CHAPTER 16
SOLUTIONS TO EXERCISES IN FUNCTIONS OF
MORE THAN ONE VARIABLE – PARTIAL DIFFERENTIATION
Exercises on 16.1
1. In the relation between P, V, and T, how does P vary with i) T, ii) V? How does
V vary with iii) T, iv) P?
Solution
RT
From P = V = P(V, T) we see that
i)
As T increases, P will also increase - in a linear fashion, or in
proportion. We say P increases linearly with T.
ii) As V increases, P will decrease - we say P is inversely proportional
to V.
iii) Solving the equation for V
RT
V= P
we see that V increases linearly with T, and
iv) V is inversely proportional to P.
2. If V increases by 10% while T remains constant, by what percentage,
approximately, does P change?
Solution
If V increases by 10% then P becomes
RT
RT
RT
P' = V + 0.1V = V(1 + 0.1)  V (1 – 0.1)
on using the binomial approximation
1
1 + x  1 – x for x small
–1–
= P – 0.1P
So
P' – P
P  – 0.1
or P decreases by approximately 10%.
Exercise 16.2
Sketch the surface representing the function
z = x 2 + y2
Solution
The easiest way to do this is to note that all values of (x, y) that lie on
the circle
x2 + y 2 = c
will give the same value, c, of z. One typical pair of values of x and y
serves to define each circle - for example (1, 1) defines the circle z = 2 =
x2 + y2 and such representative values can be used to provide a number
of circles through which the total surface can be drawn. This is
illustrated in the figure (UEM 486).
*** Figure 16.4 from book, page 486 ***
Exercise on 16.3
Find fx, fy for the following functions
i)
xy
f(x, y) = x + y
ii)
f(x,y) = e3x + cos(xy)
Solution
xy
i) Treating y as a constant in f(x, y) = x + y and differentiating with
d 1 
1
respect to x gives, using the product rule and dx x = – 2
 
x
–2–
f
y
xy
= fx = x + y –
x
(x + y)2
and putting over a common denominator gives
y2
fx =
(x + y)2
You can now do exactly the same with y exchanged for x, for practice.
However, also note that because of the symmetry between x and y in
f(x,y) one can immediately write down the result as
fy =
x2
(x + y)2
by analogy with fx.
ii) f(x,y) = e3x + cos(xy) is a classic case of function of a function rule,
even though there are two variables, which may make it look more
complicated. Keeping y constant and differentiating with respect to x
gives
fx = e3x + cos(xy)

3x + cos (xy)) = e3x + cos(xy)(3 – y sin (xy))
x (
Similarly
fy = e3x + cos(xy)

3x + cos (xy)) = – x e3x + cos(xy) sin (xy)
y (
Exercises on 16.4
1.
Find all first and second order partial derivatives of the following
functions, f(x,y), checking the equality of the mixed derivatives
i)
f(x,y) = x3y2 + 4xy4
ii)
f(x,y) = exy cos(x + y)
Solution
All that is needed in these problems is a steady hand with the differentiation
–3–
i) For f(x,y) = x3y2 + 4xy4 we have
fx = 3x2y2 + 4y4
fy = 2x3y + 16xy3
fxx =

3x2y2 + 4y4) = 6xy2 (remembering that y is regarded as a
x (
constant still)
fyy =

