Chapter 17



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

From P = = 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

V =

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

P' = = ”

on using the binomial approximation

” 1 – x for x small

= P – 0.1P

So

” – 0.1

or P decreases by approximately 10%.

Exercise 16.2

Sketch the surface representing the function

z = x2 + y2

Solution

The easiest way to do this is to note that all values of (x, y) that lie on the circle

x2 + y2 = 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) f(x, y) = ii) f(x,y) = e3x + cos(xy)

Solution

i) Treating y as a constant in f(x, y) = and differentiating with respect to x gives, using the product rule and = –

= fx = –

and putting over a common denominator gives

fx =

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 =

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) = e3x + cos(xy)(3 – y sin (xy))

Similarly

fy = e3x + cos(xy) = – x e3x + cos(xy) sin (xy)

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

i) For f(x,y) = x3y2 + 4xy4 we have

fx = 3x2y2 + 4y4

fy = 2x3y + 16xy3

fxx = = 6xy2 (remembering that y is regarded as a constant still)

fyy = = 2x3 + 48xy2

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))

= exy (xycos(x + y) – (x + y)sin(x + y))

= fxy as you can check directly.

2. Show that f(x, y) = ln satisfies the partial differential equation:

+ = 0

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) = lnwe have

fx =

and

fxx = – =

On the other hand we find by the same process that

fyy =

Direct addition then gives

fxx + fyy = + = 0

as required.

Exercises on 16.5

1. Find the total differential dz when

i) z = ln(cos(xy)) ii) z = exp(/y)

Solution

Really, these exercises amount to little more than finding the first order derivatives

i) z = ln(cos(xy))

zx = – = – 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(/y)

zx = exp(/y) and zy = – exp(/y) and so

dz = exp(/y)

= exp(/y) (ydx – xdy)

2. If z = eand x = ln t, y = t, calculate 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

= + = 2e+ 3e

= 2e+ 3e2t

= 2e+ 6te

= t2e3t2+ 6t ( t2e3t2 = (6t3+ 2t)e3t2 = 2t(3t2+ 1)e3t2

If we first substitute for x and y in terms of t we obtain

z = e= t2e3t2

on using e= x. Differentiating this with respect to t now gives

= 2te3t2 + t2e3t2 (6t) = 2t(3t2+ 1)e3t2

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) = 2xy+ 3xy at the point (2, 1)

ii) g(x, y) = (x + y)esin y at the point (0, /2)

iii) h(x, y, z) = at the points (–1, 2, 2) and (3, 2, 4)

iv) l(x, y,z) = ex2 y4 cos z at

Solution

i) The value of f(x, y) = 2xy+ 3xy at the point (2, 1) is

f(2, 1) = 2(2)(1)+ 3(2)= 4 + 12 = 16

ii) g(0, /2) = esin = =

iii) h(–1, 2, 2) = = = 3

h(3, 2, 4) = =

iv) l = e02 (– 2)4 cos = (1)(16)= 8

2. Sketch the surfaces represented by z = f(x, y) where

i) z = 1 – 3y ii) x+ y+ z= 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) x+ y+ z= 9 is the equation of a sphere with centre at the origin and with radius = 3, shown in the figure below.

*** Figure from book, 2 (ii) page 496 ***

3. Determine (z/(x, (z/(y in each case

i) z = x2 + y2 ii) z = iii) z = x3 + x2y + y4

iv) z = v) z = exy cos(3y2) vi) z = ln(1 + xy)

vii) z = e–xy (2 + 3xy) viii) z = ix) z = x3 tan–1

Solution

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

= 2x (y is constant) = 2y (x is constant)

ii) z =

zx = zy = –

iii) z = x3 + x2y + y4

zx = 3x2 + 2xy zy = x2 + 4y3

iv) z = = (x2 + y2)– 1/2

so

zx = –– 3/2 (2x) = –

and by symmetry

zy = –

v) z = exy cos(3y2)

zx = yexy cos(3y2)

zy = xexy cos(3y2) + exy (– sin (3y2)) 6y

= exy (xcos(3y2) – 6ysin (3y2))

vi) z = ln(1 + xy)

zx = and by symmetry zy =

vii) z = e–xy (2 + 3xy)

zx = (– y)e– xy (2 + 3xy) + e–xy (3y) = ye– xy (1 – 3xy)

(same as answer in book)

zy = (– x)e– xy (2 + 3xy) + e–xy (3x) = xe– xy (1 – 3xy)

viii) z =

zx =

zy = – =

=

ix) z = x3 tan–1

zx = 3x2 tan–1+ x3

= 3x2 tan–1+

zy = x3 = –

4. For each of the functions in Q3 evaluate , whenever possible.

Solution

i) z = x2 + y2

= 2x so = 2(0) = 0

= 2y so = 2(2) = 4

ii) z =

zx = does not exist at y = 0

zy = – so = – = –

iii) z = x3 + x2y + y4

zx = 3x2 + 2xy so = 0

zy = x2 + 4y3 so = 12 + 4(2)3 = 33

iv) z =

zx = – so does not exist (division by zero)

zy = – so = – = –

v) z = exy cos(3y2)

