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MATH1043+Calculus+student+study+guide
Mathematics (University of the Witwatersrand, Johannesburg)
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MATH1043
Calculus
Study Guide for MATH1043
First Year Semester 2 Course: Engineering Mathematics
School of Mathematics
University of the Witwatersrand
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Contents
1 TECHNIQUES OF INTEGRATION (TC Chapters 5.5, 5.6 and 8)
2
1.1
Simple Techniques TC 8.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.2
Differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.3
Integration by substitution TC 5.5 . . . . . . . . . . . . . . . . . . . . . . . . .
4
1.4
Indirect substitution TC 8.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
1.5
Rational functions (or Partial fractions) TC 8.4 . . . . . . . . . . . . . . . . . .
9
1.6
Integration by parts (“Product rule”) TC 8.1 . . . . . . . . . . . . . . . . . . . . 10
1.7
Mixed examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.8
Tutorial answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2 AREAS AND ARC LENGTHS, TC Chapter 11
17
2.1
General Curves in a Plane TC 11.1 . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.2
Parametric curve sketching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2.3
Another application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.4
Areas in parametric form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.5
Arc length TC 6.3, 11.2 and 11.5 . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.6
Curved surface area TC 6.4 and 11.2 . . . . . . . . . . . . . . . . . . . . . . . . 27
2.7
Tutorial answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3 ADVANCED APPLICATIONS OF DIFFERENTIATION, TC Chapter 10.2
– 10.4 & 10.8 – 10.10
31
3.1
Higher Approximation (Taylor Polynomials) TC 10.8 . . . . . . . . . . . . . . . 31
3.2
Maclaurin Series TC 10.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
3.3
The Binomial Series TC 10.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
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3.4
Taylor Series TC 10.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
3.5
Indeterminate Forms and l’Hôpital’s Rule TC 4.5 . . . . . . . . . . . . . . . . . 40
3.6
Convergence of Series I, TC 10.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
3.7
Convergence of Series II, TC 10.3& 10.4 . . . . . . . . . . . . . . . . . . . . . . 46
3.8
Tutorial answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
4 PARTIAL DIFFERENTIATION
49
4.1
Quadric Surfaces TC 12.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
4.2
Partial Differentiation TC 14.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
4.3
Implicit Differentiation TC 14.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
4.4
Second Derivatives TC 14.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
4.5
Partial Derivative Chain Rule TC 14.4 . . . . . . . . . . . . . . . . . . . . . . . 55
4.6
Differentials and First Approximations . . . . . . . . . . . . . . . . . . . . . . . 58
4.7
Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
4.8
Tutorial answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
5 DIFFERENTIAL EQUATIONS E & P Ch1
63
5.1
Introduction E&P 1.1 & 1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
5.2
Variable Separable E & P 1.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
5.3
Homogeneous E & P 1.6 (p62) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
5.4
Exact E & P 1.6 (p68) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
5.5
Linear E & P 1.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
5.6
Tutorial answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
1
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Chapter 1
TECHNIQUES OF INTEGRATION
(TC Chapters 5.5, 5.6 and 8)
Various techniques will be taught in this chapter, the difficulty will be deciding which technique
to use.
1.1
Simple Techniques TC 8.2
“Turning a product into a sum”
Recall:
• cos 2A = 2 cos2 A − 1
⇒
cos2 A = 21 (1 + cos 2A)
• cos 2A = 1 − 2 sin2 A
⇒
sin2 A = 21 (1 − cos 2A)
•
+
sin(A − B) = sin A cos B − cos A sin B
sin(A + B) = sin A cos B + cos A sin B
⇒
• Similarly
sin(A − B) + sin(A + B) = 2 sin A cos B
1
sin(A − B) + sin(A + B)
sin A cos B =
2
cos A cos B =
and
•
Example 1.1.1.
1
cos(A − B) + cos(A + B)
2
sin A sin B =
R
sin 3x cos 2x dx =
1
cos(A − B) − cos(A + B)
2
Example 1.1.2. Homework: Do the following two examples in a similar way
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1.
2.
R
R
cos 5x cos 3x dx
sin 4x sin 3x dx
Using the double angle formula we can deal with the integrals of sin2 x and cos2 x:
R
Example 1.1.3. cos2 2x dx
R
Example 1.1.4. cos4 x dx
1.1.1
Tutorial questions — Simple techniques
D1. Use trigonometric identities to simplify the following functions, and then integrate them
with respect to θ.
(a)
sin 2θ cos 2θ
(d)
(g)
sin θ sin 7θ
(b) cos 5θ cos θ
(e) cos2 θ
(c) sin 2θ cos 4θ
(f)
cos4 θ sin2 θ
(h) tan2 θ
(i)
(j)
1
1 − sin θ
sin θ sin 2θ sin 3θ.
sin4 θ
(Hint for (i): multiply top and bottom by 1 + sin θ.)
1.2
Differentials
dy
If y = f (x), then
= f ′ (x) ⇒ dy = f ′ (x)dx, when dy is called the
dx
differential of y.
Recall: ∆y = change in y, not to be confused.
In the chain rule, we are cancelling the differentials
Example 1.2.1. If y =
√
dy du
dy
=
.
dx
du dx
1 + x , find the differential of y.
Example 1.2.2. If x = 2 tan θ, find the differential of x.
1.2.1
Tutorial questions — Differentials
D2. Find the differentials of the following functions.
√
(a)
(c) arcsin(2v − 1)
1 + x2
(b) sin2 θ
(d) ex
2
(e) (sin(x3 ))1/3
(f)
3
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1.3
Integration by substitution TC 5.5
Theorem 1.3.1. If y is a function of u and u is a function of x, then
Z
Z
du
y dx = y du
dx
“we have cancelled the differentials”.
R
Example 1.3.2. : sin3 x cos x dx
R
4
Example 1.3.3. (1 + x2 ) 2x dx
R
Example 1.3.4.
x2
dx
(1 + 2x3 )2
du
appears somewhere else in the integrand (we
dx
may have to introduce a “correction factor”).
How to choose u? Choose u, such that
R√
Example 1.3.5.
R
Example 1.3.6.
1 + ex ex dx
√
x
dx
1−x
slightly different!
Example 1.3.7. Homework (give substitutions): Solve the following three integrals by using
a u substitution. Verify your answer by differentiating.
R
1. x sin x2 dx
2
u=x ,
du
= 2x,
dx
xdx = 12 du
.
R 2
y cos y 3 dy
1 du
= 3y 2 , y 2 dy = du .
u = y3,
dy
3
R
3. sinn θ cos θ dθ
du
u = sin θ,
= cos θ, du = cos θdθ
dθ
2.
Sometimes we may need to manipulate the integrand so that our choice of u becomes more
obvious.
R
Example 1.3.8. sec5 x tan x dx
R
Example 1.3.9. sin5 x dx
Example 1.3.10.
R
cos x
dx
2 − cos2 x
4
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Recall:
Example 1.3.11.
R
tan x dx =
R sin x
dx
cos x
To find the integrals of sec x and cosec x we “multiply by a form of 1” in order to get a
du
fraction which we can write as
:
u
R
Example 1.3.12. sec x dx
R
Example 1.3.13. cosec x dx
TUTORIAL
Example 1.3.14.
Example 1.3.15.
R
cot x dx
Example 1.3.16.
1.3.1
1
Z
x2 + 2x + 3
Z
x
dx
x2 + 4x + 5
dx
Definite integrals and Integration by substitution TC 5.6
R1
x(1 + x2 )10 dx Example
1.3.17.
0
u = 1 + x2 , du
= 2x, xdx = 12 du
dx
Example 1.3.18.
R π/2
π/4
sin3 5x cos 5x dx
Make sure the ORDER of the limits stays the same!
TUTORIAL
Example 1.3.19. Show that for a definite integral, the variable is a “dummy variable”.
Z b
Z b
f (u)du.
f (x)dx =
i.e.
a
a
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1.3.2
Tutorial questions — Integration by substitution
D3. Integrate with respect to x by direct substitution:
√
√
sin x
(d) ex 1 + ex
√
(a)
x
(e) cot(x)
tan x
(b) e
sec2 x
(f) sec2 x tan x
x+3
(c)
(g) x tan(x2 )
x2 + 6x + 10
(h)
(i)
x
1 + x4
sec3 x tan x.
D4. Integrate by direct substitution and some manipulation:
(a) sec4 x
(c)
(b) sec2n x sec2 x
x
x2 + 6x + 10
(d) sec x tan3 x.
T5. Show that the variable in any definite integral is a dummy variable (i.e. that
Rb
f (u)du for a general function f ) by substituting u = x.
a
Rb
a
f (x)dx =
E6. (a) Show that secn x tanm x can be integrated by a sec substitution if m is odd (put
m = 2k + 1) and by a tan substitution if n is even (put n = 2k + 2). Do not perform
the integration.
(b) Show that sinn x cosm x can be integrated by a cos substitution if n is odd and by
a sin substitution if m is odd. What technique would you use first if n and m are
both even?
R π/2
R π/2
T7. Prove that 0 f (sin θ)dθ = 0 f (cos θ)dθ for a general function f , by substituting
φ = π2 − θ .
R π/2
R π/2
E8. Prove that 0 f (sin 2θ)dθ = 0 f (sin θ)dθ for a general function f , by first substituting
φ = 2θ , next splitting the integral into one from 0 to π2 and one from π2 to π, and finally
substituting θ = φ in the first integral and θ = π − φ in the second one.
E9. The indefinite integral of ln sin θ cannot be found by any method, but the definite integral
R π/2
R π/2
from 0 to π2 can be evaluated as follows. Let I = 0 ln sin θdθ and J = 0 ln sin(2θ)dθ.
Use the double angle formula and Question 7 to show that J = π2 ln 2 + 2I, and use
Question 8 to show that J = I. Hence I = − π2 ln 2.
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1.4
Indirect substitution TC 8.3
We shall look at an indirect substitution, where
ORIGINAL variable = (trigonometric) function of NEW variable.
Recall:
• sin2 θ + cos2 θ = 1, sin2 θ = 1 − cos2 θ, cos2 θ = 1 − sin2 θ.
• 1 + tan2 θ = sec2 θ, tan2 θ = sec2 θ − 1, 1 − sec2 θ = − tan2 θ.
• how to “complete the square” method.
•
1
d
.
(arctan x) =
dx
1 + x2
•
d
1
(arcsin x) = √
.
dx
1 − x2
This method of integration is effective
when
with, amongst others, square
√ we are dealing
√
√
2
2
2
2
2
roots of quadratic functions such as a − x , a + x or x − a2 , since the square trigonometric identities and the reference right angle triangles allow us to transform the integrals into
integrals we can evaluate directly.
With x = a tan θ,
a2 + x2 = a2 + a2 tan2 θ = a2 (1 + tan2 θ) = a2 sec2 θ.
With x = a sin θ,
a2 − x2 = a2 − a2 sin2 θ = a2 (1 − sin2 θ) = a2 cos2 θ.
With x = a sec θ,
x2 − a2 = a2 sec2 θ − a2 = a2 (sec2 θ − 1) = a2 tan2 θ.
dx
R
(1 − x2 )3/2
Example 1.4.2.
R
(x2 + 1)3/2
Example 1.4.3.
R
x2 − 4
Example 1.4.1.
= (NOT arcsin x !!)
It would be useful to have a single term in the square root, so, if we let x = sin θ, then
√
of −).
1 − x2 = 1 − sin2 θ = cos2 θ. (Note: We cannot use 1 − sec2 θ = − tan2 θ since
dx
dx
(Not arctan x!!)
.
To turn quadratics into trig functions we complete the square and compare with the trig
identities to determine which would be most useful.
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dx
(use x + 1 = 2 tan θ then dx = 2 sec2 θ dθ).
(x + 1)2 + 4
−3/2
R
R
−3/2
(2x + 1)2 − 4
dx
Example 1.4.5. (4x2 + 4x − 3)
dx =
Example 1.4.4.
dx
=
x2 + 2x + 5
R
Z
Example 1.4.6.
Z
Example 1.4.7.
1.4.1
dx
(1 + 6x − 9x2 )
=
1/2
(x − 1)dx
R
[1 + 2x − x2 ]1/2
Z
dx
2 − (3x − 1)2
1/2 =
Z
√
2
cos θ dθ
3
2 1/2
[2 − 2 sin θ]
needs a DIRECT SUBSITUTION
Definite integrals and Indirect substitution
dx
Example 1.4.8. −1 √
=
x2 + 6x + 13
R1
Example 1.4.9.
R1
0
dx
x2 − 4
Z 1
dx
p
(x + 3)2 + 4
−1
(done earlier!)
WARNING: We had 2 choices.
1.4.2
Tutorial questions — Indirect substitution
D10. Integrate by completing the square if necessary and then making a trigonometric substitution:
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(a) (1 + x2 )−1/2
(d) (x2 + 2x + 2)−3/2
(b) (1 − x2 )1/2
(e) (x2 + x + 12 )−1/2
(c) (x2 − 1)−1/2
(f)
(g) (−x2 + 3x + 4)1/2 .
(x2 + x − 2)−1/2
√
1
x and secondly by completing the square. Let
D11. Integrate √x−x
2 firstly by substituting u =
θ denote the first integral and φ denote the second integral. Use trigonometric identities
(double angle formulae) to show that θ and φ differ by a constant.
D12. Integrate by combining direct and indirect substitution (which can be done in one step,
if you like).
√
√
(a)
1 + e2x (hint: first put u = ex )
e2x − 1
(c)
√
√
1 − e2x
(b)
(d) (x + 2 x + 2)−3/2 .
D13. Evaluate the following definite integrals using appropriate simple techniques or substitution methods:
Z 1
R2√
R π/4
1
(g) 0 1 + e2x dx
(a) 0 tan3 θ dθ
dx
(d)
2
Z 1
Z π/2
0 4x + 4x + 17
x
1
√
R
1
(h)
dx
dθ
(b)
2
2
(e) 1/2 1 − x dx
0 1+x
π/4 1 + cos θ
Z 1
Z 2√ 2
R 2π
x
x −1
(c) 0 sin θ cos 2θ dθ
(i)
dx.
(f)
dx
4
x
0 1+x
1
1.5
Rational functions (or Partial fractions) TC 8.4
Example 1.5.1.
Z
Example 1.5.2.
Z 2
1
Example 1.5.3.
R x5 + 2x3 + x − 1
(x2 + 1)2
dx
x2 + x − 2
.
