Now ∇g = (yz cos(xyz)−1, xz cos(xyz)−2, xy cos(xyz)−3) so ∇g (2, −1, 0) = (−1, −2, −5). Thus the tangent plane is (−1, −2, −5)·(x− 2, y + 1, z) = −x + 2 − 2y − 2 − 5z = −x − 2y − 5z = 0 or x + 2y + 5z = 0. The normal line is (2, −1, 0) + t (1, 2, 5), t ∈ R. (b) The tangent plane to the graph of z = f (x, y) defined implicitly by xy+yz+zx = 3 at (1, 1, 1) is also the tangent plane to the level set of g(x, y, z) = xy+yz+zx−3 = 0 at (1, 1, 1). This tangent plane is given by ∇g (1, 1, 1) · (x, y, z) − (1, 1, 1) = 0. Now ∇g = (y + z, x + z, y + x) so ∇g (1, 1, 1) = (2, 2, 2). The tangent plane is (2, 2, 2) · (x − 1, y − 1, z − 1) = 2x − 2 + 2y − 2 + 2z − 2 = 2x + 2y + 2z − 6 = 0 or x + y + z = 3. 7. (a) f (x, y) = x2 sin πy; p = (2, 1), v = (4, 3). We first note that ∇f = (2x sin(π y), π x2 cos(π y)) and ∇f (2, 1) = (2(2) sin π, (2)2 π cos π) = (0, −4 π). Now Dv f (p) v 12 π (0, −4 π) · (4, 3) √ = ∇f (p) · =− = . 2 2 kvk 5 4 +3 (b) f (x, y, z) = xy 3 z 2 ; p = (3, 1, 2), v = n = (2, 1, −4). Now ∇f = (y 3 z 2 , 3xy 2 z 2 , v 2xy 3 z) and ∇f (3, 1, 2) = (4, 36, 12). Hence Dv f (p) = ∇f (p) · = kvk −4 (4, 36, 12) · (2, 1, −4) p =√ . 21 22 + 1 + (−4)2 (c) f (x, y, z) = x2 + y 2 − z 2 . We want Dv f (3, 4, 5) where v is a tangent vector to the curve of intersection of the surfaces 2x2 + 2y 2 − z 2 = 25 and x2 + y 2 = z 2 which has positive x–direction. To find v we put g(x, y, z) = 2x2 + 2y 2 − z 2 − 25 and h(x, y, z) = x2 + y 2 − z 2 . Since v will be in the tangent planes of both g and h at (3, 4, 5), it must be orthogonal to both normal vectors. Now ∇g (x, y, z) = (4x, 4y, −2z), ∇g (3, 4, 5) = (12, 16, −10); ∇h (x, y, z) = (2x, 2y, −2z) and ∇h (3, 4, 5) = (6, 8, −10). Hence v will be in the direction of ∇g × ∇h = e1 e2 e3 det 12 16 −10 = −80 e1 + 60 e2 + 0 e3 . We can choose v = (4, −3, 0). 6 8 −10 ∇f (3, 4, 5)·(4, −3, 0) (6, 8, −10)·(4, −3, 0) 2 √ Hence Dv f (3, 4, 5) = = = 12 − k(4, −3, 0)k 5 16 + 9 12 + 0 = 0.
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yz xz xy 0 2x2 z . 8. f : R3 → R3 is given by f (x, y, z) = (xyz, x2 z 2 , y 3 x), so D f = 2xz 2 3 2 y 3y x 0 1 1 0 0 z y g : R3 → R4 is given by g(x, y, z) = (x + y, yz, z − 2x, x2 y), so D g = −2 0 1 2xy x2 0 1 1 0 3 2 2 0 xy xz . Now the chain rule gives and D g f (x, y, z) = −2 0 1 2x3 yz3 x2 y 2 z 2 0 D(g ◦ f )(x, y, z) = D g(f (x,y, z)) D f (x, y, z) 1 1 0 yz xz xy 3 2 2 0 xy xz 2xz 2 0 2x2 z = −2 0 1 y 3 3y 2 x 0 2x3 yz 3 x2 y 2 z 2 0 yz + 2xz 2 xz xy + 2x2 z 2x2 y 3 z 2 + x2 y 3 z 2 3x3 y 2 z 2 2x3 y 3 z = −2yz + y 3 −2xz + 3y 2 x −2xy 3 2 4 3 2 4 4 4 4 2 3 4 2 3 2x y z + 2x y z 2x yz 2x y z + 2x y z 2 yz + 2xz xz xy + 2x2 z 3x2 y 3 z 2 3x3 y 2 z 2 2x3 y 3 z . = 3 2 −2yz + y −2xz + 3y x −2xy 4x3 y 2 z 4 2x4 yz 4 4x4 y 2 z 3
To compute directly, we have g ◦ f : R3 → R4 and g ◦ f (x, y, z) = g(f (x, y, z)) = g(xyz, x2 z 2 , y 3 x) = xyz + x2 z 2 , x3 y 3 z 2 , y 3 x − 2xyz, x4 y 2 z 4 Now D(g ◦ f ) 2 2 yz + 2xz xz xy + 2x z 3 2 2 3x2 y 3 z 2 3x y z 2x3 y 3 z which is the same as we had when we used = y 3 − 2yz 3y 2 x − 2xz −2xy 4x3 y 2 z 4 2x4 yz 4 4x4 y 2 z 3 the chain rule.
