Document 6485880

CHAPTER 5
Analytic Geometry
Distance between two points: Let (a, b) and
p (x, y) be any two points in a
plane. The distance between them is d = (a − x)2 + (b − y)2. See the accompanying diagram:
y
B(a, b)
b−y
x
A(x, y) a − x
C(a, y)
2
2
2
2
2
By the Theorem
p of Pythagors, we have AB = AC +CB = (a−x) +(b−y) ,
2
2
hence AB = (a − x) + (b − y) .
Examples : 1. Find the distance
p between A(−3, 1) and
√ B(2, 4).
2
2
Soln. The distance is AB = (−3 − 2) + (1 − 4) = 34.
2. Calculate the value of x for which M(−2, 1) and N (x, −7) are equidistant
from P (1, −4).
Soln.
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M(−2, 1)
N (x, −7)
N (x, −7)
P (1, −3.7)
M P = P N implies MP 2 = P N 2 . Using the distance formula we have
(−3)2 + 52 = (x − 1)2 + (−3)2 , implying (x − 1)2 = 25, hence x − 1 = ±5. Thus
x = −4 or x = 6.
3. A(2, −1), B(−3, 4) and C(4, 5) are vertices of 4ABC. Find the perimeter of the triangle. Say whether the triangle is equilateral, isosceles, scalene or
right-angled.
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y
C(4, 5)
B(−3, 4)
x
A(2, −1)
Soln.
p
p
p
2 + 52 +
2 + 12 =
(−5)
(7)
(−2)2 + (−6)2 =
(i)√
Perimeter
is
P
=
AB+BC+CA
=
√
10 2 + 40
√
(ii) Triangle is isosceles since AB = BC = 5 2
(iii) Triangle is not right-angled since the sides do not satisfy the theorem of
Pythagoras.
4. Show that P (−1, 1), Q(0, 4), R(3, 5) and S(2, 2) are vertices of a rhombus.
(A rhombus is a 4-sided figure having that all its sides are of equal length).
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y
R(3, 5)
Q(0, 4)
S(2, 2)
P (−1, 1)
x
Soln. √
√
√
√
√
√
QR =√ 32 + 12 =√ 10; P S = 32 + 12 = 10; P Q = 12 + 32 = 10;
RS = 12 + 32 = 10. Thus the given 4-sided figure is a rhombus since all 4
sides are equal.
5. Show that M(2, −1) is the circumcentre of 4P QR having vertices P (6, 2),
Q(−1, 3) and R(−1, −5). (Show that P M = QM = RM).
Soln.
60
y
Q(−1, 3)
P (6, 2)
•M(2, −1)
x
R(−1, −5)
PM =
√
16 + 9 = 5 = QM = RM, as required.
Mid-point of a line segment
Let A(x1, y1) and B(x2, y2) be points in a plane. M is the mid-point of the
line segment if AM = BM and M lies on the same line as A and B. We want
to determine the coordinates of M in terms of those of A and B. Suppose we
use (a, b) for the coordinates of M. See diagram below:
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y
B(x2, y2)
• M(a, b)
x
A(x1, y1 )
D(x2 , y1 )
2 y1 +y2
Using congruence of triangles, we may show that (a, b) = ( x1 +x
, 2 ).
2
Examples : 1. Find the midpoint of the line segment joining A(−3, 1) to
(4, 2).
, 1+2
) = ( 12 , 32 ).
Soln. The midpoint is M = ( −3+4
2
2
2. Find x and y if (−3, 2) is the mid-point of the line segment joining (−1, 5)
and (x, y).
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y
A(−1, 5)
M(−3,
•
2)
x
B(x, y)
Soln.
Now −3 = (x − 1)/2 implies x = −5, and 2 = (5 + y)/2 implies y = −1.
Recall: If L1 : y = m1x + c1 and L2 : y = m2x + c2 , such that (i) L1 k L2 ,
then m1 = m2 , (ii) L1 ⊥ L2 , then m1m2 = −1.
Examples : 1. Show that A(−2, 3), (0, 2) and C(1, 3/2) are collinear.
Soln. Note that collinear means lying on the same line. The slope of AB is
2−3
MAB = 0−(−2)
= −1/2 and slope of BC is MBC = 3/2−2
= −1/2 = MAB , thus
1
AB k BC. Since the point B belongs to both AB and BC, we conclude that
A, B and C are collinear.
2. A(−3, −1), B(−2, 2), C(1, 3) and D(0, 0) are vertices of a quadrilateral
(four-sided figure). (a) Show that the quadrilateral is a parallelogram; (b) The
diagonals of the quad. bisect each other at right angles; (c) what kind of a
quad. is this?
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y
C(1, 3)
B(−2, 2)
D(0, 0)
x
A(−3, −1)
Soln.