4x + 1
dx
(x2 − 2x + 2)2
dx
IMPROPER deg TOP ≥ deg BOTTOM, DIVIDE.
1.5.1
Tutorial questions — Rational functions
D14. Find partial fractions for, and then integrate, the rational functions:
(a)
(b)
5x
x2 − 3x − 4
100
4
x − 3x2 − 4
9x2
x4 + 5x2 + 4
25
(d)
x4 − 5x2 + 4
(c)
(e)
(f)
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1
x3 − 3x + 2
x3 − 2x2 + 3x + 3
x4 + 2x2 + 1
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(g)
1.6
64
x4 − 6x2 − 8x − 3
.
Integration by parts (“Product rule”) TC 8.1
This technique allows us to integrate
the product rule for differentiation
R
x ex dx,
R
x sin x dx,
R
arcsin x dx etc. We start with
d
du
dv
(uv) =
v+u
dx
dx
dx
dv
d
du
=
(uv) − v
u
dx
dx
dx
Integrate both sides with respect to x
Z
Z
du
dv
u dx = uv − v dx.
dx
dx
Hopefully, the integral on the right-hand side is easier than the original on the left-hand side.
dv
need to be decided from the original integral.
dx
du
Then u is differentiated to obtain
dx
dv
is integrated to obtain v.
and
dx
u and
So we have 4 items:
Example 1.6.1.
R
du
u,
dx
dv
, v
dx
that are plugged into the RHS.
x sin 2x dx
We want to choose u and
dv
to make the integration easier.
dx
du
u = sin 2x,
= 2 cos 2x
dx 2
.
We could have chosen dv
x
= x,
v=
dx
2
This would have given us
x2
sin 2x −
2
Example 1.6.2.
Z
x2
2 cos 2x dx
2
Z
a WORSE INTEGRAL!
xe−x dx
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Example 1.6.3.
Z
(x2 + 2x) sin x dx Int. by parts needs sometimes to be done more than once!
Example 1.6.4.
R
ln x dx
we do NOT have a product, we introduce 1.
R
R
R
x
dx
Example 1.6.5. arctan x dx = 1. arctan x dx = x arctan x −
1 + x2
R
Example 1.6.6. EXTRA: arcsin x dx
R
Example 1.6.7. x arcsin x dx
Trick: Let θ = arcsin x, so x = sin θ,
R
Example 1.6.8.
1.6.1
dx
= cos θ
dθ
(ln x)2 dx
Integration by parts for Definite integrals
Z b
dv
u dx = [u v]ba −
dx
a
Example 1.6.9.
Z b
a
v
du
dx
dx
No + c needed.
R2 x
e (x + 1)dx
0
Example 1.6.10.
Example 1.6.11.
Rπ
0
(π − x) sin x dx
2 Z 2
2
x
x
ln x −
dx
x ln x dx =
2
1 2
1
1
Z 2
1.6.2
Examples where the original integral reappears
Example 1.6.12.
R
e2x sin 3x dx
Example 1.6.13.
R
sec3 θ dθ =
Example 1.6.14.
R
sinn θ dθ = In
When we have limits
R π/2
0
R
sec2 θ sec θ dθ
for n ≥ 2
π/2
Z
n − 1 π/2 n−2
− cos θ sinn−1 θ
sin
θ dθ
+
sin θ dθ =
n
n
0
0
n
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1.6.3
Repeated use of integration by parts
Example 1.6.15.
R
2x
−
dv
u dx = uv − u′
dx
Z
x e dx = x
Z
3e
Z
3 2x
2
3x2
e2x
dx
2
v+u
′′
ZZ
v−u
′′′
ZZZ
v + ···
Choose u to go to 0 with repeated differentiation. USEFUL, SAVES TIME!
Example 1.6.16.
1.6.4
R
(x5 + 7x2 ) sin 2x dx
Tutorial questions — Integration by parts
D15. Integrate by parts (using substitution as well, if necessary):
(a) (2x + 1)e3x
(d) arcsin x
(b) (3x + 2) sin 4x
(e) (x2 + 3x + 1) ln 2x
(c) x sec2 2x
(f)
cosec3 θ
(g) sec θ tan2 θ
√
1 + x2
(h)
√
(i)
x2 − 2x.
D16. Use integration by parts to evaluate the definite integrals:
R1
R1
(a) 0 e−x (x − 1) dx
(c) 0 x arctan x dx
Rπ
R π/4
(b) 0 x cos x dx
(d) 0 sin θ ln sec θ dθ.
D17. Use integration by parts twice on the integrals:
R
R
(a)
(3x2 + 2x + 1)e2x dx (c)
sin 2θ cos 3θ dθ
R −x
Re
(b)
e cos 3x dx
(d) 1 (ln x)2 dx
(e)
(f)
R1
(arcsin x)2 dx
0
R1
x(1 − x) sin nπx dx.
0
R1
D18. RUse the formula for integrating by parts several times in one line to find 0 x4 e−x dx and
17
.
x5 sin x dx. Use the fact that the first integral is positive (why?) to show that e > 2 24
R π/2
T19. (a) If In = 0 sinn x dx, evaluate I0 and I1 directly. If n ≥ 2, use integration by parts
I . Apply
(integrating sin x and differentiating sinn−1 x) to show that In = n−1
n n−2
n−1 n−3 n−5
this result repeatedly to prove that In = ( n )( n−2 )( n−4 ) · · · , ending with · · · ( 23 )1 if
n is odd, and with · · · ( 12 ) π2 if n is even. This formula should be learnt.
R π/2
(b) What can be said about 0 cosn x dx, using Question 7 above?
(c) Use the formula for In to evaluate the integrals from 0 to π2 of:
(i) sin5 x
(ii) cos6 x
(iii) cos7 x
(iv) sin8 x.
R
R
D20. Use integration by parts to express secn x dx in terms of secn−2 x dx. (Hint: integrate
R π/4
sec2 x and differentiate secn−2 x.) Use this result twice to evaluate 0 sec5 x dx.
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1.7
Mixed examples
Example 1.7.1.
R
Example 1.7.2.
R
Example 1.7.3.
R
sin θ
dθ.
3 + 4 cos θ
cos θ
dθ
2 − cos2 θ
cos θ
dθ
3 cos θ + 4 sin θ
Example 1.7.4. Homework
1.
2.
3.
4.
R
(2x + 1)3 dx.
R
e2x (2x + 1)3 dx.
R
R
dx
.
3 + x2
x
dx.
(3 + x2 )3
Example 1.7.5.
1.7.1
R
1
x3 − 1
dx
Tutorial questions — Assorted integration examples
D21. Here is a selection of functions, to test your integration technique. There is no need to
do them all at once: do a few each week, to keep in practice.
(a) e−x
(b) x−e
(c) 2x
(d) x2
(e) e2
1
4 + x2
1
(g)
4+x
x
(h)
4 + x2
4
(i)
4 − x2
4
(j) √
4 + x2
(f)
1
4−x
1
(l) √
x2 − 4
1
(m) √
4 − x2
x
(n) √
4 − x2
√
(o)
4 − x2
√
(p)
4−x
√
(q)
4 + x2
√
x2 − 4
(r)
(k) √
(s)
(t)
e
e
√
x
arcsin x
(u) eln x
(v) (ln x)3
(w) x−1 (ln x)3
(x) esin x cos x
(y) ex sin(ex )
(z) sec3 x tan x
(A) sec2 x tan x
(B) sec x tan2 x
(C) sin x cos3 x
(D) sin2 x cos2 x
(E) (1 − 2x)−3
(F)
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3x2 + 6x
x3 − 3x + 2
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3x2 − 3
x3 − 3x + 2
x5 − 11x − 6
(H)
x3 − 3x + 2
3x2 − 3x − 1
(I)
x3 − 2x2 + x − 2
(J) (4x2 + 4x + 17)−1
(G)
(K) (7 + 12x − 4x2 )−1/2
2
(L)
1 + tan θ
1.8
1 + tan2 θ
1 + tan θ
tan θ
(N)
1 + 2 cos θ
(O) sin θ sec2 θ
1
(P)
3 + 3 sin θ + 2 cos θ
3 cos θ − 2 sin θ
(Q)
3 + 3 sin θ + 2 cos θ
1
(R)
3 + 2 sin θ + 2 cos θ
(S) cos θ ln sin θ
12 − 8x
(T) √
7 + 12x − 4x2
1
√
(U)
1+ x
(M)
(V) x sec2 x
√
1+ x
(W) √
1+x
√
(X) 1 + ex
√
(Y) 1 − ex
Tutorial answers
1
1
1. (a) − 81 cos 4θ + c
(b) 18 sin 4θ + 12
sin 6θ + c
(c) 41 cos 2θ − 12
cos 6θ + c
1
1
1
sin 6θ − 16
sin 8θ + c
(e) 21 θ + 41 sin 2θ + c
(f) 38 θ − 41 sin 2θ + 32
sin 4θ + c
(d) 12
1
1
1
1
(g) 16 θ − 64 sin 4θ + 64 sin 2θ − 192 sin 6θ + c
(h) tan θ − θ + c
(i) tan θ + sec θ + c
1
1
cos 4θ − 18 cos 2θ + 24
cos 6θ + c.
(j) − 16
2
1
x
(b) sin 2θ dθ
(c) √v−v
(d) 2xex dx
(e) x2 sin(x3 )−2/3 cos(x3 ) dx
2. (a) √1+x
2 dx
2 dv
(f) 2 cosec 2φ dφ.
√
3. (a) −2 cos( x) + c
(d) 32 (1 + ex )3/2 + c
(b) etan x + c
(c) 21 ln |x2 + 6x + 10| + c
1
1
1
2
2
(g) 2 ln | sec(x )| + c
(h) 2 arctan(x2 ) + c
(e) ln | sin x| + c
(f) 2 tan x + c
(i) 31 sec3 x + c.
2r+1
P
(b) nr=0 nr tan2r+1 x +c
(c) 21 ln |x2 +6x = 10|−3 arctan(x+3)+c
4. (a) tan x+ 13 tan3 x+c
1
(d) 3 sec3 x − sec x + c.
5. Omitted.
6. Omitted.
7. Omitted.
8. Omitted.
9. Omitted.
√
√
√
10. (a) ln |x + 1 + x2 | + c
(b) 12 arcsin x + 12 x 1 − x2 + c
(c) ln |x + x2 − 1| + c
p
p
(e) ln | (2x + 1)2 + 1+(2x+1)|+c
(d) √ x+1 2 +c
(f) ln |2x+1+ (2x + 1)2 − 9|+c
(x+1) +1
√
arcsin( 2x−3
) + 14 (2x − 3) 4 + 3x − x2 + c.
(g) 25
8
5
√
11. 2 arcsin( x) + c1 = arcsin(2x − 1) + c2 .
√
√
√
√
12. (a) ln | 1 + e2x − 1| + 1 + e2x − x + c √ (b) ln |1 − 1 − e2x | + 1 − e2x − x + c
√
(c) e2x − 1 − arccos(e−x ) + c
(d) √−2( x+2)
+ c.
√
x+2 x+2
√
√
(b) 2− 2 = 1−tan π8 (c) 0
(c) 81 (arctan 34 −arctan 14 )
(d) π6 − 83
13. (a) 12 (1−ln 2)
√
√
√
√
√
(e) 3 − π3
(g) 12 ln 2
(f) 1 + e4 + ln | 1 + e4 − 1| − 2 − 2 − ln | 2 − 1|
(h) π8 .
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(c) 6 arctan( 12 x)−3 arctan x+c
x−2
−20 arctan x+c
(b) 5 ln x+2
14. (a) ln |(x+1)(x−4)4 |+c
2
25
+c
(d) 12
ln (x−2)(x+1)
(x+2)(x−1)2
1
x+2
(e) 91 ln x−1
− 3(x−1)
+c
+ 4(x+3)
+ c.
(g) ln x−3
x+1
(x+1)2
5x−2
(f) 12 ln(x2 +1)+ 12 arctan x+ 2(x
2 +1) +c
3
15. (a) 31 (2x+1)e3x − 92 e3x +c
(b) 16
sin 4x− 14 (3x+2) cos 4x+c
(c) 21 x tan 2x− 41 ln | sec 2x|+c
√
(e) ( 31 x3 + 32 x2 + x) ln 2x − ( 19 x3 + 43 x2 + x) + c
(d) x arcsin x + 1 − x2 + c
(g) 12 (sec θ tan θ − ln | sec θ + tan θ|) + c
(f) 12 (− cosec θ cot θ + ln | cosec θ − cot θ|) + c
√
√
√
√
(i) 21 (x−1) x2 − 2x− 12 ln |x−1+ x2 − 2x|+c.
(h) 12 (x x2 + 1+ln | x2 + 1+x|)+c
16. (a) −e−1
(b) −2
(c) π4 − 21
(d) − √12 + 1 − 2√1 2 ln 2.
17. (a) 21 (3x2 + 2x + 1)e2x − 41 (6x + 2)e2x + 43 e2x + c
(d) e−2
(c) 15 (3 sin 2θ sin 3θ+2 cos 2θ cos 3θ)+c
1 −x
(b) 10
e (3 sin 3x − cos 3x) + c
2
n
(e) 14 π 2 −2
(f) (nπ)
3 (1−(−1) ).
65
18. 24 − 65
, which is positive since x4 e−x > 0. Therefore e > 24
.
e
5
4
3
2
−x cos x + 5x sin x + 20x cos x − 60x sin x − 120x cos x + 120 sin x + c.
R π/2
16
8
, 5π
, 35
, 35π
.
19. (a) 0 cosn x has same value.
(b) 15
32
256
√
√
1
20. In = n−1
{tan x secn−2 x + (n − 2)In−2 }, 81 {7 2 + 3 ln( 2 + 1)}.