2x3y + 16xy3) = 2x3 + 48xy2
y (
fyx = (fy)x = (2x3y + 16xy3)x = 6x2y + 16y3
= fxy as you can check.
ii)
f(x,y) = exy cos(x + y)
fx = yexy cos(x + y) – exy sin(x + y) = exy (ycos(x + y) – sin(x + y))
fy = xexy cos(x + y) – exy sin(x + y) = exy (xcos(x + y) – sin(x + y))
fxx = yexy (ycos(x + y) – sin(x + y)) + exy (– ysin(x + y) – cos(x + y))
= exy ((y2 – 1)cos(x + y) – 2ysin(x + y))
fyy = = xexy (xcos(x + y) – sin(x + y)) + exy (– xsin(x + y) – cos(x + y))
= exy ((x2 – 1)cos(x + y) – 2xsin(x + y))
fyx = (fy)x = (exy (xcos(x + y) – sin(x + y)))x
= yexy (xcos(x + y) – sin(x + y))
+ exy (cos(x + y) – xsin(x + y) – cos(x + y))
–4–
= exy (xycos(x + y) – (x + y)sin(x + y))
= fxy as you can check directly.
2.
Show that f(x, y) = ln(x2 + y2) satisfies the partial differential equation:
2f
2f
+
= 0
x2
y2
This is called the Laplace equation in two dimensional rectangular
coordinates. It is very important in fluid mechanics, electromagnetism,
and many other areas of science and engineering, as well as being a key
equation in pure mathematics.
Solution
With f(x, y) = ln(x2 + y2) we have
fx =
2x
x + y2
2
and
fxx =
2
2x2
y2 – x2
–
=
x2 + y2 (x2 + y2)2
(x2 + y2)2
On the other hand we find by the same process that
fyy
x2 – y2
= 2
(x + y2)2
Direct addition then gives
fxx + fyy
y2 – x2
x2 – y 2
= 2
+ 2
=0
(x + y2)2
(x + y2)2
as required.
–5–
Exercises on 16.5
1.
Find the total differential dz when
ii) z = exp(x/y)
i) z = ln(cos(xy))
Solution
Really, these exercises amount to little more than finding the first order
derivatives
i) z = ln(cos(xy))
ysin(xy)
zx = – cos(xy) = – ytan(xy) and zy = – xtan(xy) and so
dz = zx dx + zy dy = – ytan(xy) dx – xtan(xy) dy
= – tan(xy)(ydx + xdy)
ii) z = exp(x/y)
1
x
zx = y exp(x/y) and zy = – 2 exp(x/y) and so
y
x 
1
dz = exp(x/y) ydx – 2dy
y


=
2.
If
z = e2x+ 3y
and
1
exp(x/y) (ydx – xdy)
y2
x = ln t,
dz
y = t2 , calculate dt from the total
derivative formula and show that it agrees with the result obtained by
substitution for x and y before differentiating.
Solution
From the total derivative we have
–6–
dz
z dx
z dy
dx
dy
2x+ 3y
2x+ 3y
=
+
=
2e
+
3e
dt
dt
dt
x dt
y dt
1
= 2e2x+3y t + 3e2x+3y 2t
2
2 1
2
2
= 2eln t + 3t t + 6teln t + 3t
2 2 2
2
2
3
2
2
2
= t t e3t + 6t  t e3t = (6t + 2t)e3t = 2t(3t + 1)e3t
If we first substitute for x and y in terms of t we obtain
2
2
2
2
z = eln t + 3t = t e3t
on using eln x = x. Differentiating this with respect to t now gives
dz
2 3t2
2
3t2
3t2
=
2te
+
t
e
(6t)
=
2t(3t
+
1)e
dt
as previously.
REINFORCEMENT EXERCISES IN FUNCTIONS OF MORE THAN ONE
VARIABLE
1.
Find the values of the following functions at the points given :i) f(x, y) = 2xy3 + 3x2 y at the point (2, 1)
ii) g(x, y) = (x + y)ex sin y at the point (0,  /2)
iii)
iv)
x2 + y2 + z2 at the points (–1, 2, 2) and (3, 2, 4)



2
l(x, y,z) = ex y4 cos z at 0, – 2, 3



h(x, y, z) =
Solution
i) The value of f(x, y) = 2xy3 + 3x2 y at the point (2, 1) is
f(2, 1) = 2(2)(1)3 + 3(2)2(1) = 4 + 12 = 16
–7–