2−(−1)
= 3 and the slope of DC is MDC = 3−0
=
(a) Slope of AB is MAB = −2−(−3)
1−0
3−2
3 = MAB , implying AB k DC. Slope of BC is MBC = 1−(−2) = 1/3 and the
slope of AD is MAD = 0−(−1)
= 1/3 = MBC , implying BC k AD. Therefore
0−(−3)
ABCD is a parallelogram. (Opposite sides are parallel).
(b) MAC = 3+1
= 1 and MBD = −2/2 = −1, implying MAC MBD = −1. Thus
1+3
AC ⊥ BD, so the diagonals intersect at right angles since we already know
that the diagonals of a parallelogram bisect each other.
(c) The√parallelogram is a rhombus . (Check that all the sides are of equal
length 10).
The Circle
A circle is a set of points that are equidistant from a fixed point called the
center of the circle. The distance from any of the points on the circle to the
center is the radius of the circle. The distance from one point to the opposite
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one through the center is the diameter of the circle. Hence the diameter is
twice the radius. See diagram below:
y
r •(xo , yo )
x
Circumference of a circle is its perimeter or length. It is given by the formula
C = 2πr where r is the radius. The area of the circle is A = πr2.
The equation of a circle of radius r centered at (x0, y0 ) is (x − x0)2 +(y − y0)2 =
r2 .
Examples : 1. (3, a) lies on the circle x2 + y 2 = 18. Find a.
Soln. Substitute the coordinates of the point (3, a) into the eqn of the circle
since it lies on the circle. Thus we have 9 + a2 = 18 implying a = ±3.
2. A chord of the circle x2 + y 2 = 50 has mid-point (6, 3). Determine the line
of which the chord is a segment.
Soln.
65
y
A
M(6,
3)
•
B
O
x
√
The circle has radius 50 which is approximately 7 units. 4OMB ≡ 4OMA
= 1/2, implying slope of AB is
(S,S,S). Thus OM ⊥ AB. Slope of OM is 3−0
6−0
−1
= −2. Therefore the eqn of the required line is y−3 = −2(x−6) = −2x+12,
1/2
implying y = −2x + 15.
3. Determine the eqn of the circle that passes through (3, 2) and (5, −4)
with its centre on the line x − y = 1.
Soln. The general eqn is of form (x − xo )2 + (y − yo )2 = r2 . Since (xo , yo ) is
the centre, it lies on the given line, implying yo = xo − 1 · · · (?). Substitute
into the circle eqn to get (x − xo )2 + (y − xo + 1)2 = r2 . Replacing (x, y)
with (3, 2) and (5, −4) respectively, we get (3 − xo )2 + (3 − xo )2 = r2 implying 2(3 − xo )2 = r2 · · · (1) and (5 − xo )2 + (3 + xo)2 = r2 = 2(3 − xo )2 by
(1). This gives xo = −2 and yo = −3 upon using (?). Substitute this into
eqn (1) to get r2 = 50. Therefore the eqn of the circle is (x+2)2 +(y +3)2 = 50.
Tangents and Normals to the Circle
66
y
L : secant
` : tangent P
O
x
A line L passing through two points on the circle is called a secant. A line
` touching the circle at only one point P is a tangent to the circle at P .
A line segment such as OP from the centre to the point P of tangency is always perpendicular to the tangent `. Thus any angle between OP and ` is 90o .
Examples: 1. Find an equation of the tangent to the circle x2 + y 2 = 10 at
(−3, 1).
Soln.
67
y
`
P
√
O
10
x
Let ` be tangent to the circle at (−3, 1). OP ⊥ `. Slope of OP is MOP =
−1
−1/3 implies slope of ` is −1/3
= 3. Therefore eqn of ` is y − 1 = 3(x + 3) =
3x + 9, implying y = 3x + 10.
2. Determine the eqn of the tangent to the circle x2 − 2x + y 2 + 4y = 5 at
(−2, −1).
Soln. We complete the squares to express the eqn in standard form: x2 − 2x +
1 − 1 +√
y 2 + 4y + 4 − 4 = 5, implying (x − 1)2 + (y + 2)2 = 10. This is a circle of
radius 10 centered (1, −2). Let ` be a line tangent to the circle at P (−2, −1).
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y
√
10
O
P
x
`
= −1/3. Thus the slope of ` is 3. Therefore
OP ⊥ `. Slope of OP is MOP −1+2
−2−1
the eqn of ` is y + 1 = 3(x + 2) = 3x + 6, implying y = 3x + 5.
3. L : y = x + 2 cuts the circle x2 + y 2 = 20 at A and B.
(a) Find A and B; (b) the length of the chord AB; (c) find M, the mid-point
of AB; (d) show that OM ⊥ AB, where O is the origin ; (e) find the tangents
at A and B; (f) find the point C of intersection of the tangents in e.