(w)
21. (a) −e−x + c
(b)
(c)
(d)
1
x−e+1 + c
1−e
1 x
2 +c
ln 2
1 3
x +c
3
2
(x) e
(z)
(A)
1
arctan x2 + c
2
(B)
(g) ln |4 + x| + c
(C)
(h)
(i)
(j)
+c
(y) − cos(ex ) + c
(e) e x + c
(f)
1
(ln x)4 + c
4
sin x
1
ln(4 + x2 ) + c
2
2+x
+c
ln 2−x
(D)
(E)
√
(F)
4 ln | x2 + 4 + x| + c
1
sec3 x + c
3
1
tan2 x + c or 21 sec2 x + c
2
1
sec x tan x − 12 ln | sec x + tan x| + c
2
− 41 cos4 x + c
1
1
x − 32
sin 4x + c
8
1
(1 − 2x)−2 + c
4
3
+c
3 ln |x − 1| − x−1
3
(k) −2(4 − x)1/2 + c
√
(l) ln |x + x2 − 4| + c
(G) ln |x − 3x + 2| + c
(n) −(4 − x2 )1/2 + c
√
(o) 2 arcsin x2 + 21 x 4 − x2 + c
(I)
(H)
(m) arcsin x2 + c
(J)
(K)
16
1 3
x + 3x + 3(x−1)
− 29 ln |x − 1|
3
− 16
ln |x + 2| + c
9
3
2
ln |x − 2x + x − 2| + arctan x + c
1
arctan 2x+1
+c
8
4
1
arcsin 2x−3
+c
2
4
(p) − 32 (4 − x)3/2 + c
√
√
(q) 21 x x2 + 4 + 2 ln | x2 + 4 + x| + c
√
√
(r) 12 x x2 − 4 − 2 ln | x2 − 4 + x| + c
√ √
(s) 2e x ( x − 1) + c
√
(t) 12 earcsin x ( 1 − x2 + x) + c
(P)
(v) x{(ln x)3 − 3(ln x)2 + 6 ln x − 6} + c
(Q) ln |3 + 3 sin θ + 2 cos θ| + c
(u)
(L) ln |1 + tan θ| − ln | sec θ| + θ + c
(M) ln |1 + tan θ| + c
(N) ln |2 + sec θ| + c
(O) sec θ + c
1 2
x +c
2
1+tan 1 θ
1
ln 5+tan 21 θ
2
2
+c
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(R) 2 arctan(2 + tan 12 θ) + c
(V) x tan x − ln | sec x| + c
√
√
√
√
(W) 2 1 + x+ x + x2 −ln( 1 + x+ x)+
c
√
√
(X) 2 1 + ex + 2 ln( 1 + ex − 1) − x + c
√
√
(Y) 2 1 − ex + 2 ln(1 − 1 − ex ) − x + c
(S) sin θ(ln(sin θ) − 1) + c
√
(T) 2 7 + 12x − 4x2 + c
√
√
(U) 2 x − 2 ln(1 + x) + c
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Chapter 2
AREAS AND ARC LENGTHS, TC
Chapter 11
2.1
General Curves in a Plane TC 11.1
So far we have used explicit equations of the form y = f (x) to describe curves in a plane.
Example 2.1.1.
y=
x2 + x − 1
,
x+3
y = 2 sin x + cos x, . . . .
For each x there is only one y.
Explicit form cannot cope with curves that have loops or doubled back on itself.
We need equation of the form
f (x, y) = constant
in an implicit form.
Example 2.1.2. (conic forms: seen in MATH1042 Chapter 4 of algebra)
(x + 1)2 (y + 2)2
−
= 1, y sin xy = 1, . . . .
4
5
dy
π
Example 2.1.3. Show that P 12
, 2 lies on the curve y sin xy = 1 and find dx
at that point.
x2 + y 2 = 4,
Another way is to express x and y in terms of a 3rd independent variable, called a parameter,
t say, or θ.
x = x(t)
called parametric equations.
y = y(t)
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Example 2.1.4. Polar eq. x = r cos θ, y = r sin θ where the parameter is θ.
y(θ)
= tan θ, then the parametric equations are polar equations (which have been done
x(θ)
in algebra).
If
Often t is used for parameter, so that
x = x(t) and y = y(t) describe the path of a particle that
at time t is at position x(t), y(t) .
In vector form we write r = r(t)
i.e.: (x, y) = x(t), y(t) = displacement vector
ṙ = (ẋ, ẏ) = velocity vector
dy
The slope of a curve given in parametric form, using the chain rule is given by dx
∴ if ẏ =
dy
dy dx
dy
dy dt
ẏ
=
·
⇒
=
·
= .
dt
dx dt
dx
dt dx
ẋ
·=
d
dt
∴
dy
ẏ
=
dx
ẋ
dy
= 0 as before, ∴ when ẏ = 0, ẋ 6= 0
Horizontal tangents occur when dx
dy
Vertical tangents occur when ẋ = 0 and ẏ 6= 0, dx
is undefined.
If we have ẋ = 0 AND ẏ = 0, we have a CUSP or KINK.
. . . the particle is at rest
Since x(t) gives the horizontal position of the particle, ẋ gives the horizontal change: negative for left and positive for right. Similarly y(t) gives the vertical position of the particle, so
ẏ gives the vertical change: negative for down and positive for up.
2.2
Parametric curve sketching

 x = t2 − 1
Example 2.2.1.
t2
 y = (2t + 3)
2
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2.2.1
Method for sketching curves in parametric form
x = x(t),
y = y(t)
• Find x − y intercepts, by putting solving y(t) = 0 and substituting the values for t into
x(t) and repeat for x(t) = 0.
• Determine behaviour of x(t) and y(t) as t → ±∞.
(unless domain is restricted).
• Find critical points by solving
ẋ(t) = 0, ẏ(t) = 0, both = 0 CUSP
v.t.
h.t.
• Make a table showing the signs of ẋ and ẏ and the direction of travel.
E.g. the 4 possible directions are:
ẋ
+ + − −
ẏ
+ − + −
Direction ր ց տ ւ
x
y
A parametric curve is
always orientated
i.e. it has direction
Examples 1 and 4 for homework (solutions will be available on SAKAI).
2
1
x = t + sin(2t) x = t + sin t
y = sin t
y = 1 + cos t
0 ≤ t ≤ 4π
Example 2.2.2.
x = t + sin t
y = 1 + cos t
Example 2.2.3.
x = cos θ
y = sin(2θ)
3
x = cos θ
y = sin(2θ)
0 ≤ θ ≤ 2π
4
x = cos t − cos(2t)
y = −2 sin t + sin(2t)
0 ≤ θ ≤ 2π
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2.3
Another application
Theorem 2.3.1. Suppose we have a POLAR CURVE, r = r(θ), with vector form r = (r cos θ, r sin θ).
Let α = angle between position vector and tangent vector at a point P on the curve: then
r
tan α = , ṙ 6= 0.
ṙ
Note: If ṙ = 0, then tan α is undefined, hence α = π2 ∴
r = constant
tangent ⊥ radius ∴ circle, i.e.
Example 2.3.2. Find the angle √α between
the tangent and the position vector to the curve
3 1
r = 2 sin θ at the point (x, y) = 2 , 2 .
Find the equation of the tangent at that point.
2.3.1
Tutorial questions — General curves in the plane
D1. For each of the following curves r = r(t) find the values of t for which the curve passes
through the given points.
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(a) r = (cos t, sin2 t) (−π < t ≤ π). Points (1, 0) and (− 21 , 34 ).
(b) r = (t2 − 1, (t − 1)2 ). Points (0, 0), (0, 4), and (3, 1).
(c) r = (t3 − t, cos πt). Points (0, 1) and (0, −1).
D2. For each of the following curves in parametric form,
(i)
find cuts on axes (y = 0 and x = 0),
(ii) find what happens to x and y as the parameter tends to ±∞ (unless the parameter
values are restricted),
(iii) find points with horizontal or vertical tangents (ẏ = 0 or ẋ = 0),
(iv) draw up a table of signs of ẋ and ẏ, including arrows to show direction of travel,
(v) plot all points found in (i) and (iii), labelling them with the parameter value,
(vi) join up the points, checking that the direction of travel agrees with the results from
(ii) and (iv).
(a) x = 6 cos θ + cos 3θ, y = 6 sin θ + sin 3θ (−π ≤ θ ≤ π). Hint: in finding intercepts
and tangents it may be helpful to remember that sin 3θ = 3 sin θ − 4 sin3 θ and
cos 3θ = −3 cos θ + 4 cos3 θ. Note that θ is not the polar co-ordinate of the point
(x, y). Why not? (The curve is the dumbbell shape of a Wankel engine.)
(b) x = cos θ(1 + 2 sin2 θ), y = 2 sin3 θ (− π2 ≤ θ ≤ π2 ). Again prove that θ is not the
polar co-ordinate of (x, y). (This is the caustic curve formed when light is reflected
in a semicircular mirror. Look inside a cylindrical coffee mug. There is a cusp that
becomes the focus if only a small part of the mirror is used.)
2
(c) x = 53 t5 − 5t3 + 12t, y = 10/(t2 + 1).
(g) x = t3 − 3t, y = 4te−t /8 .
(e) x = t3 − 3t, y = 10/(t2 − 2t + 2).
(i)
(d) x = t3 − 3t, y = 10/(t2 + 1).
(f)
x = t3 − 3t, y = 10/(t2 − 4t + 5).
2
(h) x = t3 − 3t, y = 8te−t /2 .
2
x = t3 − 3t, y = 16te−2t .
The last six curves go together in threes, and illustrate how by changing a constant one
can change a simple curve (i.e. a curve that does not cross itself) into one with a cusp,
and then into one with a loop. Think of a plane doing aerobatics: with too little power
it rises, then descends smoothly; with more power, it may stall; with still more power, it
can loop the loop.
D3. Give the vector parametric form of the polar equation r = 1 + cos θ, which represents a
cardioid. Find the vertical and horizontal tangents, and verify that there is a cusp where
θ = π. Hence sketch the curve roughly.
D4. Prove that in a logarithmic spiral (polar equation r = aebθ ) the angle between position
vector and tangent vector is constant. What happens to this angle in an Archimedean
spiral (polar equation r = a + bθ) as θ → ∞?
D5. Simplify ẏ/ẋ and then apply the method of indeterminate forms, if necessary, to find the
slopes at the cusps on the curves in Question 2(b), (e), and (h).
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2.4
Areas in parametric form
We know
Rb
a
|y|dx is the area between a curve and the x-axis from x = a to x = b.
If the curve is in parametric form x = x(t), y = y(t) then
area A =
Z t=β
t=α
|y(t)|
dx(t)
dt.
dt
(limits: integrate from “left to right”).
If we want the area between curve and y-axis we use
Z t=β
dy(t)
|x(t)|
dt.
dt
t=α
(limits: integrate from “bottom to top”).
Example 2.4.1. Find the area under one arch of the cycloid x = t − sin t, y = 1 − cos t
Example 2.4.2. Find the area of the loop of the curve x = t3 − 3t, y = 4 − t2
ALTERNATIVELY Using area under x-axis (much harder)
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2.4.1
Tutorial questions — Areas in parametric form
D6. Find the area in the first quadrant lying inside the curve x = 6 cos θ + cos 3θ, y =
6 sin θ + sin 3θ (Question 2(a)). (Hint: find the area between the curve and the x axis.
Use your sketch to obtain the θ limits, and make sure you have them in the correct order.)
D7. Find the area in the first quadrant lying inside the caustic curve x = cos θ(1 + 2 sin2 θ),
y = 2 sin3 θ (Question 2(b)). (Hint: find the area between the curve and the y axis.)
T8. Show by means of a picture that if the curve r = r(t) crosses itself when t = α and t = β,
and travels
anticlockwise
Rβ
R β around the loop thus formed, then the area of the loop is equal
to − α y dx and to α x dy. (Hint: express the area of the loop as the area between its
top and bottom boundaries, and also as the area between its left and right boundaries.)
D9. Find the area of the loop in the curve x = t3 − 3t, y = 10/(t2 + 1) (Question 2(d)). (Hint:
find the area between the loop portion of the curve and the y axis, and double it.)
D10. Find the area enclosed between the curve x = t3 − 3t, y = 10/(t2 − 4t + 5) and the line
y = 2 (Question 2(f)).
2
D11. Find the area enclosed between the x axis, the curve x = t3 − 3t, y = 4te−t /8 and the
line x = 18 (Question 2(g)).
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2.5
Arc length TC 6.3, 11.2 and 11.5
Theorem 2.5.1. If α < β then the length of arc of the curve r = r(t)
(i.e. in parametric form)
x = x(t)
y = y(t)
between the points t = α and t = β is given by
Z βp
ẋ2 + ẏ 2 dt
arc length =
α
Example 2.5.2. Find the arclengths of the curve x = 2 cos t + cos 2t and
y = 2 sin t + sin 2t between the points (3, 0) and (-1, 0).
If the curve is given EXPLICITLY, y = y(x), then we replace t by x; so
Z p
ẋ2 + ẏ 2 dt
arclength =
Z " 2 2 #1/2
dx
dy
=
dt
+
dt
dt
Z " 2 2 #1/2
dy
dx
dx
+
=
dx
dx
2 #1/2
Z "
dy
=
1+
dx
dx
arclength =
R x=b
x=a
s
1+
dy
dx
2
dx (for EXPLICIT y = y(x))
Example 2.5.3. Find the arclength of the parabola y = x2 from (0, 0) to (2, 4).
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Example 2.5.4. Verify C = 2πr for a circle. (r is constant)
Example 2.5.5. Check for a straight line y = mx + c between x = a and x = b
If the curve is given by a POLAR EQUATION r = r(θ), then
x = r cos θ
y = r sin θ
where r is a function of θ.
As before, we have s
a parametric equation with parameter θ
Z θ=β 2
dr
arclength =
+ r2 dθ
dθ
θ=α
Example 2.5.6. Find the arclength of the cardioid r = 1 + cos θ
2.5.1
Tutorial questions — Arc length
D12. Find the arc lengths of the curves below. (For parametric curves the parameter values
for the initial and final points of the arc must be determined from the given cartesian
co-ordinates.)
√
(a) x = ln(sec t + tan t), y = sec t, between the points (0, 1) and (ln(2 + 3), 2).
(b) y = ex between the points where x = 0 and x = 1.
√
(c) y = x between the points where x = 1 and x = 4.
√
√
√
(d) r = ( 3(t2 − 6t), 8t3/2 ) between the origin and the point (−9 3, 24 3).
(e) r = (cos θ + θ sin θ, sin θ − θ cos θ) between the points (1, 0) and ( π2 , 1). (This is an
involute of a circle, the profile used for gear teeth.)
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y = 21 (ex + e−x ) between the points where x = a and x = b. (This is a catenary, the
curve formed by a hanging chain. Note that y = cosh x, see Algebra Chapter 5.)
dr 2 1/2
ds
= r2 + ( dθ
)
T13. If a curve has polar equation r = r(θ), show that dθ
. (Hint: show that
dr
dr
dx dy
( dθ , dθ ) = ( dθ cos θ − r sin θ, dθ sin θ + r cos θ).)
