Module MA1132 (Frolov), Advanced Calculus
Homework Sheet 5
Each set of homework questions is worth 100 marks
Due: at the beginning of the tutorial session Thursday/Friday, 25/26 February 2016
Name:
Use Mathematica to check your answers.
1. Use appropriate forms of the chain rule to find
z = cos
x
sin 2y ;
2
∂z
∂v
where
x = 2u + 3v , y = u3 − 2v 2 .
Solution: We have
∂z
∂z ∂x ∂z ∂y
1
x
x
=
+
= − sin sin 2y × 3 + 2 cos cos 2y × (−4v)
∂v
∂x ∂v ∂y ∂v
2
2
2
3
x
x
= − sin sin 2y − 8v cos cos 2y
2
2
2
3
2u + 3v
2u + 3v
= − sin
sin 2(u3 − 2v 2 ) − 8v cos
cos 2(u3 − 2v 2 ) .
2
2
2
2. The Taylor series is given by
∞
X
f (~x) =
k1 ,...,kn
∂1k1 · · · ∂nkn f (x~o ) k1
∆x1 · · · ∆xknn ,
k
!
·
·
·
k
!
1
n
=0
(1)
where we denote
f (xo1 , . . . , xon ) ≡ f (x~o ) ,
f (x1 , . . . , xn ) ≡ f (~x) ,
xi − xoi ≡ ∆xi
(2)
k
and ∂i0 f ≡ f ; ∂ik f ≡ ∂∂xfk is the k-th partial derivative of f with respect to xi . Thus, the
i
Taylor series is an expansion in powers of ∆xi . One can write
f (~x) = f (x~o + ∆~x) .
(3)
F (t) = f (x~o + t∆~x) ,
(4)
Then, introducing the function
one sees that the Taylor expansion of F (t) in powers of t evaluated at t = 1 produces
the Taylor series of f (~x) in powers of ∆xi . Use this observation to show that the Taylor
series can be equivalently written as
n
∞
X
∂ q f (x~o )
1 X
∆xi1 · · · ∆xiq .
f (~x) =
q! i ,...,i =1 ∂xi1 · · · ∂xiq
q=0
1
q
1
(5)
Solution: The Taylor expansion of F (t) in powers of t is given by
F (t) =
∞ q q
X
t d F (0)
q!
q=0
⇒
dtq
F (1) =
∞
X
1 dq F (0)
.
q
q!
dt
q=0
(6)
Then one has
n
X
∂f (x~o + t∆~x)
dF
=
∆xi
dt
∂xi
i=1
⇒
n
X
d2 F
∂ 2 f (x~o + t∆~x)
∆xi ∆xj
=
dt2
∂x
∂x
i
j
i,j=1
n
dF (0) X ∂f (x~o )
=
∆xi ,
dt
∂xi
i=1
n
X
dF (0)
∂f (x~o )
=
∆xi ∆xj ,
dt
∂x
∂x
i
j
i,j=1
⇒
·········
n
X
dq F
∂ q f (x~o + t∆~x)
=
∆xi1 · · · ∆xiq
dtq
∂xi1 · · · ∂xiq
i ,...,i =1
⇒
q
1
n
X
dq F (0)
∂ q f (x~o )
=
∆xi1 · · · ∆xiq .
dtq
∂xi1 · · · ∂xiq
i ,...,i =1
1
q
(7)
3. Let r =
Pn
i=1
xi ei and r = |r|, where ei form an orthonormal basis of vectors in Rn . Find
∂f (r)
,
∂xi
i = 1, 2, · · · , n ,
and ∇f (r) ,
|∇f (r)|2 ,
where f is a smooth function of a single variable.
Solution: We use the chain rule to get
n
X ∂f (r)
∂r
xi
r
∂f (r)
= f 0 (r)
= f 0 (r) , ∇f (r) =
ei = f 0 (r) ,
∂xi
∂xi
r
∂xi
r
i=1
2 2
n X
∂f (r)
df (r)
2
|∇f (r)| =
=
∂xi
dr
i=1
(8)
4. Show that if z = f (x, y), x = r cos θ, y = r sin θ, then
∂ 2 z 1 ∂z
∂ 2z ∂ 2z
1 ∂ 2z
+
=
+
+
.
∂x2 ∂y 2
∂r2 r ∂r r2 ∂θ2
Solution: We have
∂z
∂z ∂r ∂z ∂θ
=
+
,
∂x
∂r ∂x ∂θ ∂x
Then
r=
p
x2 + y 2
y
θ = arctan
x
Thus
⇒
⇒
∂z
∂z ∂r ∂z ∂θ
=
+
.
∂y
∂r ∂y ∂θ ∂y
∂r
x
∂r
y
= = cos θ ,
= = sin θ ,
∂x
r
∂y
r
∂θ
y
sin θ
∂θ
x
cos θ
=− 2 =−
,
= 2 =
.
∂x
r
r
∂y
r
r
∂z
∂z sin θ
∂z
=
cos θ −
,
∂x
∂r
∂θ r
2
∂z
∂z
∂z cos θ
=
sin θ +
.