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y
L
B(2, 4)
`2
`1
M(−1,
1)
•
O
√
10
x
A(−4, −2)
Soln.
(a) To find points of intersection of L and the circle, solve their eqns simultaneously: So we have x2 + (x + 2)2 = 20, implying 2x2 + 4x − 16 = 0 =
(x + 4)(x − 2), thus x = −4 or x = 2, and y = −2 or y = 4. Therefore
A = (−4, −2)
(2, 4).
√ and B = √
(b) AB = 62 + 62 = 6 2;
(c) M = ( 2−4
, 4−2
) = (−1, 1);
2
2
= 1. This
(d) Slope of OM is MOM = −1 and slope of AB is MAB = 4+2
2+4
implies MOM MAM = −1, thus OM ⊥ AB.
(e) Let `1 and `2 be tangent at A and B respectively. So OB ⊥ `1 and OB ⊥ `2 .
= 1/2, implying slope of `1 is −2. Also slope of
Slope of OA is MOA = −2
−4
OB is MOB = 42 = 2, implying slope of `2 is −1/2. Therefore `1 is given by
y + 2 = −2(x + 4) = −2x − 8, thus y = −2x − 10 is the eqn of `1 .
Also `2 is given by y − 4 − 1/2(x − 2) = (−1/2)x + 1, thus the eqn of `2 is
y = (−1/2)x + 5.
(f) We solve y = −2x − 10 and y = (−1/2)x + 5 simultaneously. Thus
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−2x − 10 = (−1/2)x +5, implying x = −10. Hence y = (−1/2)(−10)+ 5 = 10.
Thus C = (−10, 10).
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CHAPTER 6
Differential Calculus
Calculus is a mathematics of motion and change. It is applied mainly in Sciences, Engineering and Economic Sciences.
DEF. The average gradient or slope between two points (a, b) and (x, y) is
(a)
y−b
. If y = f (x) then b = f (a), thus the average gradient is Ma = f (x)−f
.
x−a
x−a
f (a+h)−f (a)
.
Letting x = a + h, we have Ma =
h
Example: Let f (x) = x2 + 2. (i) Find the average gradient between (a, f(a))
and (a + h, f(a + h)); (ii) hence find the average gradient between (1, 3) and
(3, 11); (iii) what does the average gradient become as (a + h, f(a + h)) moves
closer and closer to (a, f(a))?
2
2 −2
(a)
= (a+h) +2−a
= 2a + h
Soln. (i) Ma = f (a+h)−f
h
h
(ii) Here a + h = 11 implies h = 3 − a and a = 1, hence h = 2. Thus
Ma = 2 × 1 + 2 = 4.
As a + h → a, h gets closer to 0. Thus Ma = 2a + h → 2a + 0 = 2a.
Example. The distance s (in metres) travelled by a body in t seconds is given
by s = 4t2 + 1. Find (a) how far the body has travelled after 2 sec and after 3
sec.; (b) average speed of the body between t = 2sec and t = 3sec; (c) average
speed of the body between t = 2sec and t = (2 + h)sec; (d) the value that the
average speed approaches as t approaches 2.
Soln.
(a) s(2) = 4 × 22 + 1 = 17m and s(3) = 4 × 32 + 1 = 37m
(b) Av speed = s(3)−s(2)
= 37−17
= 20m/s
3−2
3−2
2
2
−1
= 4(2+h) +1−4(2)
= h(16+4h)
= (16 + 4h)m/s
(c) Av speed = s(2+h)−s(2)
h
h
h
(d) As t → 2, h → t − 2 = 2 − 2 = 0, the Av speed approaches 16.
Limits
Let y = f (x) = 1/x. We want to know what y approaches when x approaches
0 from the rhs. Consider the following table:
x 1 1/10 1/100 1/1000 · · ·
So as x approaches 0, y gets larger withy 1 10
100
1000
···
out limit. We write x → 0+ to mean x approaches 0 from the rhs. We also
write y → ∞ to mean y gets larger without limit. The symbol ∞ is read as
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infinity and means arbitrarily large. −∞ means arbitrarily large in the negative direction. The symbolic statement limx→0+ y = ∞ is read as “the limit of
y as x approaches 0 from the right is equal to ∞. Since ∞ is not a number,
y has no limit as x approaches 0 from the right. Similarly limx→0 y = −∞ is
read as “the limit of y as x approaches 0 from the left is equal to −∞. Since
∞ is not a number, y has no limit as x approaches 0 from the left.
x 1 10
y 1 1/10
So limx→∞ (1/x) = 0 = limx→−∞ (1/x). See also the graph below:
y
Now let f (x) = 1/x and let x → ∞. Find limx→∞ (1/x).