(f)
D14. Find the length of the cardioid with polar equation r = 1 + cos θ. (Hint: let θ run from
−π to π.)
D15. Find the length of the curve with polar equation r = 3 + 5 cos( 4θ
) between the points
5
.
where θ = 0 and θ = 5π
8
√
√
ds
D16. If a curve has polar equation r = 2 2 + eθ + e−θ , show that dθ
= 2 + 2(eθ + e−θ ). Hence
find the length of the curve between the points where θ = 0 and θ = π.
D17. A more accurate way to represent tape being wound onto a reel (see Calculus Chapter 5
aθ
, where c is an
Question ??) is by the Archimedean spiral with polar equation r = c + 2π
ds
arbitrary constant, and dt = b (given).
ds
and hence obtain more accurate values for the angular velocity dθ
.
(a) Find dθ
dt
dr
(b) Find dr
from dθ
and dθ
.
dt
dt
E18. (Kepler’s Laws) A star of mass M is located at the origin, and a planet of mass m
moves around it in an orbit with polar equation r = 1/(1 + ǫ cos θ), where ǫ is a con)2 and its potential energy is −M mGr−1
stant.R The planet’s kinetic energy is 12 m( ds
dt
(i.e. √ M mGr−2 dr from Newton’s inverse square law of gravitation). Assuming that
dθ
= M Gr−2 , prove (using the chain rule) that
dt
(i)
dA
is constant (i.e. the radius vector sweeps out equal areas in equal times),
dt
(ii) the total energy is constant.
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2.6
Curved surface area TC 6.4 and 11.2
If you rotate a region in the plane that is bounded by the graph of a function over an interval,
it sweeps out a solid of revolution as we saw in chapter 4. If you rotate only the bounding curve
itself, it does not sweep out any interior volume, but rather a surface that surrounds the solid
and forms part of its boundary.
As an arc P Q of the curve rotates about the axis it sweeps out a band that we can approximate as a frustum of a cone. The surface is a union of bands like the one swept out by P Q
(see TC 6.4 for detailed explanation).
Using the formula for the surface area of a frustum and a Riemann Sum to sum the union
of bands, we get the formula for the Curved Surface Area of an explicitly given curve y = f (x)
rotated about the x-axis as
CSA =
Z b
2πy
a
s
1+
dy
dx
2
dx
Note the use of the formula for explicitly given arclength from the previous section.
Example 2.6.1. Find the surface area of a sphere with radius a
We use the EXPLICIT form of CSA and let y =
√
a2 − x2
Theorem 2.6.2. If a curve is rotated about the x-axis, and if s = s(t) denotes the arclength
along the curve, then the curved surface area between t = α and t = β is given by:
Z t=β
Z t=β
p
ds
2π|y(t)| dt =
CSA =
2π|y(t)| ẋ2 + ẏ 2 dt
dt
t=α
t=α
Similarly, if our parametric curve is polar,
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Z θ=β
ds
2π|y(θ)| dθ =
CSA =
dθ
θ=α
Z θ=β
√
2π|y(θ)| r2 + ṙ2 dθ
θ=α
Example 2.6.3. Find the surface area of a sphere with radius a using POLAR form of CSA:
We use x = a cos θ, y = a sin θ
Example 2.6.4. Find the CSA of the solid formed by rotating one arch of the cycloid
x = t − sin t
y = 1 − cos t about the x-axis.
Example 2.6.5. Find the CSA of the paraboloid formed by rotating y =
and x = 3 about the x-axis.
2.6.1
√
x between x = 0
Tutorial questions — Curved surface areas
D19. Find the areas of the curved surfaces formed when the arcs in Question 12 are rotated
about the x axis.
D20. A long hollow trumpet is formed by rotating the curve y = x1 between the points where
x = 1 and x = X about the x axis. Find the volume and the curved surface area. Show
that as X → ∞ the volume tends to π but the curved surface area tends to ∞.
E Is there a contradiction? Observe that in order to paint the entire inner surface it is
sufficient to pour π cubic units of paint into the trumpet.
√
D21. Consider the sphere and cylinder formed by rotating the semicircle y = a2 − x2 and the
ds
line y = a about the x axis. Show that y dx
is the same for both surfaces. Deduce that
the surface areas of the portions of the sphere and the cylinder lying between any two x
values (between −a and +a) are the same. This fact was discovered by Archimedes.
2.7
Tutorial answers
.
1. (a) r = (1, 0) at t = π; r = (− 12 , 43 ) at t = ± 2π
3
(b) r = (0, 0) at t = 1; r = (0, 4) at t = −1; r = (3, 1) at t = 2.
(c) r = (0, 1) at t = 0; r = (0, −1) at t = ±1.
√
2. (a) Cuts (±7, 0), (0, ±5). Horizontal tangents (±2, ±3 3), (0, ±5). Vertical tangents
(±7, 0). Not polar since xy 6= tan θ.
(b) Cuts (1,
(0, ±2). Horizontal tangents (1, 0) (cusp) and (0, ±2). Vertical tan√0), √
gents ( 2, ± 21 2).
, 2),
(c) Cut (0, 10). Horizontal tangent (0, 10). Vertical tangents (± 16
5
(d) Cuts (0, 52 ),
(± 38
, 5).
5
(0, 10). Horizontal tangent (0, 10). Vertical tangents (±2, 5).
√
10
(5 ± 2 3)) and (0, 5). Vertical tangent (2, 2). Cusp with slope − 10
at
(e) Cuts (0, 13
3
(−2, 10).
√
(f) Cuts (0, 2), (0, 52 (2± 3)). Horizontal tangent (2, 10). Vertical tangents (2, 1), (−2, 5).
√
(g) Cuts (0, 0), (0, ±4 3e−3/8 ). Horizontal tangents ±(2, 8e−1/2 ). Vertical tangents
±(2, −4e−1/8 ).
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√
(h) Cuts (0, 0), (0, ±8 3e−3/2 ). Cusps with slope − 83 e−1/2 at ±(2, 8e−1/2 ).
√
, −8e−1/2 ). Vertical tangents
(i) Cuts (0, 0), (0, ±16 3e−6 ). Horizontal tangents ±( 11
8
−2
±(2, −16e ).
Figure 2.1: Sketches for Question 2
√
3. r = (cos θ + cos2 θ, sin θ + sin θ cos√θ). Horizontal tangents (0, 0) (cusp) and ( 43 , ± 3 4 3 ).
Vertical tangents (2, 0) and (− 14 , ± 43 ).
4. (i) Angle = arctan 1b . (ii) Angle tends to π2 .
5. (b) 0 (e) − 10
3
(h) − 3√8 e .
.
6. 39π
4
.
7. 3π
4
8. Omitted.
√
9. 20(2π − 3 3).
10. 16 + 120 arctan 2.
11. 48(7 − 16e−9/8 ).
12. (a) ṡ = sec2 t, length =
√
(d) 27 3.
(e) 18 π 2
√
√
√
√
2
√ −1 + 1 + e2 −1− 2.
3.
(b) ln 1+e
2−1
(f) 21 (eb − ea − e−b + e−a ).
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√
√
√
√17 }.
(c) 41 {4 17−2 5+ln 4+
2+ 5
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13. Omitted.
14. ṡ = 2| cos 2θ |; length = 8.
15. ṡ = 5 + 3 cos 4θ
; length = 58 (5π + 6).
5
√
16. 2π + 2(eπ − e−π ).
√
ds
1
dθ
17. dθ
= 2π
= √4π2πb
4π 2 r2 + a2 ,
2 r 2 +a2 ,
dt
dr
= √4π2abr2 +a2 .
dt
√
1
=
18. (i) dA
M G.
(ii) KE + PE = 12 mM G(ǫ2 − 1).
dt
2
√
√
√
√
√
2
√ +e ].
19. (i) π[2 3 + ln(2 + 3)].
(ii) π[e 1 + e2 − 2 + ln 1+e
2+1
8 5
3 π
(iv) 2 35
.
(iii) π6 [173/2 − 53/2 ].
(v) 2π(3 − 41 π 2 ).
(vi) π(b − a) + π4 (e2b − e2a − e−2b + e−2a ).
√
√
√
√
20. V = π(1 − X1 ) A = π[ln( 1 + X 4 + X 2 ) − ln( 2 + 1) − 1 + X −4 + 2].
No contradiction: finite volume can cover infinite area if the layer is infinitesimally thin.
21. a.
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Chapter 3
ADVANCED APPLICATIONS OF
DIFFERENTIATION, TC Chapter
10.2 – 10.4 & 10.8 – 10.10
3.1
Higher Approximation (Taylor Polynomials) TC 10.8
Suppose we have a function, which is continuous at x = 0, and also differentiable at x = 0. We
can find a 1st approximation to f (x),using the definition of the derivative.
f (x) − f (0)
≈ f ′ (0)
x−0
so that f (x) ≈ f (0) + xf ′ (0). This represents the equation of a straight line through (0, f (0))
and the gradient f ′ (0). This line is the tangent to the curve at x = 0
We have a polynomial of degree 1, called 1st approximation, which has the same value as f at
0 and has the same 1st derivative at 0. So, near x = 0, this polynomial approximates f (x).
We shall call this polynomial P1 (x).
So P1 (x) = f (0) + xf ′ (0) where P1 (0) = f (0) and P ′ 1 (0) = f ′ (0)
Similarly we define the 2nd approximation to f (x) near x = 0 to be a polynomial of degree 2,
i.e. a quadratic with the same value, 1s t derivative and 2nd derivative at x = 0.
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Let P2 (x) = A + Bx + Cx2 . We need P2 (0) = f (0), P ′ 2 (0) = f ′ (0) and P ′′ 2 (0) = f ′′ (0).
P2 (0) = f (0) gives A = f (0)
P ′ 2 (0) = f ′ (0) gives B = f ′ (0)
P ′′ 2 (0) = f ′′ (0) gives 2C = f ′′ (0) ⇒ C = 21 f ′′ (0)
So our 2nd approximation is P2 (x) = f (0) + f ′ (0)x + 12 f ′′ (0)x2
This process can be done for a cubic polynomial.
Let P3 (x) = A + Bx + Cx2 + Dx3
P3 (0) = f (0) ⇒ A = f (0)
P ′ 3 (0) = f ′ (0) ⇒ B + 2Cx + 3Dx2 x=0 = B = f ′ (0)
P ′′ 3 (0) = f ′′ (0) ⇒ 2C + 6Dx x=0 = f ′′ (0) ⇒ C = 21 f ′′ (0)
P ′′′ 3 (0) = f ′′′ (0) ⇒ 6D = f ′′′ (0) ⇒ D = 16 f ′′′ (0)
So our 3rd approximation is P3 (x) = f (0) + f ′ (0)x + 21 f ′′ (0)x2 + 16 f ′′′ (0)x3
Note: Pn+1 (x) is Pn (x) plus an extra term.
The fourth approximate will be of the form P4 (x) = A + Bx + Cx2 + Dx3 + Ex4 :
P4 (x) = f (0) + f ′ (0)x + 12 f ′′ (0)x2 + 16 f ′′′ (0)x3 +?f ′′′′ (0)x4 ,
where ? satisfies P ′′′′ (0) = f ′′′′ (0) which implies 4 · 3 · 2E = f ′′′′ (0), so E =
1 ′′′′
f (0).
4!
dn x
= n! (calculus chap 2)
dxn
1
so P4 (x) = f (0) + f ′ (0)x + 12 f ′′ (0)x2 + 61 f ′′′ (0)x3 + f ′′′′ (0)x4 .
4!
Using factorials: recall the nth derivative of xn :
This idea can be extended to a polynomial of degree n, Pn (x), called the nth approximation to
a function f (x) near x = 0 if
• Pn (x) is of degree at most n and
• Pn (0) = f (0), Pn ′ (0) = f ′ (0), · · · , Pn (n) (0) = f (n) (0)
Hence we have:
1
1
1
1
1
Pn (x) = f (0) + f ′ (0)x + f ′′ (0)x2 + f ′′′ (0)x3 + f ′′′′ (0)x4 + · · · + f (n) (0)xn
1
2!
3!
4!
n!
n
(r)
X
f (0) r
=
x.
r!
r=0
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These are called Taylor polynomials after the English mathematician Brook Taylor (1685 1731).
Example 3.1.1. Find 1st , 2nd and 3rd approximations to the function y = sin x + cos x near
x = 0.
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Example 3.1.2. Use the above example to find the 1st , 2nd and 3rd approximations of sin(0, 1)+
cos(0, 1):
Example 3.1.3. Find a 3rd approximation to arctan x, near x = 0.
√
Example 3.1.4. Find a 4th approximation to 1 + x, near x = 0.
Example 3.1.5. Extra Q) Find a 3rd approximation to ln(1 + 2x), near x = 0.
Example 3.1.6. Find an nth approximation to ex , near x = 0.
√
√
Example 3.1.7. Use your 4th approximation to 1 + x, near x = 0, to approximate 1, 1
The error is the actual answer − approximation.
= f (x) − Pn (x)
(3.1)
When approximating a function f (x) near x = 0 using an nth approximation Pn (x), an error is
made which is f (x) − Pn (x)
Theorem 3.1.8. TAYLOR’S THEOREM: The error f (x) − Pn (x) is given by f (x) − Pn (x) =
f (n+1) (c) n+1
x , for some c, a number between 0 and x.
(n + 1)!
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Example 3.1.9. Let f (x) = ex . Find the value of c, when a 3rd approximation is used with
x = 1. (not too close to 0)
With n = 0, Taylor’s theorem is called the “Mean value Theorem”. It is better known when 0
and x are replaced by a and b.
f (1) (c)x1
1!
f (x) − f (0)
f (x) − f (0) = f ′ (c)x ⇒ f ′ (c) =
x
f (x) − P0 (x) =
If x = b, a = 0 then
f ′ (c) =
f (b) − f (a)
b−a
LHS = gradient of tangent to f (x) at x = c
RHS = gradient of secant, the line joining (a, f (a)) and (b, f (b))
So there is a point c ∈ (a, b) such that gradient of tangent at c equal to gradient of secant, i.e.
tangent k secant
Example 3.1.10. At what point on the curve y = x2 is the tangent parallel to the secant
joining the points (−1, 1) and (2, 4)?
