1. Find equations of the tangent plane and the normal line to the

Assignment 9 (MATH 214 B1)
1. Find equations of the tangent plane and the normal line to the given surface at the
specified point.
√
(a) z = 4 − x2 − 2y 2 , (1, −1, 1).
√
Solution. Let f (x, y) = 4 − x2 − 2y 2 . We have
fx =
∂f
−x
=√
∂x
4 − x2 − 2y 2
and fy =
∂f
−2y
=√
.
∂y
4 − x2 − 2y 2
It follows that fx (1, −1) = −1 and fy (1, −1) = 2. Hence, the tangent plane to the
surface at (1, −1, 1) has an equation
z − 1 = −(x − 1) + 2(y + 1) or x − 2y + z − 4 = 0.
The normal line to the surface at (1, −1, 1) has symmetric equations
x−1
y+1
z−1
=
=
.
1
−2
1
(b) x2 y 2 + xz − 2y 3 = 10, (2, 1, 4).
Solution. Let F (x, y, z) = x2 y 2 + xz − 2y 3 − 10. Then
Fx = 2xy 2 + z,
Fy = 2x2 y − 6y 2 ,
Fz = x.
It follows that
Fx (2, 1, 4) = 8,
Fy (2, 1, 4) = 2,
Fz (2, 1, 4) = 2.
Hence, the tangent plane at (2, 1, 4) has an equation
8(x − 2) + 2(y − 1) + 2(z − 4) = 0
or
4x + y + z − 13 = 0.
The normal line to the surface at (2, 1, 4) has symmetric equations
y−1
z−4
x−2
=
=
.
4
1
1
1
2. Let (a, b, c) be a point on the sphere x2 + y 2 + z 2 = 6.
(a) Find equations of the normal line of the sphere at the point (a, b, c). Show that
the normal line passes through the origin (0, 0, 0).
Solution. Let F (x, y, z) = x2 + y 2 + z 2 − 6. Then
Fx = 2x,
Fy = 2y,
Fz = 2z.
It follows that Fx (a, b, c) = 2a, Fy (a, b, c) = 2b, and Fz (a, b, c) = 2c. The normal line
to the sphere at (a, b, c) has symmetric equations
x−a
y−b
z−c
=
=
.
2a
2b
2c
For (x, y, z) = (0, 0, 0) we have
0−a
0−b
0−c
1
=
=
=− .
2a
2b
2c
2
Hence, the normal line passes through the origin (0, 0, 0).
(b) Let C be the curve of intersection of the sphere with the plane z = 2. Find
parametric equations of the line tangent to the curve C at the point (1, −1, 2).
Solution. The tangent line is orthogonal to both ∇F and k at (1, −1, 2). We have
(
)
∇F |(1,−1,2) = 2x i + 2y j + 2z k |(1,−1,2) = 2 i − 2 j + 4 k.
It follows that ∇F × k = −2 i − 2 j. Thus, the tangent line has parametric equations
x = 1 − 2t, y = −1 − 2t, z = 2,
2
−∞ < t < ∞.
3. Find all the second-order partial derivatives of the given functions.
(a) f (x, y) = ln (x2 + y 2 )
Solution. We have
Consequently,
∂f
2x
= 2
∂x
x + y2
and
∂f
2x
.
= 2
∂y
x + y2
∂2f
(x2 + y 2 )2 − 2x(2x)
2y 2 − 2x2
=
=
,
∂x2
(x2 + y 2 )2
(x2 + y 2 )2
∂2f
∂2f
2x(−2y)
−4xy
=
= 2
=
,
∂x∂y
∂y∂x
(x + y 2 )2
(x2 + y 2 )2
(x2 + y 2 )2 − 2y(2y)
2x2 − 2y 2
∂2f
=
=
.
∂y 2
(x2 + y 2 )2
(x2 + y 2 )2
(b) g(x, y) = sin (xy) cos y
Solution. We have
gx = y cos(xy) cos y,
gy = x cos(xy) cos y − sin(xy) sin y.
Hence, we obtain
gxx = −y 2 sin(xy) cos y,
gxy = gyx = cos(xy) cos y − yx sin(xy) cos y − y cos(xy) sin y,
gyy = −x2 sin(xy) cos y − x cos(xy) sin y − x cos(xy) sin y − sin(xy) cos y.
4. Let g(x, y) = xy for 0 < x < ∞ and −∞ < y < ∞.
(a) Find all the first-order and second-order partial derivatives of g.
Solution. We have gx = yxy−1 and gy = xy ln x. Moreover,
gxx = y(y − 1)xy−2 ,
gxy = xy−1 + yxy−1 ln x,
gyy = xy ln2 x.
(b) Find the linear approximation to g at the point (1, 4) and the corresponding
remainder.
Solution. We have g(1, 4) = 1, gx (1, 4) = 4 and gy (1, 4) = 0. Hence, the linear
approximation to g at the point (1, 4) is
L(x, y) = g(1, 4) + gx (1, 4)(x − 1) + gy (1, 4)(y − 4) = 1 + 4(x − 1).
The corresponding remainder E(x, y) is
]
1[
yc (yc −1)xyc c −2 (x−1)2 +2(xyc c −1 +yc xyc c −1 ln xc )(x−1)(y −4)+ xyc c (ln2 xc )(y −4)2 ,
2
where xc = 1 + c(x − 1) and yc = 4 + c(y − 4) with 0 < c < 1.
3
5. Find the second Taylor polynomial of the given function at the origin (0, 0).
(a) u(x, y) = ex+y cos(x − y).
Solution. We have
ux = ex+y [cos(x − y) − sin(x − y)] and uy = ex+y [cos(x − y) + sin(x − y)].
Moreover,
uxx = ex+y [cos(x − y) − sin(x − y) − sin(x − y) − cos(x − y)] = −2ex+y sin(x − y),
uxy = ex+y [cos(x − y) − sin(x − y) + sin(x − y) + cos(x − y)] = 2ex+y cos(x − y),
uyy = ex+y [cos(x − y) + sin(x − y) + sin(x − y) − cos(x − y)] = 2ex+y sin(x − y).
Consequently, u(0, 0) = 1, ux (0, 0) = 1, uy (0, 0) = 1, uxx (0, 0) = 0, uxy (0, 0) = 2, and
uyy (0, 0) = 0. Thus, the second Taylor polynomial of u at (0, 0) is
T2 (x, y) = 1 + x + y + 2xy.
(b) v(x, y) = ln(e − x + 2y).
Solution. We have
vx =
−1
e − x + 2y
and vy =
2
.
e − x + 2y
Moreover,
vxx =
−1
,
(e − x + 2y)2
vxy =
2
,
(e − x + 2y)2
vyy =
−4
.
(e − x + 2y)2
Consequently, v(0, 0) = ln e = 1, vx (0, 0) = −1/e, vy (0, 0) = 2/e, vxx (0, 0) = −1/e2 ,
vxy (0, 0) = 2/e2 , and vyy (0, 0) = −4/e2 . Thus, the second Taylor polynomial of v at
(0, 0) is
x 2y
x2
2xy 2y 2
T2 (x, y) = 1 − +
− 2+ 2 − 2 .
e
e
2e
e
e
4

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