100
1/100
1000
1/1000
x
Useful information about limits
(i) If f is a continuous function at xo and f (xo ) exists, then limx→xo√
f (x) =
fp(xo ). For example limx→2 1/x = 1/2; limx→1 log x = log 1 = 0; limx→2 2x2 + 4x =
2(22 ) +
√4(2) = 4. But limx→0 log x 6= log 0 since log 0 is undefined. Also
limx→−1 x does not exist.
(ii) If limx→x+o f (x) 6= limx→xo f (x), then limx→xo f (x) does not exist. If lhs
limit = rhs limit, then the limit of f (x) exists and equals the common value
of the two limits.
(iii) If f is not continuous at xo , then limx→xo f (x) 6= f (xo ). For example
1 if x < 0
let f (x) =
. Find limx→0 f (x) if it exists.
x if x ≥ 0
Soln. limx→0− f (x) = 1 while limx→0 f (x) = 0 6= limx→0− f (x). See diagram
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···
···
y
◦1
x
0
Thus limx→0 f (x) does not exist.
(iv) First cancel out any common factors in f (x) before substitution or com2 −1
puting the limit. For example : Find limx→1 xx−1
2 −1
Soln. limx→1 xx−1
= limx→1 (x−1)(x+1)
= limx→1 (x + 1) = 2 ( by letting x = 1).
x−1
x−1
Another example: limx→1 x−1
Soln. limx→1 x−1
= limx→1 1 = 1.
x−1
(v) The last example shows that limx→a c = c for any constant c. E. g.
limx→1 2 = 2.
More Examples
3 −1
.
(1) Find limx→1 xx−1
Soln. limx→1
x3 −1
x−1
(2) Let f (x) =
1, while limx→0
2
+x+1)
= limx→1 (x−1)(x
= limx→1 (x2 + x + 1) = 3
x−1
x
if x 6= 0
x
|x|
Then limx→0+ f (x) = limx→0+ |x|
= limx→0+
1
if x = 0
x
f (x) = limx→0 −x
= −1. Thus limx→0 f (x) does not exist.
x
x
2
=
+3x+2
= limx→−2 (x+1)(x+2)
= limx→−2 (x + 1) = −1
(3) limx→−2 (x x+2
x+2
√
f (x+h)−f (x)
if (a) f (x) = 2; (b) f (x) = x; (c) f (x) = x2 + 1;
(4) Find limh→0
h
(d) f (x) = 1/x; (e) f (x) = |x|, x 6= 0.
(x)
Soln. (a) limh→0 f (x+h)−f
= limh→0 (2−2)
= 0.
h
h
√
√
√
√
(x+h)−
x
(x+h)+ x
f (x+h)−f (x)
(x+h)−x
√
(b) limh→0
=
lim
(
)(
=
√ ) = limh→0 √
√
h→0
h
h
(x+h)+ x
1
√
2 x
.
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h(
(x+h)+ x)
2
2
2
(x)
−1
(c) limh→0 f (x+h)−f
= limh→0 (x+h) +1−x
= limh→0 2xh+h
= limh→0 2x+h
=
h
h
h
h
2x .
(x)
x−x−h
−h
= limh→0 1/(x+h)−1/x
= limh→0 xh(x+h)
= limh→0 xh(x+h)
=
(d) limh→0 f (x+h)−f
h
h
2
−1/x .
(x)
(|x+h|−|x|)(|x+h|+|x|)
(x+h)2 −x2 )
h(2x+h2 )
=
lim
=
lim
=
lim
=
(e) limh→0 f (x+h)−f
h→0
h→0
h→0
h
h(|x+h|+|x|)
h(|x+h|+|x|)
h(|x+h|+|x|)
1
if x > 0
(2x+h)
2x
limh→0 (|x+h|+|x|)
= limh→0 2|x|
=
−1 if x < 0
Example: Given the following graph of y = f (x), compute (a) limx→0 f (x);
(b) limx→0+ f (x); (c) limx→0 f (x):
y
6
1o
0
−1
x
-
Gradient (or Slope) of a Curve:
Let y = f (x) be as shown in the graph below:
75
y
6
Q(x + h, f(x + h))
•
• Q1
tangent at P (x, y)
P (x, y) •
curve y = f (x)
x
-
We want to find the slope of the curve at (x, y). First draw the secant
(x)
through P and Q. Therefore the slope of the secant is Msec = f (x+h)−f
.
h
Now move Q along the curve towards P. Then x + h moves closer to x. In
symbols we write x + h → x. This implies h → 0 and the secant approaches
the tangent. Now move Q even closer and closer to P; then the secant looks
more and more like a tangent. In symbols Msec → Mtan . We may also write
limh→0 Msec = Mtan ; this is read as “the limit of Msec as h approaches 0 is
(x)
Mtan ”. Thus we have established that Mtan = limh→0 f (x+h)−f
.
h
Note: In general the symbolic statement limx→a f (x) = b is read as “the limit
of f (x) as x approaches (or tends to) a is b; and means if x is moved closer
and closer to a, then f (x) moves closer and closer to b.