3.1.1
Tutorial questions — Higher approximations
D1. Find second and third approximations to the numbers in Calculus Chapter 3 Question ??.
1
(Hint: use the functions (81 + x) 4 , arctan(1 + x), ln(1 − x), arcsin(x + 21 ) near x = 0.)
√
D2. Find a second approximation to 1 + x. By putting x = −(sin2 θ)/l2 , find a second
approximation to the piston position in Calculus Chapter 3 Question ??.
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E3. Find explicitly the error in the (n − 1)th approximation to (1 − x)−1 for x near 0. (Hint:
the (n − 1)th approximation is a geometric series with n terms.) Hence show that the
number c as given in Taylor’s theorem is in this case given by c = 1 − (1 − x)1/(n+1) .
Assuming 0 < x < 1, prove that c lies between 0 and x.
T4. (a) Use Taylor’s theorem to show that if x > 0, then ex is greater than its (n + 1)th
approximation. (Hint: show that the error as given by Taylor’s theorem is positive.)
(b) Hence show that ex /xn → ∞ as x → ∞ (i.e. exponentials beat powers), by dividing
the (n + 1)th approximation by xn and then letting x → ∞.
(c) Put x = k ln t and n = 1 in the limit in (b) and hence prove that tk / ln(t) → ∞ as
t → ∞ (i.e. powers beat logarithms).
D5. (a) Draw a secant joining two points on a smooth curve, and convince yourself that the
secant is parallel to the tangent at some intermediate point.
(b) At what point on the graph y = ln x is the tangent parallel to the secant joining the
points where x = 1 and x = 2?
3.2
Maclaurin Series TC 10.8
As n increases, the approximation improves:
eg. P10 (x) is better than P9 (x)
P100 (x) is better than P99 (x), etc.
As n → ∞, and we get an exact value for f (x) near x = 0
We then get a polynomial of infinite degree, i.e. an infinite series, called Maclaurin Series, after
the Scottish mathematician Colin Maclaurin (1698 - 1746).
The Maclaurin series for F (x) near x = 0 is therefore f (x) =
3.2.1
∞ f (k) (0)
P
xk
k!
k=0
Useful Maclaurin Series
(should try to remember these)
Example 3.2.1. f (x) = ex has f (0) = 1 and f (n) (0) = 1 (seen in previous example)
Example 3.2.2. f (x) = sin x
Example 3.2.3. f (x) = cos x
NOTE: These approximations are for x near 0, the further you are from 0, the worse the approximation gets.
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3.3
The Binomial Series TC 10.10
A special case of the Maclaurin series are the BINOMIAL SERIES, these are the Maclaurin
series for f (x) of the form (1 + x)α , for α ∈ R
Let’s look at various values for α:
• if α = 0 then (1 + x)α = (1 + x)0 = 1 n
P
n
xr , a finite series, using the Binomial Theorem,
• if α = n, n ∈ N, then (1 + x)n =
r
r=0
n!
n
=
where
r
r!(n − r)!
• all other values of α:
Let’s try to find the rth derivative of f (x) = (1 + x)α
α(α − 1)(α − 2) · · · (α − r + 1)
α
α
=1
=
(r terms) with
We introduce the notation
0
r
r!
for α ∈ R, called Binomial coefficients.
α!
α(α − 1)(α − 2) · · · (α − r + 1)
=
as before.
r!
(α − r)!r!
n
= 0 if n, r ∈ N and n < r.
Exercise. Show that
r
(−1)!
−1
since −1 6∈ N
is NOT
Example 3.3.1.
5
(−1 − 5)!5!
(Calculators usually can’t cope with this...)
Note: For α ∈ N,
Example 3.3.2.
1 2
5
1
Example 3.3.3. Find the coefficient of x5 in the expansion of (1 + x) 2
3.3.1
Useful binomial series used in 2nd year to remember
Example 3.3.4. (1 + x)−1
Example 3.3.5. Replace x with −x in previous example to get an expansion for (1 − x)−1
Example 3.3.6. Find the series expansion of (1 + x)−2
Another way: Use the series expansion for (1 + x)−1 and use differentiation to find the series
expansion of (1 + x)−2
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Example 3.3.7. Use the series expansion for (1+x)−1 to find the series expansion for ln |1+x|.
Example 3.3.8. Show that
1 2
r
=
(−1)r−1 (2r − 2)!
, for n ≥ 1
22r−1 r!(r − 1)!
NOTE: The Maclaurin series for ex , sin x, cos x converge for ALL x
ln(1 + x) and (1 + x)α converge for |x| < 1, (1 + x)α converges for ALL x if α ∈ N
Example 3.3.9. (time permitting)
x2
x3
+ α(α − 1)(α − 2) + · · ·
2!
3!
∞
X
α(α − 1) · · · (α − k + 1) k
x
=
k!
k=0
(1 + x)α = 1 + αx + α(α − 1)
(3.2)
(3.3)
Let’s look at the ratio of consecutive terms an and an+1
3.3.2
Tutorial questions — Maclaurin series
T6. Find in sigma notation the full Maclaurin series for ex , sin x, and cos x. These should be
learnt.
D7. Find the terms in the Maclaurin series for tan x, sec x, and arcsin x up to and including
the term in x4 .
D8. Write down the binomial series for (1 + x)−1 and show that it is a geometric series.
By integrating the terms of the series, find in sigma notation the Maclaurin series for
ln(1 + x). (Evaluate the arbitrary constant by considering both sides when x = 0.) This
series should be learnt.
Replace x by x2 in the binomial series and integrate again, and hence find the Maclaurin
series for arctan x.
. (Hint: Write out the left hand side in full, take out − 21
= (− 14 )n 2n
D9. Show that −1/2
n
n
from each of the n factors in the top line, then multiply top and bottom by 2n n!, which
is equal to (2)(4)(6)...(2n − 2)(2n).) Deduce that
(1 − 4x)
−1/2
=
∞ X
2n
n=0
n
xn .
By replacing x by x2 and then integrating, show that
1
arcsin 2x =
2
∞ X
2n x2n+1
n=0
n
2n + 1
.
T10. By considering Maclaurin series, show that the derivative of an odd function is even and
vice versa.
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3.4
Taylor Series TC 10.8
We have looked at series and approximations near x = 0 . We shall now look at series near
x = a. These are infinite series, they are the limit of the nth approximation as n → ∞.
They are called TAYLOR SERIES about x = a.
They are of the form
∞ f (r) (a)
P
(x − a)r .
r!
n=0
This will converge to f (x) for x close to a.
It is simply the Maclaurin series where 0 is replaced by a and x is replaced by x − a.
Example 3.4.1. Find the Taylor series for ln x about x = 1. (can’t be done about x = 0.)
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3.5
Indeterminate Forms and l’Hôpital’s Rule TC 4.5
Limits of the form
∞
0
or
are called INDETERMINATE.
0
∞
For example
sin x
x→0 ln(1 + x)
lim
and
x − sin x
x→0
x2
lim
Theorem 3.5.1. (l’Hôpital’s Rule) Möbius From G. F. l’Hôpital (French 1661-1704)(result
discovered by J. Bernoulli (his teacher))
If
lim f (x) = 0
x→a
and
lim g(x) = 0
x→a
and f and g are sufficiently smooth, then
f (x)
f ′ (x)
= lim ′
.
x→a g(x)
x→a g (x)
lim
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Proof after example:
Example 3.5.2.
cos x
sin x
= lim
= cos 0 = 1
x→0
x→0 x
1
lim
as shown in Chapter 2.
NOTE: The process of l’Hôpital’s Rule may be repeated if we get the
Example 3.5.3. Find limx→0
x − sin x
.
x2
Example 3.5.4. Find limx→0
ln(1 + ax)
.
x
Example 3.5.5. (extra) limx→0
0
case again.
0
x(1 − cos x)
.
x − sin x
Useful trick for limits of the form 1∞ :
Example 3.5.6. Find
1
lim (1 + ax) x
x→0
This is not of the form
0
:
0
check of the form 1∞
Example 3.5.7. Find
πx
lim (x)sec 2
x→1
check of the form 1∞
1
∞
. For, let F (x) = x1 and G(x) = g(x)
,
L’Hôpital’s rule is also valid for limits of the form ∞
then
f (x)
G(x)
=
,
g(x)
F (x)
to which l’Hôpital’s rule can be applied since it is of the form 00 . Suppose limx→a f (x) =
limx→a g(x) = ∞, then
g ′ (x)
G(x)
G′ (x)
g(x)−2 g ′ (x)
f (x) 2
f (x)
lim
= lim
= lim ′
= lim
=
lim
x→a f ′ (x)
x→a F (x)
x→a F (x)
x→a f (x)−2 f ′ (x)
x→a g(x)
x→a g(x)
lim
from which the result follows.
Example 3.5.8. Find limx→0 x ln |x| (of the form 0 · ∞).
Example 3.5.9. Find limx→∞ x ln(1 + x1 ) (of the form ∞ · 0).
Example 3.5.10. Find limx→∞ x2 e−x (of the form ∞ · 0).
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3.5.1
Tutorial questions — Taylor series and indeterminate forms
D11. Find the first five terms (i.e. from n = 0 to 4) in the Taylor series of the following functions
about the given points.
√
(a) ln x about x = 1
(b) x−3 about x = −1
(c) x about x = 4.
T12. If y depends on x, find a second approximation to the small change ∆y caused by a small
change ∆x in x. (Hint: put a = x and x − a = ∆x, and cut off the Taylor series after the
second order terms.)
D13. Evaluate the following limits, using l’Hôpital’s Rule where appropriate.
1 + x − ex
x→0 sec x − 1
x−π
(d) lim
x→π cos 2x
ln(1 + x)
x→0
tan x
1 − cos x
(b) lim
x→0 sin(x2 )
(c)
(a) lim
lim
(e)
(f)
lim (1 + 2x)cosec x
x→0
ln x
.
x→1 sin πx
lim
D14. If money is invested at an interest rate of r% per annum, compounded k times per year,
r
then after n years P rands will amount to P (1 + 100k
)nk rands.
(a) Prove that if k → ∞ (continuous compounding), then this amount approaches
P enr/100 rands. (Hint: put h = k1 and let h → 0.)
(b) Show that the limiting rate corresponds to an effective annual rate (compounded
once per year) of 100(er/100 − 1)%. Look at Building Society advertisements.
D15. If a total mass M of compressible fluid, with density ρekx at depth x, is contained in a
kM
cylindrical flask of radius r, then it was shown that the total depth is k1 ln(1 + ρπr
2 ).
(a) Find the limit of this depth as k → 0 (i.e. as the density tends to ρ).
(b) Find the depth of a mass M of fluid with constant density ρ in the same flask, and
show that it equals the previous limiting depth.
D16. Evaluate lim n(21/n − 1). (Hint: put h = n1 and let h → 0.)
n→∞
3.5.2
Tutorial questions — Integral as the limit of a sum
Here are some additional questions, needing more advanced integration techniques.
Z 1
1
dx in the normal way.
1 + x2
0
(b) Approximate the integral by a sum, using n strips of equal width.
√
P
(c) Deduce from (a) and (b) that limn→∞ nr=1 √r21+n2 = ln(1 + 2).
(This sum cannot be expressed exactly in terms of n. If you can use a computer or
programmable calculator, evaluate the sum for n = 100 or 1000 and compare it with
the right hand side.)
D17. (a) Evaluate
√
1
E18. Use the first approximation to ln x near x = 1 to show
> ln(1 + 1r ) for r > 0. Sum
r
P∞that
this inequality for r = 1 to n, and hence show that r=1 1r diverges.
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R1
D19. (a) Evaluate h ln x dx in the normal way, assuming h > 0. By letting h → 0 (and
R1
remembering that powers beat logs), show that 0 ln x dx = −1.
(b) Approximate the integral by a sum, using n strips of equal width.
(c) Deduce from (a) and (b) that n(n!)−1/n → e as n → ∞. (Hint: take exponentials of
the sum and the integral.)
accurate result is Stirling’s formula, which says
√ A more
n −n
that the ratio of n! and 2πn n e tends to 1 as n → ∞.
(d) Verify Stirling’s formula with a calculator, using the largest value of n! it can cope
with.
3.6
Convergence of Series I, TC 10.2
Convergence
and divergence of series were introduced in MATH1042 Algebra Chapter 3. A
P
series
an is said to converge to the number S if the partial sums a1 + a2 + · · · + aN tend to
S as N → ∞. Roughly speaking, this means that the more terms you add on, the closer you
get to the number S, which is sometimes called the sum to infinity.
Example 3.6.1. Consider the series
N
X
an .
n=1
st
1 partial sum =a1
2nd partial sum =a1 + a2
3rd partial sum =a1 + a2 + a3
..
.
N th partial sum =a1 + a2 + a3 + · · · + aN
Now, if the partial sums tend to S as N → ∞, that is
a1 + a2 + a3 + · · · + aN → S as N → ∞,
then
P
an is said to converge to S and S is called the sum to infinity.
Note:
1010
1. When you are taking a partial sum, even for N 10 , you are omitting infinitely many
terms, i.e. more than you are taking, so convergence is determined by the tail of the series.
2. A partial sum must always be finite, but the sum to infinity might not be.
A series that does not converge is said to diverge. We shall describe some tests to determine
whether or not a series converges. They are important because not many sums to infinity can
be found exactly. We often use a computer to approximate a sum to infinity by the partial
sum for say N = 100 or N = 1000. This is called truncating the series, but it is meaningless
if the series does not converge. Even if the series converges it is valuable to know how fast it
converges, so we can estimate the truncation error, which is the difference between the sum to
infinity and the partial sum.
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Definition 3.6.2. The truncation error for a convergent series is the absolute value of the
difference between the sum to infinity and the partial sum:
S−
N
X
an =
n=1
∞
X
.
n=N +1
Note: In testing for convergence it is only the infinite tail of the series that matters. We can
ignore any number of terms at the beginning, because a finite sum must have a finite value and
therefore cannot affect the overall convergence or divergence, though it obviously does affect
the sum to infinity if it exists.
3.6.1
The divergence test
A necessary condition for convergence is that the nth term, an , must tend to 0, that is
an → 0 as n → ∞.
We illustrate the idea as follows:
The partial sums must approach the value S as N → ∞, not necessarily from below, so
the difference between any two successive partial sums must go to zero. But this difference is
precisely aN .
Theorem 3.6.3. If
∞
X
n=1
converges, then an → ∞.
