MATH 2004 Homework Solution
Han-Bom Moon
Homework 8 Model Solution
Section 14.5 ∼ 14.6.
14.5.2 Use the Chain Rule to find
dz
where
dt
x = 5t4 ,
z = cos(x + 4y),
dz
dt
=
1
y= .
t
∂z dx ∂z dy
+
∂x dt
∂y dt
1
= − sin(x + 4y)20t + (−4) sin(x + 4y) − 2
t
4
=
−20t3 + 2 sin(x + 4y)
t
4
4
=
−20t3 + 2 sin 5t4 +
t
t
3
14.5.5 Use the Chain Rule to find
dw
where
dt
w = xey/z ,
dw
dt
=
=
=
=
=
x = t2 ,
y = 1 − t,
z = 1 + 2t.
∂w dx ∂w dy ∂w dz
+
+
∂x dt
∂y dt
∂z dt
xy x y/z
y/z
e · 2t + e · (−1) + − 2 ey/z · 2
z
z
x 2xy
t2
2t2 (1 − t) (1−t)/(1+2t)
y/z
2t − − 2 e
= 2t −
−
e
z
z
1 + 2t
(1 + 2t)2
2t(1 + 2t)2 − t2 (1 + 2t) − 2t2 (1 − t) (1−t)/(1+2t)
e
(1 + 2t)2
2t + 5t2 + 8t3
e(1−t)(1+2t)
(1 + 2t)2
14.5.10 Use the Chain Rule to find
∂z
∂z
and
where
∂s
∂t
s
x= ,
t
z = ex+2y ,
1
t
y= .
s
MATH 2004 Homework Solution
Han-Bom Moon
∂z ∂x ∂z ∂y
+
∂x ∂s ∂y ∂s
t
1 2t x+2y
x+2y
x+2y 1
− 2 =
− 2 e
= e
· + 2e
t
s
t
s
2t
s
1 2t
=
− 2 et+ s
t
s
∂z
∂s
=
∂z
∂t
=
∂z ∂x ∂z ∂y
+
∂x ∂t
∂y ∂t
s
s
2 x+2y
x+2y
x+2y 1
e
= e
− 2 + 2e
· = − 2+
t
s
t
s
s
2t
s
2
et+ s
=
− 2+
t
s
14.5.11 Use the Chain Rule to find
∂z
∂z
and
where
∂s
∂t
z = er cos θ,
∂z
∂s
=
r = st,
θ=
p
s2 + t2 .
∂z ∂r ∂z ∂θ
+
∂r ∂s ∂θ ∂s
s
s
= er cos θ · t + (−er sin θ) √
= t cos θ − √
sin θ er
s2 + t2
s2 + t2
p
p
s
=
t cos( s2 + t2 ) − √
sin( s2 + t2 ) est
2
s + t2
∂z
∂t
=
∂z ∂r ∂z ∂θ
+
∂r ∂t
∂θ ∂t
t
t
= e cos θ · s + (−e sin θ) √
= s cos θ − √
sin θ er
s2 + t2
s2 + t2
p
p
t
2
2
2
2
=
s cos( s + t ) − √
sin( s + t ) est
s2 + t2
r
r
14.5.23 Use the Chain Rule to find
∂w ∂w
,
where
∂r ∂θ
w = xy + yz + zx,
when r = 2, θ =
x = r cos θ,
y = r sin θ,
z = rθ,
π
.
2
∂w
∂r
∂w ∂x ∂w ∂y ∂w ∂z
+
+
∂x ∂r
∂y ∂r
∂z ∂r
= (y + z) cos θ + (x + z) sin θ + (y + x)θ
=
2
MATH 2004 Homework Solution
Han-Bom Moon
π
π
π
π
r = 2, θ = ⇒ x = 2 cos = 0, y = 2 sin = 2, z = 2 · = π
2
2
2
2
π
π
π
∂w = (2 + π) cos + (0 + π) sin + (2 + 0) = 2π
∂r r=2,θ= π
2
2
2
2
∂w ∂x ∂w ∂y ∂w ∂z
+
+
∂x ∂θ
∂y ∂θ
∂z ∂θ
= (y + z)(−r sin θ) + (x + z)r cos θ + (y + x)r
∂w
∂θ
=
∂w π
π
= (2 + π)(−2 sin ) + (0 + π)2 cos + (2 + 0)2 = −2π
∂r r=2,θ= π
2
2
2
14.5.26 Use the Chain Rule to find
u = xety ,
∂u ∂u ∂u
,
,
where
∂α ∂β ∂γ
x = α2 β,
y = β 2 γ,
t = γ 2 α,
when α = −1, β = 2, γ = 1.