Examples : (1) Let f (x) = x2 . Find the equation of the tangent to the
curve at P (3, 9). Find also the eqn of the normal to the curve at P (3, 9).
(x)
(x+h)2 −x2
Soln. Slope of the tangent is Mtan = limh→0 f (x+h)−f
=
lim
=
h→0
h
h
h(2x+h)
= limh→0 (2x + h) = 2x. At P (3, 9), Mtan = 2 × 3 = 6. Thus
limh→0 h
the eqn of the tangent is y − 9 = 6(x − 3) = 6x − 18. Therefore y = 6x − 9.
= −1/6.
The normal is perp to the tangent; hence its slope is Mn = M−1
tan
So the eqn of the normal is y − 9 = (−1/6)(x − 3) = −x/6 + 1/2, implying
y = (−1/6)x + 19/2.
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Instantaneous speed:
The instantaneous speed of an object at any time t is given by v(t) = limh→0 s(t+h)−s(t)
h
where s = s(t) is the eqn of motion of the object.
Example 2: A body travels s metres in t sec, where s(t) = 2t2. Find the speed
of the body at t = 3sec.
2
2
Soln. s(t) = 2t2 implies the speed is v = limh→0 s(t+h)−s(t)
= limh→0 2(t+h)h −2t =
h
= limh→0 (4t + 2h) = 4t. Therefore s(3) = 4 × 3 = 12m/s.
limh→0 h(4t+2h)
h
The Derivative: Differentiation
(x)
Recall: If y = f (x), then Mtan = limh→0 f (x+h)−f
. If the right hand side
h
dy
(read as
(rhs) exists, then it is called the derivative of f at x, written as dx
f (x+h)−f (x)
df (x)
0
0
.
dydx) or f (x) or dx . Thus f (x) = limh→0
h
√
dy
at x = 8.
Example: If f (x) = 2x, find dx
√
√
√
√ √
√
√
2(x+h)− 2x
( (x+h)− x)( (x+h)+ x)
f (x+h)−f (x)
dy
0
√
Soln. dx = f (x) = limh→0
=
lim
=
2
lim
=
√
h→0
h→0
h
h
h( (x+h)+ x)
√
√
√
2 limh→0 √(x+h)−x√ = 2 limh→0 √ h √ = 2 limh→0 √ 1 √ =
√
√2
2 x
h( (x+h)+ x)
1
√ . Therefore dy |x=8
dx
2x
h(
(x+h)+ x)
(
(x+h)+ x)
√1
2×8
=
=
= 1/4.
Note: If we say differentiate f (x) with respect to x we mean find f 0 (x) or
df
df
or dx
. dx
is called the derivative of f with respect to x.
Rules for Differentiation
dy
dx
df
(1) If f (x) = k, a constant, then dx
= 0.
f (x+h)−f (x)
df
= limh→0 k−k
= 0.
P roof. dx = limh→0
h
h
dy
n
(2) If y = x , n a rational number, then dx = nxn−1 . (Note: A rational number
is a number of the form
√ p/q where p and q are whole numbers
√ (integers)). For
example 2/3; -4 and 9 are all rational numbers. But 3 is not a rational
number.)
(x)
dy
(3) If y = f (x) ± g(x), then dx
= dfdx
± dg(x)
.
dx
df (x)
dy
(4) If y = kf (x), then dx = k dx .
√
2
+ 3/x6 − 3x5 + 10. Find f 0 (x).
Examples: (1) Let f (x) = x√
Soln. f (x) = x2 + 3 × x−6 − 3x5/2 + 10
f 0 (x) = 2x − 6 × 3x−7 −
√ implies
√
(5/2)−1
7
(3/2)
3(5/2)x
+ 0 = 2x − 18/x − (5/2) 3x
.
dy
(2) Find dx from first principles ( i.e from definition ) if y = f (x) = (1/2)x3 .
77
(x)
Soln. f 0 (x) = limh→0 f (x+h)−f
= (1/2) limh→0
h
2
2
h(3x +3xh+h )
= (1/2)3x2 .