Note of caution: Theorem 3.6.3 does not say that
possible for a series to diverge when an → 0.
P∞
n=1 an
converges if an → 0. It is
In a more useful form we have the divergence test:
If lim an 6= 0, then
n→∞
∞
X
an is divergent.
n=1
Note: No conclusion about convergence or divergence can be drawn if limn→∞ an = 0.
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Example 3.6.4. Determine whether
∞
X
r
is convergent or divergent.
r+1
r=1
∞
X
1
is convergent or divergent.
Example 3.6.5. Determine whether
ln 1 +
r
r=1
There is another important general result which applies to series with real or complex terms,
see TC 10.5.
P
P
P
Theorem 3.6.6. If
|an | converges, then
an converges and we say
an is absolutely
convergent.
This result seems obvious, since |a1 + a2 + · · · | ≤ |a1 | + |a2 | + · · · , but the proof is rather
subtle and will be omitted.
3.6.2
The ratio test, TC 10.5
The ratio test enables us to deal with fast converging or diverging series. Usually the terms
involve exponentials or factorials.
The simplest examples
are geometric series in which the ratio between successive terms is
P∞ r−1
n
= brbrn−1 = r, a constant. Furthermore, if |r| < 1 the
constant, that is, if r=1 b , then an+1
an
series converges, and if |r| ≥ 1 the series diverges.
The ratio test applies to series that behave like geometric series in that the ratio (in modulus)
between successive terms tends to a constant.
Ratio Test: Suppose
then the series
L = 1.
P
an+1
→ L as n → ∞,
an
an converges if L < 1 and diverges if L > 1. There is no conclusion if
The conclusion is basically the same as for geometric series except that the borderline L = 1
gives no conclusion. For this reason the ratio test is useless unless n appears in an exponent or
factorial.
Note: Since this test works with moduli, it can be applied to series with positive, negative
or non-real terms.
n
P
7π
Example 3.6.7. Determine whether ∞
n
converges or diverges.
n=1
22
Example 3.6.8. Determine whether the following series are convergent or divergent?
1.
∞
X
(−1)n
n=1
2.
∞
X
√
n
n
n=1
Example 3.6.9. Determine those values of x for which
∞
X
r(4x2 )r is convergent.
r=1
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3.6.3
Tutorial questions — Convergence of series I
D20. Use one of the tests above to determine, if possible, whether the following series converge
(C) or diverge (D):
X n
X1
X 2n
(a)
(d)
(f)
2−2n
2n
n
n
X 3n
X √n
X n!
(b)
√
(e)
(g)
n2
1+ n
nn
X (−3)n
(c)
n!
3.7
Convergence of Series II, TC 10.3& 10.4
3.7.1
P -series, TC 10.3
We now deal with series for which the ratio test fails, i.e. for which the limit is 1, which are
series involving powers or logs of n. Unfortunately we are restricted to series of positive terms.
The partial sums cannot oscillate: since the terms are all positive, it follows that adding on
an extra term must always increase the value of the partial sum, so the partial sums are always
increasing. So there are only two possibilities: either the partial sums go off to infinity, or they
approach a limiting value.
Thus we have the following lemma:
P
Lemma 3.7.1. If
an is a series with positive terms, then either it converges, or its partial
sums tend to infinity.
The simplest series of this type are where an is a power of n, say an = n1p where p is constant,
which we call p-series.
Theorem 3.7.2 (The p-series test). If p is constant, then the series
p ≤ 1 and converges if p > 1.
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X 1
np
diverges if
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Figure 3.1
3.7.2
Comparison tests, TC 10.4
The p-series are used as a set of standards by which other series of positive terms can be
compared. There are two tests for comparing series having positive terms, each saying essentially that is such a series converges, then so does any such series with smaller terms, and if it
diverges, then so does any series with larger terms.
Theorem 3.7.3 (Comparison test). If 0 ≤ an ≤ bn for all n sufficiently large, then:
P
bn converges, then
an converges,
P
P
(ii) if
an diverges, then
bn diverges.
(i) if
P
Example 3.7.4. Consider
X
1
n3 + n
Example 3.7.5. Determine whether the following series are convergent or divergent.
1.
2.
∞
X
5
5n − 1
n=1
∞
X
1
n=0
3.7.3
n!
Tutorial questions — Convergence of series II
D21. Use the comparison test to determine whether the following series converge (C) or diverge
(D):
X 1
2n n!
X
1
(b)
n2 + 30
(a)
(c)
(d)
X cos2 n
3
2
X
n
r
n+4
n4 + 4
(e)
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∞
X
n=2
√
1
n−1
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3.8
Tutorial answers
1. (a) 3, 0092164; 3, 0092167
(b) π4 +0, 004975;
(d) π6 − 0, 056773, π6 − 0, 056837.
√
2
4
2
2. 1 + x ≈ 1 + x2 − x8 ; y ≈ cos θ − sin2l θ − sin8l3 θ .
π
+0, 00497508
4
(c) −0, 105;
−0, 1053
n
x
3. Error = 1−x
.
n+1
2
n+2 c
n+1
x
x
e
x
x
4. ex = 1 + x + x2! + · · · + (n+1)!
+ x(n+2)!
> (n+1)!
, so xen > (n+1)!
, which → ∞ as x → ∞.
5. (b) ( ln12 , − ln(ln 2)).
P
P∞
P∞
xn
n x2n+1
n x2n
6. ex = ∞
n=0 n! , sin x =
n=0 (−1) (2n+1)! , cos x =
n=0 (−1) (2n)! .
3
2
4
3
7. tan x = x + x3 + · · · , sec x = 1 + x2 + 5x
+ · · · , arcsin x = x + x6 + · · · .
24
P∞
P
n x2n+1
n+1 xn
,
arctan
x
=
8. ln x = ∞
(−1)
n=0 (−1) 2n+1 .
n=1
n
9. Omitted.
P
P
10. If f (x) =
an xn , then f ′ (x) =
nan xn−1 . If all the values of n (with non-sero coefficients) are odd, then all the values of n − 1 are even, and vice versa.
3
2
4
+ (x−1)
− (x−1)
+ ···
11. (a) (x − 1) − (x−1)
2
3
4
(b) −1 − 3(x + 1) − 6(x + 1)2 − 10(x + 1)3 − 15(x + 1)4 + · · ·
2
3
4
(d) 0
(e) e2
− (x−4)
+ (x−4)
− 5(x−4)
+ ···
(c) 2 + (x−4)
4
64
29
214
12. ∆y ≈ y ′ ∆x + 21 y ′′ ∆x2 .
13. (a) 1
(b) 12
(c) −1
(f) − π1 (use Taylor series about x = 1).
rate
14. effective
= er/100 − 1.
100
M
15. Limit = ρπr
2 = depth.
16. ln 2.
17. (a) ln(1 +
√
2).
P
18. x − 1 > ln x, so 1r > ln(1 + 1r ), and nr=1 1r > ln(n + 1) (telescoping series).
P
19. (b) n1 nr=1 ln nr = n1 ln(n!) − ln n. (c) n1 ln(n!) − ln n → −1. Change the sign, and take
exponentials.
20. (a) C
clusion
(b) D
(g) C
(c) C
(d) no conclusion (at this stage)
21. (a) C
(b) C
(c) C
(d) C
(e) D
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(e) D
(f) no con-
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Chapter 4
PARTIAL DIFFERENTIATION
We will look at functions of the form z = f (x, y), functions of more than one variable.
Example 4.0.1. Examples you may have encountered already in first year engineering are:
1. P = f (x, y), P = pressure, depending on position.
2. T = f (x, y), T = temperature, depending on position.
3. T = f (x, t), T = temperature along a copper wire, depending on its position x at time t.
z = f (x, y) gives height at various positions in a 3D model. These can be plotted on contour
maps, where all points with the same height are plotted on a curve.
4.1
Quadric Surfaces TC 12.6
Recall from Algebra (chapter 5) we looked at the following curves:
x2 y 2
x2 y 2
Parabola y 2 = 4ax, Ellipse 2 + 2 = 1 and Hyperbola 2 − 2 = 1
a
b
a
b
Quadric surfaces z = Q(x, y) have an equation of the form:
z = Ax2 + Bxy + Cy 2 + Dx + Ey + F
By rotating the axes and slightly shifting the origin we get z = λu2 + µv 2 , called the canonical
form.
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There are 5 possibilities:
• λ > 0, µ > 0 (both positive)
Let λ =
1
u2 v 2
1
,
µ
=
,
then
z
=
+ 2.
a2
b2
a2
b
For each z = k, where k is constant we get
u2 v 2
+ 2 = k which is an ellipse.
a2
b
This shape is called an elliptic paraboloid, or “CUP”.
c2 v 2
c2
v2
When u is constant, say u = c, then z = 2 + 2 , i.e. z − 2 = 2 is a parabola. Similarly
a
b
a
b
for when v is constant.
Vertical cross sections in both the u and v directions are parabola ⌣ in shape.
• λ < 0, µ < 0 (both negative)
1
1
u2 v 2
Let λ = − 2 , µ = − 2 , then z = − 2 − 2 .
a
b
a
b
z = k ellipse elliptic paraboloid, or “CAP”
u = k parabola.
v = k parabola.
Vertical cross sections in both the u and v directions are parabola ⌢
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• λ > 0, µ < 0
Let λ =
1
u2 v 2
1
,
µ
=
−
,
then
z
=
− 2.
a2
b2
a2
b
For each z = k, we get
u2 v 2
− 2 = k which is a hyperbola.
a2
b
k2 v2
− 2 is a parabola ⌢
a2
b
u2 k 2
Similarly for when v is constant z = 2 − 2 is a parabola ⌣
a
b
When u is constant, say u = k, then z =
This shape is called an hyperbolic paraboloid, or ”SADDLE”.
• λ > 0, µ = 0 [or λ = 0, µ > 0] (one positive, one zero)
Let λ =
u2
1
,
then
z
=
only.
a2
a2
√
For z constant, we get u2 = za2 , u = ±a k, i.e. 2 lines.
For v constant, z =
u2
a parabola ⌣
a2
This shape is a parabolic cylinder.
If λ = 0, µ > 0
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• λ < 0, µ = 0 [or λ = 0, µ < 0] (one negative, one zero)
u2
1
Let λ = − 2 , then z = − 2 .
a
a
Also a parabolic cylinder.
Example 4.1.1. Identify:
1. z = 2x2 + 4x + y 2 − 2y + 3
2. z = −x2 + 2x + 2y 2 + 8y
4.1.1
Tutorial questions — Quadric surfaces
D1. Identify the surfaces below by completing the square and shifting the origin.
(a) z = −x2 − y 2 + 2x − 4y (b) z = x2 − 3y 2 − 4x + 6y (c) z = x2 + 2y 2 − 2x + 4y.
Identify the surfaces below by rotating the axes. (Use your working, or the solutions,
from Algebra Chapter 5 Question ??).
√
(e) z = 17x2 + 12xy + 8y 2 (f) z = 4x2 + 12xy + 9y 2 .
(d) z = x2 − 2 3xy − y 2
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4.2
Partial Differentiation TC 14.3
We look at slope of tangents to surfaces (and not curves). We look at sections of the surface,
where we get curves. We differentiate as before keeping all other variables constant. We call
∂z
∂z
or
if z = f (x, y). zx and zy , or fx and fy can also
these partial derivatives, denoted by
∂x
∂y
be used.
Definition 4.2.1. If z = f (x, y) then zx =
∂f
f (x + ∆x, y) − f (x, y)
(x, y) = fx (x, y) = lim
.
∆x→0
∂x
∆x
Differentiate as usual, keeping y constant.
Similarly zy =
f (x, y + ∆y) − f (x, y)
∂f
(x, y) = fy (x, y) = lim
∆y→0
∂y
∆y
Example 4.2.2. z = 3(x − 2)2 + (y + 1)2 = 3x2 − 12x + 12 + y 2 + 2y + 1
Example 4.2.3. z = x2 y sin(x + y 2 )
Example 4.2.4. z =
4.3
x
x2 + y 2
Implicit Differentiation TC 14.3
Example 4.3.1. x tan(y + z) = yz
Example 4.3.2. (in handout): Let x = uv 2 , y = u2 v, where u, v are functions of x. Differentiate each equation with respect to x and solve for ux and vx , using Cramer’s rule.
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4.4
Second Derivatives TC 14.3
∂ ∂z
∂ 2z
=
= zxx
∂x2
∂x ∂x ∂ 2z
∂ ∂z
=
= zyy
2
∂y
∂y ∂y

∂
∂z
∂ 2z

=
= zxy 



∂y∂x
∂y
∂x


Mixed partial
⇑
⇑
⇑⇑
derivatives
2nd 1st 1st 2nd 


2

∂
∂z
∂ z

= zyx 
=

∂x∂y
∂x
∂y
We will assume zxy = zyx .
—————————————————————————
Example 4.4.1. If z = e2x cos 3y, show that zxy = zyx and that 9zxx + 4zyy = 0.
Example 4.4.2. (in handout): If z = sin(xy 2 ), find zxx , zxy , zyx , zyy . Verify that zxy = zyx .
4.4.1
Tutorial questions — Partial differentiation
∂z
∂z
and ∂y
if
D2. Find ∂x
(a) z = x arctan(x + y 2 )
(b) z sin z = x2 arcsin 2y (use implicit differentiation)
(c) ex+y+2z = xyz (take logs, then differentiate implicitly).
D3. For each of the functions z below, find zxx , zxy , zyx , zyy , and verify that zxy = zyx .
(a) z = cosh x cos y
(b) z = ln(xy 2 )
(c) z = exy
(d) z = arctan( xy ).
(The hyperbolic cosine function cosh is defined in Algebra Chapter 5.)
D4. If x = r cos θ and y = r sin θ, prove that
∂y 2
∂x 2
∂y 2
∂x 2
+
= 1 and
+
= r2 .
∂r
∂r
∂θ
∂θ
D5. In each of the examples below, u and v are functions of x and y satisfying two equations.
∂v
Find ∂u
and ∂x
by implicitly differentiating both equations partially with respect to x and
∂x
then solving for ux and vx . (There are two equations in two unknowns — use Cramer’s
Rule.) Then repeat, replacing x by y.
(c) eu+v = cos(x+y) and ln(uv) = sin(xy).
