MA2712 – Problems
2016 | October | 18
Martins Bruveris
4
The Chain Rule
32. Suppose that a duck is swimming along a curve x = 3 + 2t, y = 2 − t 2 , while the water temperature
dT
is given by the formula T = e x (x + y). Find
in two ways:
dt
(a) by the chain rule and
(b) by expressing T in terms of t and differentiating.
∂T
∂T
and
, because T (x, y) is a function of two variables. After
∂x
∂y
dT
subsituting T (t) is a function of one variable and therefore we write
.
dt
(a) By the chain rule the temperature of the duck is given by
Solution. Note that we write
dT
∂ T dx ∂ T d y
=
+
.
dt
∂ x dt
∂ y dt
We have
∂T
= e x (1 + x + y)
∂x
∂T
= ex ,
∂y
and after substitution
∂T
= e3+2t 6 + 2t − t 2
∂x
∂T
= e3+2t .
∂y
The derivatives of x(t) and y(t) are
dy
= −2t .
dt
dx
=2
dt
Therefore
∂ T dx ∂ T d y
+
= e3+2t 12 + 2t − 2t 2 ,
∂ x dt
∂ y dt
by the chain rule.
(b) If we express T as a function of t, we get
T (t) = e3+2t 5 + 2t − t 2 ,
and differentiating it leads to
dT
= e3+2t 2 5 + 2t − t 2 + 2 − 2t
dt
= e3+2t 12 + 2t − 2t 2 .
We see that both method lead to the same result.
MA2712
2016 | October | 18
Problems
33. Suppose that a bird flies along a helical curve x = 2 cos t, y = 2 sin t, z = 3t. The bird suddenly
encounters a weather front, so that the barometric pressure is varying rather wildly from point to
3 x 2z
atm.
point as P(x, y, z) =
10 y
(a) Use the chain rule to determine how the pressure is changing at t =
π
min.
4
(b) Using the linear approximation, determine the approximate pressure at t =
π
+ 0.01 min.
4
Solution. Note that we can ignore the physical units for most of the calculations, but we will use
them to state the answer.
(a) The partial derivatives of P(x, y, z) are
∂P
3 xz
=
∂x
5 y
∂P
3 x 2z
=−
∂y
10 y 2
∂P
3 x2
=
,
∂z
10 y
and the derivative of the bird’s trajectory is
dx
= −2 sin t
xt
dy
= 2 cos t
xt
dz
= 3.
xt
The chain rule now gives
dP
9t cos t
9t cos2 t
3 cos2 t
=
· (−2) sin t −
·
2
cos
t
+
·3
dt
5 sin t
10 sin2 t
5 sin t
18t
9t cos3 t 9 cos2 t
=−
cos t −
+
.
5
5 sin2 t 5 sin t
Evaluated at t =
π
we obtain
4
p 
‹
dP π
9 2
3π
=−
−
+ 1 ≈ −1.726 . . . .
dt 4
10
4
Therefore at
π
atm
min the barometric pressure changes at a rate of −1.726
.
4
min
(b) First we evaluate
π 9πp2
P(
=
≈ 0.9996 . . .
4
40
The linear approximation tells us that
P
π
π dP π
+ 0.01 ≈ P
+
· 0.01
4
4
dt 4
≈ 0.9996 − 1.726 · 0.01 ≈ 0.9823 .
The problem is stated in physical units, so it is natural to enquire, whether the numbers make sense.
π
At t = min the barometric pressure is 0.9996 atm, which is average sea-level pressure. How much
4
does pressure change over time? The highest recorded barometric presssure is about 1.08 atm and
the lowest pressures occur during typhoons and are about 0.86 atm. Hence a rate of change of
atm
−1.726
would indicate an extremely rapid decrease.
min
Martins Bruveris
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MA2712
2016 | October | 18
Problems
34. Use the chain rule to determine an expression for the indicated derivative.
(a) Let z =
Æ
x 2 + y 2 + 2x y 2 , where x and y are functions of u. Find an expression for
(b) Let u = sin (a + ln b), where a and b are functions of t. Find an expression for
dz
.
du
du
.
dt
Solution.