(1/2) limh→0
h
(x+h)3 −x3
h
= (1/2) limh→0
x3 +3x2 h+3xh2 +h3 −x3
h
Applications of Derivatives
Tangents and Normals
Examples. (1) Find the tangent and the normal to the curve y = −x2 + 2x + 3
at x = 3.
dy
Soln. Slope of the tangent to the curve is Mtan = dx
= −2x + 2 = −2(3) + 2 =
−4. Thus the eqn of the tangent is y − y0 = −4(x − 3) = −4x + 12. When
x = 3, y0 = −32 + 2(3) + 3 = 0. Hence the tangent is given by y = −4x + 12.
Normal has slope Mn = (−1)/(−4) = 1/4. Thus eqn of normal is y − 0 =
(1/4)(x − 3) = 1/4 − 3/4.
(2) (a) Find the eqn of the tangent to the curve y = −x2 + 3x if the tangent
is parallel to the line y = x + 2.
(b) Find the point at which the tangent to the curve y = −x2 + 3x is parallel
to the tangent to the curve y = 2x2 − 3x.
− 2x + 3 = 1 since parallel lines have the same
Soln. (a) Slope Mtan = dy
dx
slope and the slope of y = x + 2 is 1. Therefore x = 1 and y = −1 + 3 = 2.
The point of tangency is (1, 2) and the eqn of the tangent is y − 2 = 1(x − 1),
implying y = x + 1.
(b) parallel lines have the same slope, thus −2x + 3 = 4x − 3, giving x = 1
and y = 2 by using y = −x2 + 3x.
(3) Show that the curves y = x2 and y = −x2 + 1/2 intersect at right angles.
Soln. Let M1 be the slope of y = x2 and M2 be the slope of y = −x2 + 1/2.
We must show that m1m2 = −1 at the point of intersection of the curves.
= 2x and m2 = dy
= −2x. For the point of intersection, solve
Now m1 = dy
dx
dx
2
2
x = −x + 1/2 which gives x = ±(1/2).
At P1 (1/2, 1/4): m1m2 = 2(1/2)(−2)(1/2) = −1, and
At P1 (−1/2, 1/4): m1m2 = 2(−1/2)(−2)(−1/2) = −1. Thus the curves intersect at right angles.
Curve Sketching
Let y = f (x), x ∈ [a, b] have its graph shown below:
78
=
y
•
•
a
a1
a2
a3 a4
•
a5
b
x
As we move from left to right, the graph (or function) increases from x = a
to x = a1 and from x = a3 to x = a5. We say the region of increase of f is
[a, a1] ∪ [a3, a5]. Similarly the region of decrease of f is [a1, a3] ∪ [a5, b].
Turning Points: The points (a1, f(a1)) and (a5, f(a5)) are called the turning
(or critical) points of the graph (or f ) because the graph at these points changes
from going up (or down) to going down (or up). That is, the graph changes
from increasing (or decreasing) to decreasing (or increasing). At such points
the graph has a horizontal tangent, or has no tangent. Thus at turning points
f 0 (x) = 0 or f 0 (x) does not exist. The points a and b are the end − points of
f.
Around a turning point of f , the graph is concave up or down. The turning
point at which the graph is concave up is a local minimum point, while the
turning point at which the graph is concave down is a local maximum point.
The value of f (x) at such points is a local minimum or maximum respectively.
The largest value of f (x) on [a, b] is the absolute maximum of f , while the
least value of f (x) on [a, b] is the absolute minimum of f . From our diagram,
Absolute max f (x) = f (a1 ), while Absolute min f (x) = f (b). Abs max/min
may occur at critial points or end-points of f .
Also, in the region of increase of f , the slope of a tangent to the graph of f
is positive (+ve), thus f 0 (x) > 0. Similarly in the region of decrease of f , the
79
slope of a tangent to the graph of f is negative (-ve), thus f 0 (x) < 0.
Points such as (a2, f(a2 )) and (a4, f(a4 )) at which the graph of f changes from
concave down to up (or up to down) , are called points of inf lection. At such
points we usually have f 00(x) = 0.
Determining which critical points give local maxima and minima:
If f 0 (x) changes from +ve to -ve at (x0, f(x0 )), then f has a local maximum
atx0 . The graph of f around this point is concave down. If f 0 (x) changes from
-ve to +ve at (x1, f(x1 )), then f has a local minimum atx1. The graph of f
around this point is concave up.
Examples : (1) Discuss and draw the graph of y = f (x) = 2x3 + 3x2 − 12x − 9.
Soln. y-intercept: y = −9.
Critical points: solve f 0 (x) = 0. Thus 6x2 + 6x − 12 = 0 = 6(x + 2)(x − 1),
implying x = −2 or x = 1 . f (−2) = 11 and f (1) = −16. Thus the turning
points are (−2, 11) and (1, −16).