ii) g(0,  /2) = 0 + 2 e0 sin 2 = 2(1)(1) = 2


iii) h(–1, 2, 2) =
h(3, 2, 4) =


iv) l 0, – 2, 3

2.
(– 1)2 + 22 + 22
32 + 22 + 42
=
9 =3
= 29

1

2
 = e0 (– 2)4 cos 3 = (1)(16)2 = 8
 

Sketch the surfaces represented by z = f(x, y) where
i) z = 1 – 3y
ii) x2 + y2 + z2 = 9
Solution
i) Since z = 1 – 3y does not depend on x, its profile will be the same
for all values of x. If we sketch it for say x= 0 then we simply get the line
z = 1 – 3y in the yz plane. The total surface is then obtained by
translating this line parallel to itself in the positive and negative x
directions. The figure shows this on the positive side of the xz plane
with z positive.
*** Figure from book, 2 (i) page 496 ***
ii)
x2 + y2 + z2 = 9 is the equation of a sphere with centre at the
origin and with radius 9 = 3, shown in the figure below.
*** Figure from book, 2 (ii) page 496 ***
3.
Determine z/x, z/y in each case
x
i) z = x2 + y2
ii) z = y
1
iv) z =
v) z = exy cos(3y2)
2
2
x +y
x2 + y2
vii) z = e–xy (2 + 3xy)
viii) z =
1+y
Solution
–8–
iii) z = x3 + x2y + y4
vi) z = ln(1 + xy)
x 
ix) z = x3 tan–1y
 
This is really nothing more than a marathon exercise in differentiation, and you
need no more skills than we covered in Chapter 8 - along with an ability to keep
focused on whether it is x or y you are differentiating with respect to!
i) z = x2 + y2
z
= 2x (y is constant)
x
z
= 2y (x is constant)
y
x
ii) z = y
1
zx = y
zy = –
x
y2
iii) z = x3 + x2y + y4
zx = 3x2 + 2xy
iv)
z =
1
2
x +y
2
zy = x2 + 4y3
= (x2 + y2)– 1/2
so
1
x
zx = –2 (x2 + y2) – 3/2 (2x) = – 2
(x + y2)3/2
and by symmetry
zy = –
v)
y
(x + y2)3/2
2
z = exy cos(3y2)
zx = yexy cos(3y2)
zy = xexy cos(3y2) + exy (– sin (3y2)) 6y
=
vi)
exy (xcos(3y2) – 6ysin (3y2))
z = ln(1 + xy)
–9–
y
x
zx = 1 + xy and by symmetry zy = 1 + xy
vii) z = e–xy (2 + 3xy)
zx = (– y)e– xy (2 + 3xy) + e–xy (3y) = ye– xy (1 – 3xy)
(same as answer in book)
(– x)e– xy (2 + 3xy) + e–xy (3x) = xe– xy (1 – 3xy)
zy =
viii)
z =
zx =
x2 + y2
1+y
2x
1+y
zy =
(remember y and therefore
2y
1 x2 + y2
– 2
=
(1 + y)3/2
1+y
=
1 + y is a constant)
4y(1 + y) – x2 – y2
2(1 + y)3/2
3y2 + 4y – x2
2(1 + y)3/2
x 
ix) z = x3 tan–1y
 
1
x 
1
zx = 3x2 tan–1y + x3
2 y
 
1 + (x/y)  
yx3
x
= 3x2 tan–1y + 2
 
x + y2
1
x4
 x
zy = x
–  = – 2
x + y2
1 + (x/y)2  y2
3
– 10 –
4.
z (0, 0)
x
For each of the functions in Q3 evaluate
,
z (1, 2)
y
whenever possible.
Solution
i) z = x2 + y2
z
z (0, 0)
= 2x so
x
x
= 2(0) = 0
z
z (1, 2)
= 2y so
y
y
= 2(2) = 4
x
ii) z = y
1
zx = y
zy = –
does not exist at y = 0
x
z (1, 2)
2 so y
y
=–
1
1
2 =–4
2
iii) z = x3 + x2y + y4
zx = 3x2 + 2xy so
zy = x2 + 4y3 so
iv)
zy = –
v)
=0
= 12 + 4(2)3 = 33
1
z =
zx = –
z (1, 2)
y
z (0, 0)
x
x2 + y2
x
(x + y2)3/2
2
y
(x + y2)3/2
2
so
so
z (0, 0)
x
does not exist (division by zero)
z (1, 2)
y
= –
z = exy cos(3y2)
– 11 –
2
2
2 3/2 = –
(1 + 2 )
5 5
2
z (0, 0)
x
zx = yexy cos(3y2) so
=0
zy = exy (xcos(3y2) – 6ysin (3y2))
so
z (1, 2)
y
vi)
= e2 (cos(12) – 12sin (12))
z = ln(1 + xy)
y
zx = 1 + xy
x
zy = 1 + xy
so
so
z (0, 0)
x
z (1, 2)
y
=0
1
1
= 1+2 =3
vii) z = e–xy (2 + 3xy)
zx = ye– xy (1 – 3xy) so
z (1, 2)
y
zy = xe– xy (1 – 3xy) so
viii)
z (0, 0)
x
=0
= 1e– 2 (1 – 6) = – 5e– 2
x2 + y2
z =
1+y
zx =
zy =
3y2 + 4y – x2
2(1 + y)3/2
2x
so
1+y
so
z (0, 0)
x
z (1, 2)
y
=
=0
3(2)2 + 4(2) – 12
2(1 + 2)3/2
=
19
6 3
x 
ix) z = x3 tan–1y
 