(a) The chain rule states that
dz
∂ z dx
∂z dy
=
+
.
du ∂ x du ∂ y du
The partial derivatives of z = z(x, y) are
y
∂z
=p
+ 4x y .
∂y
x2 + y2
x
∂z
=p
+ 2 y2
2
2
∂x
x +y
Hence
1
dz
=p
du
x2 + y2

‹
dy
dy
dx
dx
x
+y
+ 2 y2
+ 4x y
.
du
du
du
du
(b) The chain rule states that
du ∂ u d a ∂ u d b
=
+
.
dt
∂ a dt ∂ b dt
The partial derivatives of u = u(a, b) are
∂u
1
= cos(a + ln b) .
∂b
b
∂u
= cos(a + ln b)
∂a
Hence

‹
du
da 1 d b
= cos(a + ln b)
+
.
dt
dt
b dt
d
35. For the following functions, calculate the derivative
( f ◦ σ) using the chain rule.
dt
ۮ
Š
(a) f (x, y) = (x 2 + y 2 ) ln
x 2 + y 2 ; σ(t) = (e t , e−t ).
(b) f (x, y) = x e x
2
+ y2
; σ(t) = (t, −t).
(c) f (x, y, z) = x + y 2 + z 3 ; σ(t) = (cos t, sin t, t).
(d) f (x, y, z) = e x−z ( y 2 − x 2 ); σ(t) = (t, e t , t 2 ).
Solution. We will write σ = (σ1 , σ2 ) for the components of the curve σ. Thus if σ(t) = (e t , e−t ),
then σ1 (t) = e t and σ2 (t) = e−t .
(a) The partial derivatives of f are
€
Š
Æ
Æ
f x = x 2 ln x 2 + y 2 + x 2 + y 2
€
Š
Æ
Æ
f y = y 2 ln x 2 + y 2 + x 2 + y 2 ,
and
σ 0 (t) = e t , −e−t .
Thus
€
Š
p
p
d
( f ◦ σ) = e t 2 ln e2t + e−2t + e2t + e−2t · e t
dt
€
Š
p
p
+ e−t 2 ln e2t + e−2t + e2t + e−2t · −e−t
Š
p
p
€
= e2t − e−2t 2 ln e2t + e−2t + e2t + e−2t .
Martins Bruveris
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MA2712
2016 | October | 18
Problems
(b) The partial derivatives of f are
f x = (1 + 2x 2 )e x
2
+ y2
f y = 2x y e x
2
+ y2
,
and
σ 0 (t) = (1, −2) .
Thus
2
2
2
d
( f ◦ σ) = (1 + 2t 2 )e2t · 1 − 2t 2 e2t · (−1) = 1 + 4t 2 e2t .
dt
(c) The partial derivatives of f are
fx = 1
fz = 3z 2 ,
fy = 2y
and
σ 0 (t) = (− sin t, cos t, 1) .
Thus
d
( f ◦ σ) = 1 · (− sin t) + 2 sin t · cos t + 3t 2 · 1 = − sin t + 2 sin t cos t + 3t 2 .
dt
(d) The partial derivatives of f are
f x = e x−z ( y 2 − 2x − x 2 )
fz = −e x−z ( y 2 − x 2 ) .
f y = 2 y e x−z
and
σ 0 (t) = (1, e t , 2t) .
Thus
2
2
2
d
( f ◦ σ) = e t−t e2t − 2t − t 2 · 1 + 2e t e t−t · e t − e t−t e2t − t 2 · 2t
dt
2
2
= (3 − 2t)e3t−t − (2t + t 2 − 2t 3 )e t−t .