zx = yexy cos(3y2) so = 0

zy = exy (xcos(3y2) – 6ysin (3y2))

so

= e2 (cos(12) – 12sin (12))

vi) z = ln(1 + xy)

zx = so = 0

zy = so = =

vii) z = e–xy (2 + 3xy)

zx = ye– xy (1 – 3xy) so = 0

zy = xe– xy (1 – 3xy) so = 1e– 2 (1 – 6) = – 5e– 2

viii) z =

zx = so = 0

zy = so = =

ix) z = x3 tan–1

zx = 3x2 tan–1+ so does not exist

zy = – so = – = –

5. Determine all first order partial derivatives

i) w = x2 + 2y2 + 3z2 ii) w =

iii) w = xyz iv) w = x cos(x + yz)

v) w = exy 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 wy = 4y wz = 6z

ii) w = = (1 – x2 – y2 – z2)– 1/2

wx = – – 3/2 (– 2x) =

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 =

wz =

iii) w = xyz

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)

wx = yexy ln(x + y + z) + exy

= exy

wy = xexy ln(x + y + z) + exy

= exy

wz =

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

ii) z =

zx = zy = –

So

zxx = = 0

zyy = =

zxy = = –

iii) z = x3 + x2y + y4

zx = 3x2 + 2xy zy = x2 + 4y3

So

zxx = = 6x + 2y

zyy = = 12y2

zxy = = 2x

iv) z = = (x2 + y2)– 1/2

zx = – and zy = –

So

zxx = = – –

= – – =

=

zyy = =

zxy = = –

=

v) z = exy cos(3y2)

zx = yexy cos(3y2) zy = exy (xcos(3y2) – 6ysin (3y2))

So

zxx = = y2exy cos(3y2)

zyy =

= 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 =

= exy cos(3y2) + xyexy cos(3y2) – 6y2exy sin(3y2)

= exy (cos (3y2) + xy cos (3y2) – 6 y2sin (3y2))

vi) z = ln(1 + xy)

zx = zy =

So

zxx = = –

zyy = = –

zxy = = – =

vii) z = e–xy (2 + 3xy)

zx = ye– xy (1 – 3xy) zy = xe– xy (1 – 3xy)

So

zxx = =

= – ye– xy (y – 3xy2) + e– xy (– 3y2) = e– xy (– y2 + 3xy3 – 3y2)

= e– xy (3xy3 – 4y2)

Similarly, we can see from symmetry that

zyy = = e– xy (3x3y – 4x2)

zxy = =

= – xe– xy (y – 3xy2) + e– xy (1 – 6xy)

= e– xy (– xy + 3x2y2 + 1 – 6xy)

= e– xy (3x2y2 – 7xy + 1)

viii) z =

zx = zy =

So

zxx = =

zyy = = –

=

=

after a bit of tidying up.

zxy = = =

= –

ix) z = x3 tan–1

zx = 3x2 tan–1+ zy = –

So

zxx =

= 6xtan–1+ 3x2 + – 2

and you can regard it as an algebra test to show that this reduces to

= 6x tan–1+

zyy = =

For zxy it is probably easier in this case to evaluate zyx, ie do the y–derivative first. Then

zxy = = – + =

7. Determine all second order partial derivatives for Q5 i), iii), iv).

Solution

i) w = x2 + 2y2 + 3z2

wx = 2x wy = 4y wz = 6z

So

wxx = = 2

wyy = = 4

wzz = = 6

and

wxy = wxz =wyz = 0

iii) w = xyz

wx = yz wy = xz wz = xy

So

wxx = = 0

wyy = = 0

wzz = = 0

and

wxy = = z

wxz = = y

wyz = = x

iv) w = x cos(x + yz)

wx = cos(x + yz) – xsin(x + yz)

wy = – xzsin(x + yz)

wz = – xysin(x + yz)

So

wxx = = – 2sin(x + yz) – xcos(x + yz)

wyy = = – xz2cos(x + yz)

wzz = = – xy2cos(x + yz)

wxy = = – zsin(x + yz) – xzcos(x + yz)

wxz = = – ysin(x + yz) – xycos(x + yz)

wyz = = – xsin(x + yz) – xyzcos(x + yz)

8. Show that T(x, t) = ae–b2t cos bx, where a and b are arbitrary constants, satisfies the equation

=

Solution

With T(x, t) = ae– b2t cos bx we have

= – ab2e– b2t cos bx

= – abe– b2t sin bx

= – ab2e– b2t cos bx =

9. Determine dz for the functions

i) z = x2 – 3y2 ii) z = 3x2y3

iii) z = ln(x2 + y2) iv) z = cos(x + y)

v) z = x2e–xy

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

= 2x = – 6y

So

dz = dx + dy = 2xdx – 6ydy

ii) z = 3x2y3

dz = dx + dy = 6xy3 dx + 9x2y2dy

iii) z = ln(x2 + y2)

dz = dx + dy = dx + dy

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 = 1 + sin t and y = 3 cos t–1 evaluate 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

= + = (6x + 2y) + (2x – 2y)

= (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

= 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

= 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.

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