yx3
z (0, 0)
zx = 3x tan y + 2
2 so x
 
x +y
2
–1x 
– 12 –
does not exist
x4
z (1, 2)
zy = – 2
2 so y
x +y
5.
14
1
=– 2
2 =–5
1 +2
Determine all first order partial derivatives
i) w = x2 + 2y2 + 3z2
ii) w =
iii) w = xyz
v) w = e
xy
1
1 – x2 – y2 – z2
iv) w = x cos(x + yz)
ln(x + y + z)
Solution
Again, little more than standard plodding differentiation - except you might
use symmetry sometimes to simplify the calculations. And, of course, we have
three independent variables to contend with now!
i) w = x2 + 2y2 + 3z2
wx = 2x
ii) w =
1
2
2
2
1–x –y –z
wy = 4y
wz = 6z
= (1 – x2 – y2 – z2)– 1/2
1
x
wx = – 2 (1 – x2 – y2 – z2) – 3/2 (– 2x) =
2
(1 – x – y2 – z2)3/2
There is no need to repeat the calculations for y and z since the form of
the results will clearly be the same and we will get
wy =
y
(1 – x – y2 – z2)3/2
wz =
z
(1 – x – y2 – z2)3/2
2
2
iii) w = xyz
– 13 –
wx = yz
wy = xz
wz = xy
iv) w = x cos(x + yz)
wx = cos(x + yz) – xsin(x + yz)
wy = – xzsin(x + yz)
wz = – xysin(x + yz)
v) w = exy ln(x + y + z)
1
wx = yexy ln(x + y + z) + exy x + y + z
1


= exy yln(x + y + z) + x + y + z


1
wy = xexy ln(x + y + z) + exy x + y + z (or use symmetry in x and y)
1


= exy xln(x + y + z) + x + y + z


exy
wz = x + y + z
6.
Determine all second order partial derivatives for the functions in Q3.
Solution
More tedious practice in differentiation, using the results obtained in Q3. And,
of course, remember that zxy = zyx
i) z = x2 + y2
zx = 2x and zy = 2y
So, differentiating a second time,
zxx = (zx)x = (2x)x = 2
and similarly zyy = 2 and zxy = 0
– 14 –
x
ii) z = y
1
zx = y
zy = –
x
y2
So
zxx =
zyy =
zxy =
 1
  =0
x y
2x
  x
– 2 = 3

y  y 
y
1
 1
y = – 2
y  
y
iii) z = x3 + x2y + y4
zx = 3x2 + 2xy
So
zxx =

3x2 + 2xy) = 6x + 2y
x (
zyy =
zxy =
iv)
z =
1
2
x +y
2
zx = –
zy = x2 + 4y3

x2 + 4y3) = 12y2
y (

3x2 + 2xy) = 2x
y (
= (x2 + y2)– 1/2
x
(x + y2)3/2
2
and zy = –
y
(x + y2)3/2
2
So
zxx =
x
1
x(– 3/2)
 

– 2
=– 2
– 2
(2x)


2
3/2
2
3/2
x  (x + y ) 
(x + y )
(x + y2)5/2
– 15 –
1
3x2
– (x2 + y2) + 3x2
=– 2
–
=
(x + y2)3/2 (x2 + y2)5/2
(x2 + y2)5/2
2x2 – y2
= 2
(x + y2)5/2
zyy
y
 

=
– 2
=

y  (x + y2)3/2
zxy =
x
x(– 3/2)
 

– 2
=– 2
(2y)