∂u ∂x ∂u ∂y
∂u ∂t
+
+
∂x ∂α ∂y ∂α
∂t ∂α
ty
ty
= e 2αβ + xte · 0 + xyety γ 2 = (2αβ + xyγ 2 )ety
∂u
∂α
=
α = −1, β = 2, γ = 1 ⇒ x = (−1)2 · 2 = 2, y = 22 · 1 = 4, t = 12 · (−1) = −1
∂u = (2 · (−1) · 2 + 2 · 4 · 12 )e−1·4 = 4e−4
∂α α=−1,β=2,γ=1
∂u
∂β
∂u ∂x ∂u ∂y
∂u ∂t
+
+
∂x ∂β
∂y ∂β
∂t ∂β
ty 2
ty
= e α + xte 2βγ + xyety · 0 = (α2 + 2xtβγ)ety
=
∂u = ((−1)2 + 2 · 2 · (−1) · 2 · 1)e−1·4 = −7e−4
∂β α=−1,β=2,γ=1
∂u
∂γ
∂u ∂x ∂u ∂y
∂u ∂t
+
+
∂x ∂γ
∂y ∂γ
∂t ∂γ
ty
ty 2
= e · 0 + xte β + xyety · 2γα = (xtβ 2 + 2xyγα)ety
=
∂u = (2 · (−1) · 22 + 2 · 2 · 4 · 1 · (−1))e−1·4 = −24e−4
∂γ α=−1,β=2,γ=1
14.5.39 The length `, width w, and height h of a box change with time. At a certain
instant the dimensions are ` = 1 m and w = h = 2 m, and ` and w are increasing
at a rate of 2 m/s while h is decreasing at a rate of 3 m/s. At that instant find the
rates at which the following quantities are changing.
3
MATH 2004 Homework Solution
Han-Bom Moon
(a) The volume:
Volume = V = `wh
From the conditions above,
`(0) = 1, w(0) = 2, h(0) = 2,
d` dw dh = 2,
= 2,
= −3.
dt t=0
dt t=0
dt t=0
dV dt t=0
∂V d` ∂V dw ∂V dh
d`
dw
dh
dV
=
+
+
= wh + `h
+ `w
dt
∂` dt
∂w dt
∂h dt
dt
dt
dt
d` dw dh = wh +`h
+`w
= 2·2·2+1·2·2+1·2·(−3) = 6 (m3 /s)
dt t=0
dt t=0
dt t=0
(b) The surface area:
Surface area = A = 2`w + 2`h + 2wh
dA
∂A d` ∂A dw ∂A dh
d`
dw
dh
=
+
+
= (2w + 2h) + (2` + 2h)
+ (2` + 2w)
dt
∂` dt ∂w dt
∂h dt
dt
dt
dt
d` dw dh dA = (2w + 2h) + (2` + 2h)
+ (2` + 2w)
dt dt
dt dt t=0
t=0
t=0
t=0
2
= (2 · 2 + 2 · 2)2 + (2 · 1 + 2 · 2)2 + (2 · 1 + 2 · 2)(−3) = 10 (m /s)
(c) The length of a diagonal:
Diagonal = D =
p
`2 + w 2 + h2
dD
∂D d` ∂D dw ∂D dh
=
+
+
dt
∂` dt
∂w dt
∂h dt
w
h
`
d`
dw
dh
+√
+√
=√
`2 + w2 + h2 dt
`2 + w2 + h2 dt
`2 + w2 + h2 dt
dD `
d` w
dw h
dh =√
+√
+√
dt t=0
`2 + w2 + h2 dt t=0
`2 + w2 + h2 dt t=0
`2 + w2 + h2 dt t=0
=√
12
1
2
2
·2+ √
·2+ √
· (−3) = 0 (m/s)
2
2
2
2
2
2
+2 +2
1 +2 +2
1 + 22 + 2 2
14.5.43 One side of a triangle is increasing at a rate of 3 cm/s and a second side is
decreasing at a rate of 2 cm/s. If the area of the triangle remains constant, at what
rate does the angle between the sides change when the first side is 20 cm long,
π
the second side is 30 cm, and the angle is ?