Local mx/min:
We draw up a table that shows how f 0 (x) changes signs around the turning
points:
2
1
(x − 1)
- 0 +++
(x + 2)
0 +++ + +++
(x − 1)(x + 2) +++ 0 0 +++
Thus region of increase is given by f 0 > 0, implying x < −2 or x > 1
Region of decrease is given by f 0 < 0, implying −2 < x < 1
Local max is at (−2, 11) and the graph of f is concave down around this point;
Local min is at (1, −16) and the graph of f is concave up around this point.
Point of inflection: Solve f 00(x) = 0, implying 12x + 6 = 0, giving x = −1/2
and y = −5/2. So the point of inflection is (−1/2, −5/2)
Sketch:
80
y
(−2, 11)
•
x
•(−.5, −2.5)
•
(1, −16)
(2) Discuss and sketch the curve y = x3 − 4x2 + 4x.
Soln. y = x(x2 − 4x + 4) = x(x − 2)2 . x-intercepts are x = 0 and x = 2 while
the y-intercept is y = 0.
Turning Points: Solve y 0 = 0. Thus 3x2 − 8x + 4 = 0 = (3x − 2)(x − 2),
implying x = 2/3 or x = 2. When x = 2/3, y = 2/3(4/3)2 = 32/27. When
x = 2, y = 0.
Increase/Decrease:
2/3
2
Thus region of increase of f is
(3x − 2)
+ + +++
(x − 2)
- 0 +++
(x − 1)(x + 2) +++ 0
- 0 +++
x < 2/3 or x > 2.
Region of decrease of f is 2/3 < x < 2.
Local max is at (2/3, 32/27) and the graph is concave down around this point.
Local min is at (2, 0) and the graph is concave up around this point.
Point of inflection: Solve f 00 (x) = 0. Thus 6x − 8 = 0, giving x = 4/3 and
y = 4/3(−2/3)2 = 16/27. The graph is shown below:
81
y
(2/3, 32/27)
•
• (4/3, 16/27)
•
(0, 0)
(2, 0)
x
Application of Maxima/Minima Theory
Examples : (1) The length of a rectangle is x mm while its width is (200 − x)
mm. Find the dimensions of a rectangle of maximum area.
Soln.
x
200 − x
Area A = x(200 − x) = 200x − x2. Max may occur at critical points of
82
A. For critical poits, solve A0 (x) = 0; thus 200 − 2x = 0, giving x = 100.
Since A = 200x − x2 is a parabola concave down, its absolute max = its local
max. Therefore A is a maximum at x = 100. Thus the rectangle of max area
must have length = 100 mm and width = 200 - 100 = 100 mm. Its area is
100 × 100 = 10000mm2 .
(2) The sum of two positive numbers is 12. Find the numbers when their
product is a maximum.
Soln. Let the numbers be x and y, then x + y = 12, implying y = 12 − x.
Thus the product is P = x(12 − x) = 12x − x2. Since P = 12x − x2 is a
concave down parabola, absolute max of P = local max of P and this occurs
at a critical point of P . For critical points, solve P 0 (x) = 0. Thus 12 − 2x = 0,
implying x = 6. Therefore the required numbers are x = 6 and y = 6.
(3) The edges of a rectangular box of square cross section are to be x mm,
x mm and (240 − 2x)mm respectively. Find x so that the volume of the box
is a maximum, and find this volume of the box.
x
x
240 − 2x
Soln. Volume V = x2(240 − 2x) = 240x2 − 2x3 . For critical points of V , solve
V 0 (x) = 0, thus 480x − 6x2 = 6x(80 − x), implying x = 0 or x = 80.
0
80
Thus for max volume, x = 80; hence max
(6x)
- 0 + + +++
(80 − x) + + + 0
V 0 (x)
- 0 + 0
volume = 6400(240 - 160) = 512 000.
(4) The radius of the base of the solid metal cylinder is x mm while its height
is (300 − x)mm. Find x so that the volume of the cylinder is a maximum.
Soln.
83
•
x
y=h
V = area of base × height = πx2y = πx2(300 − x) = 300πx2 − πx3. Critical
points of V are given by V 0 (x) = 0 = 600x − 3x2 = 3x(200 − x); thus x = 0 or
x = 200.
0
(3x)
(200 − x)
V 0 (x)
- 0
+ +
- 0
200
+
+
+
+
0
0
+++
-
Thus max volume is when x = 200mm.
(5) A piece of cardboard 80mm by 50mm has square corners of side x mm
cut out. The edges are then folded up so as to form a box, without a lid, of
depth x mm. Find x so that the volume of the box is a maximum.
Soln.
84
80 − 2x
50 − 2x
x
x
80 − 2x
x
50 − 2x
Volume V = x(50 − 2x)(80 − 2x) = 4000x − 260x2 + 4x3 . Critical points of
V are given by V 0 (x) = 0 = 4000 − 520x + 12x2 = 2(2000 − 260x + 6x2 ), giving
x = 10 or x = 100/3.