2
3/2
y  (x + y ) 
(x + y2)5/2
=
v)
2y2 – x2
(by symmetry)
(x2 + y2)5/2
3xy
(x + y2)5/2
2
z = exy cos(3y2)
zx = yexy cos(3y2)
So
zxx =
zyy =
zy = exy (xcos(3y2) – 6ysin (3y2))

yexy cos(3y2)) = y2exy cos(3y2)
x (
 xy
e (xcos(3y2) – 6ysin (3y2)))
y (
= xexy (xcos(3y2) – 6ysin (3y2)) + yexy (– 6yxcos(3y2) – 6sin (3y2) –
36y2cos(3y2))
= exy (x2 cos (3y2) – 12xy sin (3y2) – 6 sin (3y2) – 36 y2cos (3y2))
zxy =

yexy cos(3y2))
y (
= exy cos(3y2) + xyexy cos(3y2) – 6y2exy sin(3y2)
= exy (cos (3y2) + xy cos (3y2) – 6 y2sin (3y2))
– 16 –
vi)
z = ln(1 + xy)
y
zx = 1 + xy
x
zy = 1 + xy
So
zxx
y2
  y 
=

 =–
x 1 + xy 
(1 + xy)2
zyy =
zxy =
x2
  x 
1 + xy = –
y 

(1 + xy)2
1
xy
1
  y 
1 + xy = 1 + xy –
=
2
y 

(1 + xy)2
(1 + xy)
vii) z = e–xy (2 + 3xy)
zx = ye– xy (1 – 3xy)
So
zxx =

ye– xy (1 – 3xy)) =
x (
zy = xe– xy (1 – 3xy)
 – xy
e
(y – 3xy2))
x (
= – ye– xy (y – 3xy2) + e– xy (– 3y2) = e– xy (– y2 + 3xy3 – 3y2)
= e– xy (3xy3 – 4y2)
Similarly, we can see from symmetry that
zyy =
zxy =

xe– xy (1 – 3xy)) = e– xy (3x3y – 4x2)
y (

 – xy
ye– xy (1 – 3xy)) =
e
(y – 3xy2))
y (
y (
= – xe– xy (y – 3xy2) + e– xy (1 – 6xy)
= e– xy (– xy + 3x2y2 + 1 – 6xy)
= e– xy (3x2y2 – 7xy + 1)
– 17 –
x2 + y2
z =
1+y
viii)
3y2 + 4y – x2
zy =
2(1 + y)3/2
2x
1+y
zx =
So
zxx =
zyy
  2x 
=
x  1 + y
2
1+y
6y + 4
3 3y2 + 4y – x2
 3y2 + 4y – x2
=

 =
– 

y  2(1 + y)3/2 
2(1 + y)3/2 2  2(1 + y)5/2 
=
2(6y + 4)(1 + y) – 3(3y2 + 4y – x2)
4(1 + y)5/2
=
3y2 + 8y + 3x2 + 8
4(1 + y)5 /2
after a bit of tidying up.
zxy =
2x
  2x 

 1
=
2x(1 + y)– 1/2) =
– 
(


3/2
y  1 + y
y
(1 + y)  2
=–
x
(1 + y)3/2
x 
ix) z = x3 tan–1y
 
yx3
x 
zx = 3x2 tan–1y + 2
 
x + y2
zy = –
x4
x2 + y2
So
zxx
yx3 
  2
–1x 
=
3x tan y + 2

x 
 
x + y2
1
3x2y
yx3

= 6xtan y + 3x 1/(1 + (x/y) ) y + 2
– 2 2
(2x)
 
x + y2
(x + y2)2
–1x 
2
2
– 18 –
and you can regard it as an algebra test to show that this reduces to
2x4y + 6x2y3
x
= 6x tan–1y +
 
(x2 + y2)2
x4 
2x4y
 
– 2

=
y  x + y2
(x2 + y2)2
zyy =
For zxy it is probably easier in this case to evaluate zyx, ie do the y–
derivative first. Then
zxy
7.
x4 
– 4x3
 
=
–
 = – 2
x  x2 + y2
x + y2
x4 (2x)
– 2x5 – 4x3y2
+ 2
=
(x2 + y2)2
(x + y2)2
Determine all second order partial derivatives for Q5 i), iii), iv).
Solution
i) w = x2 + 2y2 + 3z2
wx = 2x
wy = 4y
wxx =