36. In this problem we will use the chain rule to rederive results from single-variable calculus.
(a) Apply the chain rule to the function u = x yz, where x = x(t), y = y(t) and z = z(t) are
functions of t to get a rule for differentiating a product of three functions of one variable.
(b) Use the chain rule and the function f ( y, z) = y z to find
d
(x x ).
dx
Solution.
(a) The chain rule gives
du
∂ u d x ∂ u d y ∂ u dz
=
+
+
.
dt
∂ x dt ∂ y dt
∂ z dt
The partial derivatives are
∂u
= yz
∂x
∂u
= xz
∂y
∂u
= xy,
∂z
and hence
dy
du
dx
dz
=
yz + x
z+xy
,
dt
dt
dt
dt
which is the product rule for a product of three factors. A more common form would be
( f gh)0 = f 0 gh + f g 0 h + f gh0 .
Martins Bruveris
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MA2712
2016 | October | 18
Problems
(b) If we consider y(x) = x and z(x) = x as functions of x, then
x x = f (x, x) ,
where f ( y, z) = y z . Using the chain rule we have
∂ f d y ∂ f dz
d
(x x ) =
+
.
dx
∂ y dx
∂ z dx
To compute the partial derivatives of f we make the following observation: keeping z fixed
we can differentiate y z like a polynomial; keeping y fixed we can write y = eln y and thus
y z = ez ln y . Therefore the partial derivatives of f ( y, z) are
∂f
= y z ln y ,
∂z
∂f
= z y z−1
∂y
and both
dy
dz
= 1 and
= 1. Thus
dx
dx
d
(x x ) = x x x−1 + x x ln x = x x (1 + ln x) .
dx
37. Let F (u, v) be a function of two variables. Suppose that u = x + y and v = x y. Express
terms of u- and v-derivatives of F .
Solution. We first compute
∂ 2F
in
∂ x∂ y
∂F
∂F
and
,
∂x
∂y
∂F
∂F
∂F
=
+y
∂x
∂u
∂v
∂F
∂F
∂F
=
+x
.
∂x
∂u
∂v
Note that these formulas are valid for all functions F (u, v). In particular we can replace F by
∂ ∂F
∂
=
∂x∂y
∂x

∂F
∂F
+x
∂u
∂v
‹
∂ 2F
∂F
∂ 2F
+
y
+
+x
=
∂ u2
∂ v∂ u ∂ v
=
∂F
,
∂y
∂ 2F
∂ 2F
+y
∂ u∂ v
∂ v2
∂ F ∂ 2F
∂ 2F
∂ 2F
+
+
(x
+
y)
+
x
y
.
∂ v ∂ u2
∂ u∂ v
∂ v2

‹
∂
∂F
When calculating
x
we had to apply the product rule. Differentiating the first factor gives
∂x
∂v
∂
(x) = 1 and for the second factor we use the result from above, which was obtained using the
∂x
chain rule.
38. Let u(x, y, z) be a function of three variables. Suppose that x = pq2 r, y = pq3 and z = p3 . Express
∂ 2u
in terms of x-, y- and z-derivatives of u.
∂ p∂ r
Martins Bruveris
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MA2712
2016 | October | 18
Problems
Solution. We start with the first derivatives
∂u
∂u
∂u
∂u
= q2 r
+ q3
+ 3p2
∂p
∂x
∂y
∂z
∂u
∂u
= pq2
.
∂r
∂x
We do not need the q-derivative in this problem. Then
2
2
2
∂
∂ 2u
2 ∂u
2
2 ∂ u
3 ∂ u
2 ∂ u
=
pq
+ pq q r
+q
+ 3p
∂ p∂ r
∂p
∂x
∂ x2
∂ y∂ x
∂ z∂ x
= q2
2
2
∂u
∂ 2u
5 ∂ u
3 2 ∂ u
+ pq4 r
+
pq
+
3p
q
.
∂x
∂ x2
∂ x∂ y
∂ x∂ z
39. Cartesian (rectangular) coordinates (x, y) of a point can be expressed in terms of polar coordinates
(r, θ ) using the expressions
x = r cos θ
y = r sin θ .