(a) u2 − v 2 = xy and uv = x2 + y 2
(b) u cos v = yex and u sin v = xey
∂v
∂v
= ∂y
and ∂u
= − ∂x
(these are
T6. Suppose u and v are functions of x and y such that ∂u
∂x
∂y
called the Cauchy-Riemann equations).
(a) Using the fact that the mixed partial derivatives are equal, show that uxx + uyy = 0
and vxx + vyy = 0. (These are examples of Laplace’s equation).
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(b) Suppose ρ and α are the polar co-ordinates of the point with cartesian co-ordinates
(ux , vx ), i.e. suppose ux = ρ cos α and vx = ρ sin α. Use the Cauchy-Riemann
equations to express uy and vy in terms of ρ and α, and hence show that the matrix
1
ux uy
p
u2x + vx2 vx vy
is a rotation matrix (see Algebra Chapter 4 Question ??).
D7. Show that the following pairs of functions u and v satisfy the Cauchy-Riemann equations
ux = vy and uy = −vx .
4.5
y
v = arctan( )
x
−y
v= 2
.
x + y2
(a) u = ex cos y,
v = ex sin y
(c) u = 21 ln(x2 + y 2 ),
(b) u = x2 − y 2 ,
v = 2xy
(d) u =
x
x2 + y 2
,
Partial Derivative Chain Rule TC 14.4
Theorem 4.5.1. Suppose z is a function of x and y and both x and y are functions of u.
Then
dz
∂z dx ∂z dy
=
+
du
∂x du ∂y du
If z is a function of x and y and x and y are both functions of u and v.
∂z
∂z ∂x ∂z ∂y
=
+
∂u
∂x ∂u ∂y ∂u
∂z ∂x ∂z ∂y
∂z
=
+
Or
∂v
∂x ∂v ∂y ∂v
Then
———————————————————————————————–
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Example 4.5.2. If z = x2 + y 2 , and x = tan 2u, y = sec 3u, find
dz
.
du
Example 4.5.3. If z = xy 2 , and x = uv + 2u, y = u2 − 3v, find
∂z
.
∂v
———————————————————————————————–
Example 4.5.4. (extra) Use the chain rule to rederive the product rule. Suppose z = uv where
u and v are functions of x.
———————————————————————————————–
Example 4.5.5. Repeat for the Quotient rule (extra).
z=
u
v
———————————————————————————————–
Example 4.5.6. z = ex sin y , x = sec u, y = ln u, find
dz
.
du
————————————————
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Example 4.5.7. The volume of a cone is increasing at a rate of π cm3 / sec. The radius is
increasing at a rate of 1 cm/ sec. Find the rate of change of the height when r = 1cm and
h = 2cm.
V = 31 πr2 h,
dr
dV
= π,
=1
dt
dt
dh
=? When r = 1, h = 2
dt
4.5.1
Tutorial questions — Chain rule for partial derivatives
D8. If z = cos(xey ), where x = tan 2t and y = ln 3t, find dz
in two different ways:
dt
(a) by using the chain rule for partial derivatives
(b) by substituting for x and y and then differentiating with respect to t.
Verify that your answers agree.
D9. A rectangle has vertices (0, 0), (x, 0), (0, y), and (x, y) where x = 2 cos t and y = sin t for
0 ≤ t ≤ π2 .
(a) Express its area A and perimeter P in terms of x and y.
(b) Use partial differentiation to find the rate of change of A and P with respect to t
when t = π3 .
(c) Find the maximum values of A and P . For what values of t do they occur?
(Note that this problem can be done without partial differentiation.)
D10. If z is a function of x and y, where x = r cos θ and y = r sin θ, show that
1 ∂z 2
∂z 2
∂z 2
∂z 2
+ 2
=
+
.
∂r
r ∂θ
∂x
∂y
and
D11. (a) In the belt problem (Calculus Chapter 1 Question ??), in which cos θ = R−r
x
∂θ
∂L
L = 2{x sin θ + rθ + R(π − θ)}, show that ∂x = 2{sin θ + (x cos θ + r − R) ∂x }. (Hint:
R and r are kept constant, but θ depends on x.)
= 2 sin θ. (Hint: substitute for cos θ.)
(b) Deduce that ∂L
∂x
∂L
(c) Similarly show that ∂R
= 2(π − θ).
E12. If z is a function of x and y, which are both functions of u and v, show by applying the
∂z
(not z itself) that
chain rule to ∂x
∂ ∂z ∂ 2 z ∂x
∂ 2 z ∂y
=
+
= zxx xu + zxy yu .
∂u ∂x
∂x2 ∂u ∂y∂x ∂u
∂
∂
∂
Obtain similar formulae for ∂v
(zx ), ∂u
(zy ), and ∂v
(zy ).
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4.6
Differentials and First Approximations
We saw that if z is a function of x and y and these are both functions of u only, then
∂z dx ∂z dy
dz
=
+
du
∂x du ∂y du
⇒ dz = zx dx + zy dy
gives the differential dz.
∆f = ∆z ≈ zx ∆x + zy ∆y
We have the similar approach as in Chapter 3. We can get a 1st approximation to the change
in z coming from small changes in x and y.
∆f = f (x + ∆x, y + ∆y) − f (x, y) ∴ f (x + ∆x, y + ∆y) = f (x, y) + ∆f.
So,
f (x + ∆x, y + ∆y) ≈ f (x, y) + (zx ∆x + zy ∆y)
where x and y are values where we can evaluate the function such that ∆x and ∆y are
small.
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Example 4.6.1. Estimate
p
Let z = x2 + y 2 = f (x, y)
4.6.1
p
4, 022 + 2, 992
Tutorial questions — Differentials and first approximations
D13. Find first approximations to the expressions below, and compare them with accurate
values
obtained by using your calculator.
p
3
(b) {(1, 98)6 + (2, 01)4 + 1}1/4
(c) ln(2, 02 − e0,03 ).
(a) (5, 01)2 + 1, 97
D14. Use the results of Question 11 to find the change in the length of the fan belt if x
decreases by 2, 0 mm and R increases by 1, 0 mm, if the original values were R = 100 mm,
r = 30 mm, and x = 140 mm.
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4.7
Inverse Functions
Theorem 4.7.1. Suppose x and y are functions of u and v and vice versa then the matrices
xu xv
ux uy
and
yu yv
vx vy
are inverses of each other.
∂x
∂u
6= ( )−1 (y and v a constant).
∂x
∂u
—————————————————————————Note:
Example 4.7.2. If u = x + y, v = 4xy, find ux , uy , vx , vy and then find xu , xv , yu , yv in terms
of x and y.
———————————————————————–
Example 4.7.3. (in Handout)
Let x = sin u cos v and y = sin u + cos v, find ux .
————————————————————————
4.7.1
Tutorial questions — Inverse functions
D15. For each of the pairs of functions below,
(i) find the partial derivatives xu , xv , yu , yv ,
(ii) express u and v in terms of x and y,
(iii) find the partial derivatives ux , uy , vx , vy ,
(iv) check that the product of the matrices
matrix.
xu xv
yu yv
and
ux uy
vx vy
is the identity
(Hint for (d): x2 + y 2 = 1/(u2 + v 2 ).)
√
√
(a) x = u + v, y = u − v
(b) x = au + bv,
u
(c) x = e cos v,
y = cu + dv
y = eu sin v
(d) x = u/(u2 + v 2 ),
y = −v/(u2 + v 2 ).
D16. Suppose x = u sec v and y = u tan v, where u > 0 and − π2 < v < π2 .
(a) Differentiatex and ypartially with respect to u and v, and hence write down the
xu xv
matrix A =
.
yu yv
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(b) Use the adjoint-determinant method to find A−1 .
p
(c) Show that u = x2 − y 2 and v = arcsin( xy ).
(d) Differentiate u and v partially with respect to x and y, and verify that
A−1 , as found in part (b).
ux uy
vx vy
=
E17. If u and v satisfy the Cauchy-Riemann equations with respect to x and y (see Question 6
above), use the fact that the matrices of partial derivatives are inverses of each other to
show that x and y satisfy the Cauchy-Riemann equations with respect to u and v.
4.8
Tutorial answers
1. (a) Circular paraboloid
(b) Hyperbolic paraboloid
(c) Elliptic paraboloid
perbolic paraboloid
(e) Elliptic paraboloid
(f) Parabolic cylinder.
2. (a)
∂z
x
= arctan(x + y 2 ) + 1+(x+y
2 )2
∂x
(b)
2x arcsin 2y
∂z
= sin
∂x
z+z cos z
(c)
z(1−x)
∂z
= x(2z−1)
∂x
(c) zxx = y 2 exy ,
2xy
∂z
= 1+(x+y
2 )2
∂y
∂z
2x2
=√
∂y
1−4y 2 (sin z+z cos z)
z(1−y)
∂z
= y(2z−1)
.
∂y
3. (a) zxx = cosh x cos y,
(b) zxx = − x12 ,
(d) Hy-
zyy = − cosh x cos y,
zyy = − y22 ,
zyy = x2 exy ,
(d) zxx = (x22xy
,
+y 2 )2
zxy = 0,
zxy = − sinh x sin y,
zxx + zyy 6= 0
zxy = exy (xy + 1),
zyy = (x−2xy
2 +y 2 )2 ,
zxx + zyy = 0
y 2 −x2
zxy = (x2 +y2 )2 ,
zxx + zyy 6= 0
zxx + zyy = 0.
4. Omitted.
4yv+ux
4ux−vy
4uy−vx
4xv+uy
5. (a) ux = 2(u
2 +v 2 ) , uy = 2(u2 +v 2 ) , vx = 2(u2 +v 2 ) , vy = 2(u2 +v 2 ) .
(b) ux = ex y cos v + ey sin v, uy = ex cos v + xey sin v,
vx = u1 {−ex y sin v + ey cos v}, vy = u1 {−ex sin v + xey cos v}.
u
u
{e−u−v sin(x+y)+vy cos(xy)}, uy = − u−v
{e−u−v sin(x+y)+vx cos(xy)},
(c) ux = − u−v
v
v
−u−v
−u−v
vx = u−v {e
sin(x + y) + uy cos(xy)}, vy = u−v {e
sin(x + y) + ux cos(xy)}.
6. Omitted.
7. Omitted.
= sin(xey )ey {−2 sec2 2t − xt }
(b) dz
= −3(2t sec2 2t + tan 2t) sin(3t tan 2t).
8. (a) dz
dt
dt
√
√
dP
π
π
9. dA
=
−1
and
=
1
−
2
.
Maximum
A
=
1
when
t
=
.
Maximum
P
=
2
3
at
t
=
5
dt
dt
3
4
when t = arctan 21 .
10. Omitted.
∂θ
1
11. ∂x
= x tan
,
θ
∂θ
1
= − x sin
.
∂R
θ
∂
zx = zxx xv + zxy yv ,
12. ∂v
∂
z = zxy xu + zyy yu ,
∂u y
∂
z = zxy xv + zyy yv .
∂v y
13. (a) z + ∆z ≈ 3, 0025926 (true value 3, 002594 . . . ).
(b) z + ∆z ≈ 2, 9674 (true value
2, 96779 . . . ).
(c) z + ∆z ≈ −0, 01 (true value −0, 0105 . . . ).
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√
mm.
14. ∆L ≈ 2(π − θ)∆R + 2 sin θ∆x; ∆L ≈ −2 3 + 4π
3
ux uy
y y
x y
xu xv
1
1
1
2
2
2
2
15. (a) u = 2 (x + y ), v = 2 (x − y ),
,
= 2xy
.
=
vx vy
yu yv
x −x
x −y
d −b
xu xv
ux uy
a b
ay−cx
dx−by
1
.
,
=
= ad−bc
(b) u = ad−bc , v = ad−bc ,
−c a
yu yv
vx vy
c d
y
1
2
2
,
(c) u
x = 2 ln(x
+y ), v =arctan
xu xv
ux uy
x −y
x y
1
,
=
.
= x2 +y2
yu yv
y x
vx vy
−y x
−y
x
(d) u = x2 +y
2 , v = x2 +y 2 ,
2
2
ux uy
v − u2
2uv
xu xv
v − u2 −2uv
1
,
.
=
= (u2 +v2 )2
2uv
v 2 − u2
vx vy
−2uv v 2 − u2
yu yv
xu xv
ux uy
sec v u sec v tan v
u sec v −u tan v
1
16. (a)
.
(b)
=
.
=u
− sin v
1
vx vy
yu yv
tan v
u sec2 v
17. Omitted.
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Chapter 5
DIFFERENTIAL EQUATIONS E & P
Ch1
5.1
Introduction E&P 1.1 & 1.2
A differential equation (d.e. for short) is any equation that involves derivatives or differentials
(ordinary or partials).
These occur frequently in Science and engineering.
The order of a d.e. is the order of the highest derivative.
Here are a few examples:
dN
= −kN, radioactive decay, order 1, ordinary.
dt
d2 x
= −ωx, simple harmonic motion, mechanics, order 2, ordinary.
dt2
(x2 + y)dy − (sin x)dx = 1, order 1, ordinary.
∂ 2u
∂u
k 2 =
, the heat equation, order 2, partial.
∂x
∂t
dy
+ sin ty = t, order 1, ordinary.
dt
We are interested in solving a d.e., this means finding the function that satisfies the d.e..
Example 5.1.1. Solve for y the d.e.
d2 y dy
−
− 2y = 0.
dt2
dt
In this chapter, we will learn how to find y.
For now, we shall just check that y = 3e−t + 4e2t is a solution.
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In fact, more generally, we can check that y = Ae−t + Be2t is also a solution, where A and
B are two arbitrary constants. Check this for yourself!
d2 y dy
−
− 2y = 0 has general solution
dt2
dt
y = Ae−t + Be2t . It has 2 arbitrary constants. Whereas y = 3e−t + 4e2t is a particular solution,
where A = 3 and B = 4.
In the above example, we say the second order d.e.
The order of the d.e. corresponds to the number of arbitrary constants in the general
solution.
dn y
If the d.e. is of the form n = f (x), the general solution y can easily be found by integrating
dx
n times.
Example 5.1.2. Solve (at this stage it means “find the general solution”)
d2 y
= cos x.
dx2
Now, if we are given some extra information (often called boundary conditions, or initial
conditions) we can use this information to find the arbitrary constants. This will give the
particular solution.
d2 y
Example 5.1.3. Find the particular solution to the above d.e.
= cos x, given that y = 0
dx2
dy
= 0 when x = 0.
and
dx
In this chapter, we shall concentrate on solving first order ordinary differential equations.