2x = 2
x ( )
wyy =

4y = 4
y ( )
wzz =

6z = 6
z ( )
So
and
wxy = wxz =wyz = 0
iii) w = xyz
– 19 –
wz = 6z
wx = yz
wy = xz
wxx =

yz = 0
x ( )
wyy =

xz = 0
y ( )
wzz =

xy = 0
z ( )
wxy =

yz = z
y ( )
wxz =

yz = y
z ( )
wyz =

xz = x
z ( )
So
wz = xy
and
iv) w = x cos(x + yz)
wx = cos(x + yz) – xsin(x + yz)
wy = – xzsin(x + yz)
wz = – xysin(x + yz)
So
wxx =

cos(x + yz) – xsin(x + yz)) = – 2sin(x + yz) – xcos(x + yz)
x (
wyy =

– xzsin(x + yz) ) = – xz2cos(x + yz)
y (
wzz =

– xysin(x + yz) ) = – xy2cos(x + yz)
z (
– 20 –
8.
wxy =

– xzsin(x + yz) ) = – zsin(x + yz) – xzcos(x + yz)
x (
wxz =

– xysin(x + yz)) = – ysin(x + yz) – xycos(x + yz)
x (
wyz =

– xysin(x + yz)) = – xsin(x + yz) – xyzcos(x + yz)
y (
2
Show that T(x, t) = ae–b t cos bx, where a and b are arbitrary constants,
satisfies the equation
T
2T
=
t
x2
Solution
2
With T(x, t) = ae– b t cos bx we have
T
2
= – ab2e– b t cos bx
t
T
2
= – abe– b t sin bx
x
2T
T
2 – b2t
cos bx =
2 = – ab e
t
x
9.
Determine dz for the functions
i) z = x2 – 3y2
iii) z = ln(x2 + y2)
v) z = x2e–xy
ii)
z = 3x2y3
iv)
z = cos(x + y)
Solution
To find the total derivative, all we need is the first order derivatives of the
function. We will do the first question in detail and the rest in outline, for you
to fill in the gaps.
i) For z = x2 – 3y2 we have
– 21 –
z
= 2x
x
z
= – 6y
y
So
dz =
z
z
dx +
dy = 2xdx – 6ydy
x
y
ii) z = 3x2y3
dz =
iii)
z
z
dx +
dy = 6xy3 dx + 9x2y2dy
x
y
z = ln(x2 + y2)
dz =
z
z
2x
2y
dx +
dy = 2
dy
2 dx + 2
x
y
x +y
x + y2
iv) z = cos(x + y)
dz = – sin(x + y)dx – sin(x + y)dy
v) z = x2e–xy
dz = (2xe–xy – x2ye–xy)dx – x3e–xydy = xe–xy((2 – xy)dx – x2dy
10.
If z = 3x2 + 2xy – y2 and x and y vary with time t according to
x
dz
= 1 + sin t and y = 3 cos t–1 evaluate dt directly and by using the total
derivative (chain) rule.
Solution
If z = 3x2 + 2xy – y2 and x = 1 + sin t , y = 3 cos t – 1 then
dz
z dx
z dy
dx
dy
=
+
=
(6x
+
2y)
+
(2x
–
2y)
dt
dt
dt
x dt
y dt
= (6x + 2y) cos t – 3(2x – 2y) sin t
= (6(1 + sin t) + 2(3 cos t – 1)) cos t – 3(2(1 + sin t) – 2(3 cos t – 1)) sin t
– 22 –
= 6(cos2 t + 4cos t sin t – sin2 t – 2 sin t) + 4 cos t
= 6(cos 2t + 2sin 2t – 2 sin t) + 4 cos t
= 6cos 2t + 12sin 2t – 12 sin t + 4 cos t
On the other hand, substituting first for x and y in terms of t in the
expression for z we have
z = 3x2 + 2xy – y2
= 3(1 + sin t)2 + 2(1 + sin t)(3 cos t – 1) – (3 cos t – 1)2
= 3 sin2 t – 9 cos2 t + 3 sin 2t + 4 sint + 12 cos t
and then differentiation with respect to t gives
dz
dt = 6 sin t cos t + 18 cos t sin t + 6 cos 2t + 4 cos t – 12 sin t
= 12 sin 2t + 6 cos 2t – 12 sin t + 4 cos t
agreeing with the previous result.
– 23 –