Suppose that a function is given in terms of cartesian coordinates by u = f (x, y).
(a) Express
∂u
∂u
∂u
∂u
and
in terms of
and
.
∂r
∂θ
∂x
∂y
(b) Show that
u2x + u2y = u2r +
1 2
u .
r2 θ
(c) Express
1
1
u r r + u r + 2 uθ θ
r
r
in cartesian coordinates.
Solution. The derivative of (x, y) with respect to (r, θ ) is
∂x
= −r sin θ
∂θ
∂y
= r cos θ .
∂θ
∂x
= cos θ
∂r
∂y
= sin θ
∂r
(a) We can do this by simply applying the chain rule
∂u
∂u
∂u
= cos θ
+ sin θ
∂r
∂x
∂y
∂u
∂u
∂u
= −r sin θ
+ r cos θ
.
∂θ
∂x
∂y
(b) We start with the right hand side of the sought after identity,
u2r +
2 1
2
1 2
u
=
cos
θ
u
+
sin
θ
u
+
−r
sin
θ
u
+
r
cos
θ
u
x
y
x
y
θ
r2
r2
2
2
2
= cos θ + sin θ u x + cos2 θ + sin2 θ u2y = u2x + u2y .
Please note the difference between u2x and (u2 ) x and u x x . By u2x we mean the square of the
function u x , in other words u2x = (u x )2 . On the other hand (u2 ) x is the x-derivative of the
square u2 and one has (u2 ) x = 2uu x . Finally u x x is the second partial derivative. These are in
general different functions.
Martins Bruveris
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MA2712
2016 | October | 18
Problems
1
1
(c) In order to express u r r + u r + 2 uθ θ in cartesian coordinates we need expressions for the
r
r
second derivatives. First we have
∂ 2u
∂
∂u
∂ 2u
∂ 2u
=
(cos θ )
+ cos θ cos θ
+ sin θ
∂ r2
∂r
∂x
∂ x2
∂ y∂ x
2
∂
∂u
∂ u
∂ 2u
+
(sin θ )
+ sin θ cos θ
+ sin θ
∂r
∂y
∂ x∂ y
∂ y2
= cos2 θ
2
∂ 2u
∂ 2u
2 ∂ u
+
2
sin
θ
cos
θ
+
sin
θ
.
∂ x2
∂ x∂ y
∂ y2
Next is the second θ -derivative,
∂ 2u
∂
∂u
∂ 2u
∂ 2u
=
(−r sin θ )
− r sin θ −r sin θ
+ r cos θ
∂ θ2
∂θ
∂x
∂ x2
∂ y∂ x
2
∂
∂u
∂ u
∂ 2u
+
(r cos θ )
+ r cos θ −r sin θ
+ r cos θ
∂θ
∂y
∂ x∂ y
∂ y2
= −r cos θ
2
∂u
∂u
∂ 2u
∂ 2u
2
2
2 ∂ u
− r sin θ
+ r 2 sin2 θ
−
2r
sin
θ
cos
θ
+
r
cos
θ
.
∂x
∂y
∂ x2
∂ x∂ y
∂ y2
The required sum is
1
1
u r r + u r + 2 uθ θ =
r
r
1
= cos2 θ u x x + 2 sin θ cos θ u x y + sin2 θ u y y +
cos θ u x + sin θ u y
r
1
+ 2 −r cos θ u x − r sin θ u y + r 2 cos2 θ u x x − 2r 2 sin θ cos θ u x y + r 2 cos2 θ u y y
r
2
= cos θ + sin2 θ u x x + cos2 θ + sin2 θ u y y = u x x + u y y .
1
1
Therefore u r r + u r + 2 uθ θ becomes u x x + u y y in cartesian coordinates.
r
r
40. Compute the following derivative matrices and evaluate them at the given points.
(a)
∂ (x, y)
, where x = u sin v, y = euv ; evaluate at (0, 1).