6
Let x be the length of the first side, y be the length of the second side, and let θ be
the angle between them. then from the assumption,
dy π dx = 3,
= −2.
x(0) = 20, y(0) = 30, θ(0) = ,
6 dt t=0
dt t=0
The area of the triangle is
1
A := xy sin θ.
2
4
MATH 2004 Homework Solution
Because the area is constant,
By the Chain Rule,
dA 0=
=
dt t=0
Han-Bom Moon
dA
= 0.
dt
∂A dx ∂A dy ∂A dθ +
+
∂x dt t=0 ∂y dt t=0 ∂z dt t=0
y
dx x
dy 1
dθ sin θ
+ sin θ
+ xy cos θ
2
dt t=0 2
dt t=0 2
dt t=0
30
π
20
π
1
π dθ sin · 3 +
sin · (−2) + · 20 · 30 cos
2
6
2
6
2
6 dt t=0
√ dθ 25
+ 150 3
2
dt =
=
=
t=0
dθ 25
1
√ = − √ (rad/s)
⇒
=−
dt t=0
300 3
12 3
14.5.45 If z = f (x, y), where x = r cos θ and y = r sin θ,
(a) find
∂z
∂z
and
.
∂r
∂θ
∂z ∂x ∂z ∂y
∂z
∂z
∂z
=
+
=
cos θ +
sin θ
∂r
∂x ∂r
∂y ∂r
∂x
∂y
∂z
∂z ∂x ∂z ∂y
∂z
∂z
=
+
=
(−r sin θ) +
r cos θ
∂θ
∂x ∂θ
∂y ∂θ
∂x
∂y
(b) Show that
∂z
∂x
2
+
∂z
∂y
2
=
∂z
∂r
2
1
+ 2
r
∂z
∂θ
2
.
2
2
∂z
∂z
1 ∂z
∂z
cos θ +
sin θ + 2
(−r sin θ) +
r cos θ
∂x
∂y
r
∂x
∂y
2
2
∂z
∂z ∂z
∂z
2
=
cos θ + 2
cos θ sin θ +
sin2 θ
∂x
∂x ∂y
∂y
!
2
2
∂z
∂z
∂z
1
∂z
+ 2
r2 sin2 θ − 2
r2 sin θ cos θ +
r2 cos θ
r
∂x
∂x ∂y
∂y
2
2
∂z ∂z
∂z
∂z
2
2
cos θ + sin θ +2
sin2 θ + cos2 θ
=
(cos θ sin θ − cos θ sin θ)+
∂x
∂x ∂y
∂y
2 2
∂z
∂z
=
+
∂x
∂y
14.6.6 Find the directional derivative of f (x, y) = ex cos y at (0, 0) in the direction indiπ
cated by the angle θ = .
4
π
π
1 1
Direction vector u: hcos , sin i = h √ , √ i
4
4
2 2
∇f = hfx , fy i = hex cos y, −ex sin yi
5
MATH 2004 Homework Solution
Han-Bom Moon
∇f (0, 0) = he0 cos 0, −e0 sin 0i = h1, 0i
1 1
1
Du f (0, 0) = ∇f (0, 0) · u = h1, 0i · h √ , √ i = √
2 2
2
14.6.8
(a) Find the gradient of f (x, y) =
y2
.
x
∇f = hfx , fy i = h−
y 2 2y
, i
x2 x
(b) Evaluate the gradient at P = (1, 2).
∇f (1, 2) = h−
22 2 · 2
,
i = h−4, 4i
12 1
√
1
(c) Find the rate of change of f at P in the direction of the vector u = (2i+ 5j).