We shall consider 4 types namely: VARIABLE SEPARABLE, HOMOGENEOUS, EXACT and
LINEAR.
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5.2
Variable Separable E & P 1.4
A d.e. is said to be variable separable if it can be written in the form
M (x)dx = N (y)dy.
To solve this we integrate the left hand side with respect to x and the right hand side with
respect to y. Note, the d.e. may first have to be put in the required form, you first have to
separate the variables.
Example 5.2.1. Solve
sin x
dy
=
.
dx
cos y
Example 5.2.2. Solve
dy
= 2xy.
dx
Example 5.2.3. Solve xy
dy
= 1 − x2 .
dx
dy
−1
= (3x + 2)−2 . Solution: y = 3(3x+2)
+ c.
Example 5.2.4. Homework: Solve dx
Example 5.2.5. Homework: Find the particular solution to
1
x = 0 then y = 1. Solution: y = 1−x
.
dy
= y 2 given that y(0) = 1 or if
dx
Example 5.2.6. Find the particular solution of the d.e.
(1 + 4x2 )dy = (1 − y 2 )dx
given that y = 3 when x = 1/2.
Example 5.2.7. Solve y(x + 1)
dy
= x2 + 2.
dx
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Example 5.2.8. The rate of decay of a radioactive substance is proportional to the quantity
Q which remains. Initially, Q = 2g. If it takes 8 hours for the quantity to reduce to 50% (called
the half life), how long will it take for the quantity to reduce from 2g to 0.1g?
dQ
= kQ, where k < 0 because of decay.
dt
Solution: Proportional, hence
Example 5.2.9. Homework: Solve y − x
5.2.1
dy
dy
Ax
= 1 + x2 . Solution: y = 1 + 1+x
.
dx
dx
Tutorial questions — Variables separable
D1. Find the general solution of the differential equations
√
dy
= sec2 x tan y
(c) y 1 − y dy = arcsin x dx
(a) dx
(d) ex+y dy = xy −1 dx.
(b) (1 + 3x2 ) dy = (1 − 3y 2 ) dx
Find the particular solution of (b) such that y =
√
3 when x = 1.
= kb y(b − y) and verify that the general solution is
D2. Solve the differential equation dy
dt
the logistic curve y = b/(1 + Re−kt ), where R is the arbitrary constant. (See Calculus
Chapter 5 Question ??.) Show that the solution can be rewritten in the form y = 21 b{1 +
sinh θ
tanh( 21 kt − α)}, where α = 12 ln R and the hyperbolic tangent is defined by tanh θ = cosh
θ
(see Algebra Chapter 5).
5.3
Homogeneous E & P 1.6 (p62)
A differential equation is said to be homogeneous if it is of the form
M (x, y)dx + N (x, y)dy = 0,
where M (x, y) and N (x, y) are functions in x and y, with the SAME total degree. i.e. the total
sum of the exponents of x and y for every term is the same.
5
Example 5.3.1. (y 2 x2 + xy 3 + xy )dx − (x4 + y 4 + x3/2 y 5/2 )dy = 0 has the same total degree 4.
3
Example 5.3.2. (y 2 + xy + 1)dx = (x2 + y + xy )dy has not, because of 1 and y.
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5.3.1
Solving a homogeneous d.e.
A homogeneous differential equation can be written in the form
M (x, y)
dy
= f (y/x) =
,
dx
N (x, y)
or if we put y = vx then
dy
can be written as a function of v only
dx
dy
= f (v).
dx
Total degree k =⇒ for everyxm y n , m + n = k, so xm (vx)n = xm+n v n = xk v n
Therefore, every term will have xk in the numerator and denominator and it will cancel, leaving
a function in v only.
Example 5.3.3. Let (4x + 5y)dx + (x − 4y)dy = 0. The total degree is 1 everywhere.
Example 5.3.4. (in handout): Find the particular solution of (x2 + 3y 2 )dx + (3x2 + y 2 )dy = 0,
given that y = 1 when x = 0.
Example 5.3.5. Solve (x +
p
dy
= y.
y 2 − xy) dx
Example 5.3.6. Homework: Solve (−2y 3/2 + 3x3/2 )dx + x(x1/2 + 6y 1/2 )dy = 0.
Solution: 4y 3/2 + yx1/2 + 3x3/2 = Ax1/2 , where y = vx.
Example 5.3.7. Homework in handout: Solve (x + y + xey/x )dx − xdy = 0.
Solution: x(1 + e−y/x ) = A with y = vx.
R dv
−v
.
Hint: When you reach 1+e
v , multiply top and bottom by e
5.3.2
Tutorial questions — Homogeneous equations
D3. Find the general solution of the differential equations
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(d) (x2 − y 2 ) dx + x(x + 2y) dy = 0
(a) (3x + 4y) dx + y dy = 0
(b) 2y 2 dx + (x2 + y 2 ) dy = 0
E(e) ln(1 +
(c) 2y 2 dx + (2x2 + y 2 ) dy = 0
y2
y
) dx − 2 arctan( ) dy = 0.
2
x
x
Find the particular solution of (b) such that y = 1 when x = 1.
5.4
Exact E & P 1.6 (p68)
5.4.1
Total differential
In order to fully understand EXACT differential equations we first need to look at the partial
chain rule done in Chapter 9.
Suppose f is a function of 2 variables x and y, where x and y are themselves functions of
u. Then by the partial chain rule we saw that
∂f dx ∂f dy
df
=
+
.
du
∂x du ∂y du
If we cancel the differential du we obtain
df =
∂f
∂f
dx +
dy,
∂x
∂y
which is called the total differential of f .
5.4.2
Definition of an exact d.e.
A d.e. is said to be exact if it is of the form
M (x, y)dx + N (x, y)dy = 0
and if there is a function f (x, y) such that the left hand side of this d.e. is the total differential
of f . i.e.,
M (x, y)dx + N (x, y)dy = df
∂f
∂f
M (x, y)dx + N (x, y)dy =
dx +
dy.
∂x
∂y
Equating coefficients we get
(
M (x, y) = fx
N (x, y) = fy .
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Now if we differentiate M (x, y) partially with respect to y and N (x, y) with respect to x,
we obtain:
My (x, y) = fxy
and
Nx (x, y) = fyx
and if we use the fact that the mixed derivatives fxy and fyx are equal we get
My (x, y) = Nx (x, y),
which is called the TEST FOR EXACTNESS. This checks whether the d.e. is exact or not.
This must be done before attempting to solve the d.e.
df
= 0 then f (x, y) must be a constant. So the final general solution is
Finally, if df = 0 or du
of the form f (x, y) = c.
5.4.3
Solving an exact d.e.
We have all the information on how to solve an exact d.e. from the previous subsection.
• We first make sure the d.e. is of the form M (x, y)dx + N (x, y)dy = 0.
• We state what M (x, y) and N (x, y) are equal to.
• We perform the test for exactness and check that My (x, y) = Nx (x, y), if it is then we
process, otherwise we stop and try to relable the d.e.(separable, homogeneous or linear).
• Our aim is now to find f (x, y).
• We write M (x, y) = fx and integrate with respect to x to get f , treating y as a constant.
The constant of integration will be some function of y, say ψ1 (y).
• We do the same thing for N (x, y) = fy . We integrate with respect to y, treating x as a
constant. The constant of integration will be some function of x, say ψ2 (x).
• We end up with two versions of f (x, y), we amalgamate them, making sure equal terms
are not repeated.
• Once f (x, y) is found, the final general solution is just f (x, y) = c.
Example 5.4.1. Solve (6x + 9y + 11)dx + (9x − 4y + 3)dy = 0.
Example 5.4.2. Solve (x2 + y 2 )dx + (2xy + cos y)dy = 0.
Example 5.4.3. in handout: Solve (sin y + yex )dx + (x cos y + ex )dy = 0.
Example 5.4.4. Homework in handout: Solve (12x − 4y + 2)dx + (18y − 4x + 1)dy = 0.
Solution: 6x2 − 4xy + 2x + 9y 2 + y = c.
Example 5.4.5. Solve (tan y + y sec x tan x)dx + (x sec2 y + sec x)dy = 0.
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5.4.4
Tutorial questions — Exact equations
D4. First test for exactness and then find the general solution of the differential equations
below.
(a) xy{2 sin(x + y) + x cos(x + y)} dx + x2 {sin(x + y) + y cos(x + y)} dy = 0
(b) (1 + yexy ) dx + (2y + xexy ) dy = 0
(c) (2x + 3y + 1) dx + (3x + 4y − 1) dy = 0
y2
y
(d) ln(1 + 2 ) dx − 2 arctan( ) dy = 0.
x
x
Find the particular solution of (b) such that y = 1 when x = 0.
D5. If ux = vy and uy = −vx (Cauchy-Riemann equations), show that the differential equations u dx − v dy = 0 and v dx + u dy = 0 are exact.
5.5
Linear E & P 1.5
A d.e. is said to be LINEAR if it can be written in the form:
dy
+ P (x)y = Q(x),
dx
where P (x) and Q(x) are functions of x only. Note there is only one term in y only, no y 2 etc...
The form may be disguised, for example (P (x)y − Q(x))dx + dy = 0 is a linear d.e..
dy
+ 2y cot x = cos x is linear where
Example 5.5.1. The equation dx
P (x) = 2 cot x and Q(x) = cos x.
5.5.1
Solving a linear d.e.
We aim to transform the linear d.e. into a d.e. that can easily be solved. We first find an
expression called the integrating factor, denoted by µ. It is, for a reason that will be explained
R
µ = e P (x)dx .
Note, that when integrating P (x) no constant is added, this constant gets cancelled later on in
the process. Check it for yourself!
Take the original linear d.e. and multiply each term by µ to obtain
R
e P (x)dx
dy
+ P (x)y = Q(x)
dx
dy
+ µ P (x)y = µ Q(x)
µ
dx
R
R
dy
+ e P (x)dx P (x)y = e P (x)dx Q(x).
dx
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Now, let’s look at the left hand side. This is in fact the derivative of a product, by the product
rule we have
R
d R P (x)dx dy R P (x)dx
=
+ ye P (x)dx P (x).
ye
e
dx
dx
Thus
R
d R P (x)dx ye
= e P (x)dx Q(x),
dx
or
d
(yµ) = µ Q(x).
dx
We integrate with respect to x:
ye
R
R
=
Z
e P (x)dx Q(x) dx,
yµ =
Z
µ Q(x) dx.
P (x)dx
or
Do not cancel µ, it is not a constant that can go outside the integral!.
Finally we solve for y.
dy
+ 2y cot x = cos x.
Example 5.5.2. Solve dx
dy
Example 5.5.3. Solve dx
+ 2xy = x3 .
dy
− y tan x = tan x + sec x.
Example 5.5.4. in handout: Solve dx
dy
+ y = ln x, given that x = 1 and y = 10.
Example 5.5.5. Solve (x + 1) dx
y
dy
+ √1−x
Example 5.5.6. Solve dx
2 = 10x, given that y = 0 when x = 0.
5.5.2
Tutorial questions — Linear equations
D6. Find the general solution of the differential equations
y
sin x
dy
+2 =
dx
x
x
dy
− y sec x = ex cos x
(b)
dx
(a)
dy
4y
+ 2
= x2 + 1
dx x − 1
dy
y
(d)
+√
= 2x
dx
1 − x2
(c)
(e)
(f)
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dy
− 2xy = x3
dx
dy
xy
= 1.
−
dx 1 + x2
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Find the particular solution of (b) such that y = 1 when x = 0.
D7. The differential equation y ′ + P (x)y = Q(x)y n is called a Bernoulli equation. Show that it
has an integrating factor of the form y −n µ(x). (Hint: multiply through by y −n µ(x), then
use the exactness condition to solve for µ(x).) Write down the solution of the equation
in terms of µ(x).
5.6
Tutorial answers
√
√
√3y = arctan( 3x) + c
1. (a) ln | sin y| = tan x + c or sin y = Aetan x
(b) 21 ln 1+
1− 3y
√
(c) 52 (1 − y)5/2 − 23 (1 − y)3/2 = x arcsin x + 1 − x2 + c
(d) ey (y − 1) = −e−x (x + 1) + c.
√
√
√3y = arctan( 3x) + 1 ln 2 − π .
Particular: 21 ln 1+
2
3
1− 3y
kt
Abe
b
2. y = 1+Abe
kt or y = 1+Ce−kt .
2y
−2x
+ A(= x+y
+ B)
3. (a) (3x + y)3 = A(x + y)
(b) ln |y| = x+y
x+y
2x+y
(c) ln |y| = 2 arctan( x ) + c(= −2 arctan y + d).
(d) y 2 + xy + x2 = Ax
(e) x ln[1 + ( xy )2 ] − 2y arctan xy = k.
−2x
Particular: ln |y| = x+y
+ 1.
4. (a) ∂M
= ∂N
= x(2 − xy) sin(x + y) + x(x + 2y) cos(x + y). Solution x2 y sin(x + y) = k.
∂y
∂x
= ∂N
= exy (1 + xy). Solution x + exy + y 2 = k.
(b) ∂M
∂y
∂x
= ∂N
= 3. Solution x2 + 2y 2 + x + 3xy − y = k.
(c) ∂M
∂y
∂x
(d) ∂M
= ∂N
= x22y
. x ln[1 + ( xy )2 ] − 2y arctan xy = k.
∂y
∂x
+y 2
Particular: x + exy + y 2 = 2.
5. Omitted.
6. (a) IF = x2 . Solution x2 y = sin x − x cos x + c.
y cos x
1
cos x
(b) IF = sec x+tan
= 1+sin
. Solution 1+sin
= ex + 12 ex (cos x − sin x) + c
x
x
x
2 1 3
x−1 2
8
(c) IF = x+1
. Solution y x−1
= 3 x − 2x2 + 9x − 16 ln |x + 1| − x+1
+ c.
x+1
√
(d) IF = earcsin x . Solution y = 15 (2x 1 − x2 + 4x2 − 2) + ce− arcsin x .
2
2
(e) IF = e−x . Solution y = − 12 (x2 + 1) + cex .
√
√
(f) IF = (1 + x2 )−1/2 . y = 1 + x2 (ln | 1 + x2 + x| + c).
y cos x
Particular: 1+sin
= ex + 12 ex (cos x − sin x) − 12 .
x
7. Omitted.
THE END OF THE COURSE!
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