∂ (u, v)
(b)
π π
∂ (x, y, z)
, where x = r sin ϕ cos θ , y = r sin ϕ sin θ , z = r cos ϕ; evaluate at 4, ,
.
∂ (r, θ , ϕ)
4 4
(c)
∂ (u, v)
, where u = x yz, v = x + y + z; evaluate at (1, 2, 3).
∂ (x, y, z)
Solution. It is important to remember the order of partial derivatives in the derivative matrix.
(a) We have
‚
sin v
∂ (x, y)
=
∂ (u, v)
veu
u cos v
Œ
‚
Œ
sin 1 0
∂ (x, y)
(0, 1) =
.
∂ (u, v)
1
0
ue v
(b) We have

sin ϕ cos θ
∂ (x, y, z) 
=  sin ϕ sin θ
∂ (r, θ , ϕ)
cos ϕ
Martins Bruveris
−r sin ϕ sin θ
r sin ϕ cos θ
0
r cos ϕ cos θ


r cos ϕ sin θ  .
−r sin ϕ
Page 7 / 9
MA2712
2016 | October | 18
Problems
p
p
π π
2
2
π
π
To evaluate at 4, ,
we use that sin =
and cos =
and hence
4 4
4
2
4
2


1
−1 1
2
∂ (x, y, z)
π π


1
1 .
4, ,
=  12
p
∂ (r, θ , ϕ)
4 4
2
0 −2
2
(c) We have
∂ (u, v)
=
∂ (x, y, z)
‚
yz
xz
xy
1
1
1
Œ
‚
Œ
6 3 2
∂ (u, v)
(1, 2, 3) =
.
∂ (x, y, z)
1 1 1
41. Differentiating under the integral. Under mild continuity restrictions, it is true that if
F (x) =
b
Z
g(t, x) dt ,
a
0
then F (x) =
b
Z
g x (t, x) dt. Using this fact and the chain rule we can find the derivative of
a
F (x) =
f (x)
Z
g(t, x) dt
a
by letting
G(u, x) =
Z
u
g(t, x) dt ,
a
where u = f (x). Find the derivatives of the following functions:
(a) F (x) =
x2
Z
p
t 4 + x 3 dt .
0
(b) F (x) =
Z
1
x ln(t 2 + x 2 ) dt .
x
Solution. With G(u, x) defined as above we have
F (x) = G( f (x), x)
and we can find the derivative of F using the chain rule,
F 0 (x) = Gu ( f (x), x) f 0 (x) + G x ( f (x), x) .
We know from the problem statement, that
G x (u, x) =
Z
u
g x (t, x) dt ,
a
and what we need know is the derivative Gu (u, x). But this we obtain from the fundamental theorem
of calculus, which states that
d
du
Martins Bruveris
Z
u
g(t, x) dt = g(u, x) ,
a
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MA2712
2016 | October | 18
Problems
or in our notation, Gu (u, x) = g(u, x). It means that differentiating an integral with respect to the
upper limit results in the function we are integrating. Putting it all together we obtain
F 0 (x) = Gu ( f (x), x) f 0 (x) + G x ( f (x), x)
Z f (x)
= g( f (x), x) f 0 (x) +
g x (t, x) dt .
a
Now we are ready to attempt the problems.
(a) F (x) =
Z
x2
p
t 4 + x 3 dt. Then
0
0
F (x) =
(b) We rewrite F (x) as F (x) =
Z
p
x8
+
x3
3
· 2x +
2
p
0
1
2
x ln t + x
2
x2
Z
dt = −
t4 + x3
dt .
x
Z
x ln t 2 + x 2 dt. Then
1
x
F 0 (x) = −x ln x 2 + x 2 −
x
Z
1
= −x ln 2x 2 +
Z
1
x
Martins Bruveris
x2
t2
2x 2
dt
t2 + x2
2x
dt .
+ x2
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