3
√
√
2 5
−8 + 4 5
Du f (1, 2) = ∇f (1, 2) · u = h−4, 4i · h ,
i=
3 3
3
14.6.15 Find the directional derivative of f (x, y, z) = xey + yez + zex at the given point
(0, 0, 0) in the direction of v = h5, 1, −2i.
|v| =
p
√
52 + 12 + (−2)2 = 30
The unit vector to the direction of v:
u=
v
5
1
2
= h√ , √ , −√ i
|v|
30 30
30
∇f = hfx , fy , fz i = hey + zex , xey + ez , yez + ex i
∇f (0, 0, 0) = he0 + 0e0 , 0e0 + e0 , 0e0 + e0 i = h1, 1, 1i
5
1
2
4
Du f (0, 0, 0) = ∇f (0, 0, 0) · u = h1, 1, 1i · h √ , √ , − √ i = √
30 30
30
30
14.6.24 Find the maximum rate of change of f (x, y, z) =
direction in which it occurs.
x+y
at (1, 1, −1) and the
z
1 1 x+y
∇f = hfx , fy , fz i = h , , − 2 i
z z
z
1+1
1 1
∇f (1, 1, −1) = h ,
.−
i = h−1, −1, −2i
−1 −1
(−1)2
The maximum rate of change occurs to the direction of ∇f (1, 1, −1) = h−1, −1, −2i.
1
1
2
(The unit vector in this direction is h− √ , − √ , − √ i.) In this case, the maxi6
6
p6
√
mum rate of change is |∇f (1, 1, −1)| = (−1)2 + (−1)2 + (−2)2 = 6.
6
MATH 2004 Homework Solution
Han-Bom Moon
14.6.35 Let f be a function of two variables that has continuous partial derivatives
and consider the points A(1, 3), B(3, 3), C(1, 7), and D(6, 15). The directional
−−→
derivative of f at A in the direction of the vector AB is 3 and the directional
−→
derivative at A in the direction of AC is 26. Find the directional derivative of f at
−−→
A in the direction of the vector AD.
−−→
AB = h3, 3i − h1, 3i = h2, 0i
−−→
The unit vector u to the direction of AB is i.
−→
AC = h1, 7i − h1, 3i = h0, 4i
−→
The unit vector v to the direction of AC is j.
If ∇f (1, 3) = ha, bi,
3 = Du f (1, 3) = ∇f (1, 3) · u = ha, bi · i = a
26 = Dv f (1, 3) = ∇f (1, 3) · u = ha, bi · j = b
⇒ ∇f (1, 3) = h3, 26i
−−→
AD = h6, 15i − h1, 3i = h5, 12i
p
√
−−→
|AD| = 52 + 122 = 169 = 13
−−→
5 12
The unit vector w to the direction of AD is h , i.
13 13
Dw f (1, 3) = ∇f (1, 3) · w = h3, 26i · h
14.6.43
5 12
327
, i=
13 13
13
(a) Find an equation of the tangent plane to xyz 2 = 6 at (3, 2, 1).
A point on the tangent plane: (3, 2, 1)
f (x, y, z) = xyz 2
∇f = hfx , fy , fz i = hyz 2 , xz 2 , 2xyzi
∇f (3, 2, 1) = h2 · 12 , 3 · 12 , 2 · 3 · 2 · 1i = h2, 3, 12i
A normal vector: h2, 3, 12i
An equation of the tangent plane:
2(x − 3) + 3(y − 2) + 12(z − 1) = 0
or
2x + 3y + 12z = 24
7
MATH 2004 Homework Solution
Han-Bom Moon
(b) Find equations of the normal line to xyz 2 = 6 at (3, 2, 1).
A point on the normal line: (3, 2, 1)
A direction vector: ∇f (3, 2, 1) = h2, 3, 12i
Symmetric equations of the normal line:
x−3
y−2
z−1
=
=
2
3
12
14.6.51 Show that the equation of the tangent plane to the ellipsoid
at the point (x0 , y0 , z0 ) can be written as
x2
y2
z2
+
+
=1
a2
b2
c2
xx0 yy0 zz0
+ 2 + 2 = 1.