∂y
∂r
∂θ r
(9)
(10)
(11)
Differentiating these formulae one gets
∂ 2z
∂z sin θ ∂r
∂ ∂z
∂z sin θ ∂θ
∂ ∂z
cos θ −
+
cos θ −
,
=
∂x2
∂r ∂r
∂θ r
∂x ∂θ ∂r
∂θ r
∂x
2
∂ z
∂ 2 z sin θ ∂z sin θ
=
+
cos θ
cos θ −
∂r2
∂r∂θ r
∂θ r2
2
∂ z
∂z
∂ 2 z sin θ ∂z cos θ sin θ
−
cos θ −
sin θ − 2
−
,
∂θ∂r
∂r
∂θ r
∂θ r
r
∂ 2z
∂z cos θ ∂r
∂ ∂z
∂z cos θ ∂θ
∂ ∂z
sin θ +
+
sin θ +
,
=
∂y 2
∂r ∂r
∂θ r
∂y ∂θ ∂r
∂θ r
∂y
2
∂ z
∂ 2 z cos θ ∂z cos θ
=
−
sin θ
sin θ +
∂r2
∂r∂θ r
∂θ r2
2
∂z
∂ 2 z cos θ ∂z sin θ cos θ
∂ z
sin θ +
cos θ + 2
−
.
+
∂θ∂r
∂r
∂θ r
∂θ r
r
(12)
Collecting the terms one gets the formula.
5. Consider the surface
z = f (x, y) = ln
1 2/3 p
e 3 8x2 − 6xy 2 − y 3 + 32 − 12 sin(2x − y)
2
.
(a) Find an equation for the tangent plane to the surface at the point P = (1, 2, z0 )
where z0 = f (1, 2).
(b) Find points of intersection of the tangent plane with the x-, y- and z-axes.
(c) Sketch the tangent plane, and show the point P = (1, 2, z0 ) on it.
(d) Find parametric equations for the normal line to the surface at the point P =
(1, 2, z0 ).
(e) Sketch the normal line to the surface at the point P = (1, 2, z0 ).
Solution:
(a) We first find
8x2 − 6xy 2 − y 3 + 32 − 12 sin(2x − y)|P = 8 ,
z0 =
2
,
3
and then simplify
1 2/3 p
3
2
2
3
z = ln
e
8x − 6xy − y + 32 − 12 sin(2x − y)
2
1
2
=
ln(8x2 − 6xy 2 − y 3 + 32 − 12 sin(2x − y)) + − ln 2 .
3
3
Then, we compute the partial derivatives at P
∂
4
z|P = − .
∂x
3
3
(13)
∂
z|P = −1 .
∂y
The tangent plane equation is given by
z=
2 4
4
− (x − 1) − 1(y − 2) = 4 − x − y .
3 3
3
(b) (3, 0, 0), (0, 4, 0) , (0, 0, 4)
(c) The tangent plane is the one through the points in (b).
(d) The normal line to the surface (and the tangent plane) is given by
2
4
r = i + 2j + k + t i + j + k .
3
3
(e) The normal line is perpendicular to the plane.
6. Show that the equation of the plane that is tangent to the cone
x2 y 2 z 2
+ 2 − 2 =0
a2
b
c
at (x0 , y0 , z0 ) can be written in the form
x0
y0
z0
x + 2y − 2z = 0.
2
a
b
c
Solution: Consider the function F (x, y, z) =
∇F = 2
x2
a2
+
y2
b2
−
z2
.
c2
The gradient ∇F
x
y
z
e1 + 2 2 e2 − 2 2 e3
2
a
b
c
(14)
is normal to the level surfaces of F , and therefore to tangent planes to the level surfaces
of F . Thus, the equation of the plane tangent to the cone at (x0 , y0 , z0 ) can be written
in the form
1
(r − r0 ) · ∇F (x0 , y0 , z0 ) = 0 .
(15)
2
Explicitly one gets
xx0 yy0 zz0
x0 x0 y0 y0 z0 z0
+ 2 − 2 − ( 2 + 2 − 2 ) = 0,
2
a
b
c
a
b
c
(16)
which proves the formula.
7. Prove: If the surfaces z = f (x1 , . . . , xn ) and z = g(x1 , . . . , xn ) intersect at P = (xo1 , . . . , xon , z o ),
and if f and g are differentiable at (xo1 , . . . , xon ), then the normal lines at P are perpendicular if and only if
n
X
∂f (xo , . . . , xo ) ∂g(xo , . . . , xo )
1
i=1
n
∂xi
1
∂xi
4
n
= −1 .
Solution: Consider the functions F (x1 , . . . , xn , z) = f (x1 , . . . , xn ) − z and G(x1 , . . . , xn , z) =
g(x1 , . . . , xn ) − z. The normal lines to the level surfaces of F and G are parallel to ∇F
and ∇G. Since
n
n
X
X
∂g
∂f
∇F =
ei − ez , ∇G =
ei − ez ,
(17)
∂xi
∂xi
i=1
i=1
one gets that ∇F and ∇G are perpendicular if and only if
∇F · ∇G = 1 +
5
n
X
∂f ∂g
= 0.
∂x
∂x
i
i
i=1
(18)