a2
b
c
A point on the plane: (x0 , y0 , z0 )
x2 y 2 z 2
+ 2 + 2
a2
b
c
2x 2y 2z
∇f = hfx , fy , fz i = h 2 , 2 , 2 i
a b c
2x0 2y0 2z0
∇f (x0 , y0 , z0 ) = h 2 , 2 , 2 i
a
b
c
2x0 2y0 2z0
x0 y0 z0
A normal vector: ∇f (x0 , y0 , z0 ) = h 2 , 2 , 2 i or h 2 , 2 , 2 i
a
b
c
a b c
An equation of the tangent plane:
f (x, y, z) =
y0
z0
x0
(x − x0 ) + 2 (y − y0 ) + 2 (z − z0 ) = 0
2
a
b
c
x0 x x20 y0 y y02 z0 z z02
− 2 + 2 − 2 + 2 − 2 =0
a2
a
b
b
c
c
2
2
xx0 yy0 zz0
x0 y0
z02
+
+
=
+
+
a2
b2
c2
a2
b2
c2
Because (x0 , y0 , z0 ) is on the ellipsoid,
xx0 yy0 zz0
+ 2 + 2 = 1.
a2
b
c
14.6.55 Are there any points on the hyperboloid x2 − y 2 − z 2 = 1 where the tangent
plane is parallel to the plane z = x + y?
The tangent plane at (x0 , y0 , z0 ) is parallel to the plane z = x + y when the normal vector ∇f (x0 , y0 , z0 ) is parallel to h1, 1, −1i, or equivalently, ∇f (x0 , y0 , z0 ) =
ch1, 1, −1i for some constant c.
f (x, y, z) = x2 − y 2 − z 2
∇f = hfx , fy , fz i = h2x, −2y, −2zi
∇f (x0 , y0 , z0 ) = h2x0 , −2y0 , −2z0 i = ch1, 1, −1i
8
MATH 2004 Homework Solution
Han-Bom Moon
⇒ 2x0 = c, −2y0 = c, −2z0 = −c
c
c
c
⇒ x0 = , y 0 = − , z 0 =
2
2
2
Because (x0 , y0 , z0 ) is on the hyperboloid,
1 = x20 − y02 − z02 =
c 2
2
c 2 c 2
c2
− −
−
=−
2
2
4
The right hand side is always non-positive. Therefore there is no solution and
there is no such point.
14.6.59 Where does the normal line to the paraboloid z = x2 + y 2 at the point (1, 1, 2)
intersect the paraboloid a second time?
f (x, y, z) = x2 + y 2 − z = 0
∇f = h2x, 2y, −1i
∇f (1, 1, 2) = h2, 2, −1i
A direction vector: h2, 2, −1i
A point on the line: (1, 1, 2)
A parametric equation of the normal line:
r(t) = h1, 1, 2i + th2, 2, −1i = h1 + 2t, 1 + 2t, 2 − ti
f (r(t)) = (1 + 2t)2 + (1 + 2t)2 − (2 − t) = 0
⇒ 2(1 + 4t + 4t2 ) − 2 + t = 0 ⇒ 8t2 + 9t = 0 ⇒ t = 0, −
9
8
Another intersection point:
9
5 5 25
r(− ) = h− , − , i
8
4 4 8
14.6.62 Show that the pyramids cut off from the first octant by any tangent planes to
the surface xyz = 1 at points in the first octant must all have the same volume.
f (x, y, z) = xyz
∇f = hyz, xz, xyi
∇f (x0 , y0 , z0 ) = hy0 z0 , x0 z0 , x0 y0 i
The tangent plane at (x0 , y0 , z0 ):
y0 z0 (x − x0 ) + x0 z0 (y − y0 ) + x0 y0 (z − z0 ) = 0
xy0 z0 − x0 y0 z0 + yx0 z0 − x0 y0 z0 + zx0 y0 − x0 y0 z0 = 0
xy0 z0 + yx0 z0 + zx0 y0 = 3
9
MATH 2004 Homework Solution
Han-Bom Moon
because x0 y0 z0 = 1.
The intersection of tangent plane and x-axis:
y=z=0⇒x=
3
y0 z0
x=z=0⇒y=
3
x 0 z0
x=y=0⇒z=
3
x 0 y0
The intersection with y-axis:
The intersection with z-axis:
The area of the base of the pyramid:
1
3
3
9
9
·
·
=
=
2
2 y0 z0 x 0 z0
2z0
2x0 y0 z0
The height of the pyramid:
3
x 0 y0
The volume of the pyramid:
1 9
1
3
9
9
· area of the base · height = ·
·
=
=
3
3 2z0 x0 y0
2x0 y0 z0
2
Therefore it is independent from the choice of a point (x0 , y0 , z0 ) on the surface.
10