10
(2x − 20)
(3x − 100)
V 0 (x)
Thus max
- 0 +
- + 0 volume when
100/3
+
+++
0
+
0
+
x = 10.
85
(6) A firm manufacturing concrete asbestos products wishes to wall-in 200m2
exhibition space in the form of a rectangular plot bordering on a road. If the
cost of the ornamental walling along the road is treble that along the other
three sides, find the dimensions of the display area which will cost the company
the least.
Soln.
x
y
road
Perimeter P = 2x + 2y and xy = 200 implies y = 200/x. Cost C: Suppose
1 m costs r rand along three sides, then the total cost C = xr + 2yr + 3xr =
4xr + (400/x)r. Critical points of C are given by C 0(x) = 0 = 4r − (400r)/x2 ,
implying 4x2 = 400, hence x = 10.
-10
(x − 10) - (x + 10) - 0
C 0(x)
+ 0
200/10 = 20 m.
10
- 0
+ +
- 0
+++
+
+
Thus min C when x = 10 m and y =
(7) 4ABC is isosceles with hight 80mm. P QRS is a rectangle with P on
AB and Q on AC. The height of the rectangle is x mm. Find the dimensions
of the rectangle of maximum area.
Soln.
86
A
80 − x
T
P
Q
x
B
S
U
80
R
C
4P T A is similar to 4BDA, implying the corresponding sides of these triangles are in proportion. Thus 40/(y/2) = 80/(80 − x) implying 40(80 − x) =
80(y/2). Thus y = 80 − x. Area A = xy = x(80 − x) = 80x − x2 is a concave
down parabola. Thus its maximum occurs only at the turning point. The
turning point is given by A0 (x) = 0 = 80 − 2x, giving x = 40. Thus the
rectangle must have length x = 40mm and width y = 80 − 40 = 40mm.
Rate of Change
dy
If y = f (x), then the rate of change of f or y with respect to x is dx
= f 0 (x).
Examples : (1) The profit P (in rands) from the sale of x m of cable produced
at a factory is given by P = 5x − 200.
(a) What is the rate of change of the profit with respect to the length of a
cable being produced?
(b) What will the rate of change of profit be when (i) 100 m of cable is produced? (ii) 200 m of cable is produced?
Soln. (a) The rate of change is dP
= 5 rands per meter.
dx
dP
(b) (i) The rate of change is dx = 5 rands per meter.
= 5 rands per meter.
(ii) The rate of change is dP
dx
(2) The mass P of a bacteria culture varies with time t, according to the
eqn P = 500 + 200t + 15t2 , (t in minutes). Find how fast the mass is growing
after 5 min.
= 200 + 30t . Thus at t = 5, the rate is
Soln. The rate of growth is dP
dt
200 + 30(5) = 350 units per min.
(3) A radar device tells an operator that a motor-bike is moving according to
87
the eqn s = (1/15)t3 − 2t2 + 20t meters, (t in seconds). The bike is approaching
a stop sign.
(a) Find how fast the bike was travelling when first detected.
(b) Find (i) how long it took the bike to stop after first detected, (ii) the distance covered while being tracked until it stopped.
(c) Find the acceleration of the bike after 5 seconds.
Soln. (a) Speed = v = ds
= (1/5)t2 − 4t + 20. t = 0 when first detected, and
dt
the speed in this case is v(0) = 20 m/s.
(b) (i) When the bike stops, ds
= 0 = t2 − 20t + 100 = (t − 10)2 , implying
dt
t = 10 sec is the time it took the bike to stop after first detected.
(ii) s(10) = 1000/15 − 2(100) + 20(10) = 200/3 m.
= (2/5)t − 4, implying a(5) = 2 − 4 = −2 m/s2 .
(c) Acceleration = a = dv
dt
The negative sign shows that the speed was reducing.
Calculus of Motion.
Let s = f (t) be an eqn of motion of an object with t representing the time
taken. s gives the position of the object at any time t. ds
is the velocity v of
dt
ds
the object at any time t whle | dt | is the speed of the object. Velocity gives the
speed and direction of the object. When v is negative, this indicates motion
opposite the forward direction. The acceleration a = dv
is the rate of change
dt
of the velocity at any time t.
Example. A stone is thrown vertically upwards. Its height s meters, at any
time t in sec is given by s = 20t − 5t2 .
(a) Find the height of the stone after t = 1 sec; 2 sec; 3 sec.
(b) At what time will the stone hit the ground?
(c) Find the velocity v at t =1; 2; 3 sec;
(d) Find the maximum height reached by the stone.
(e) Find the acceleration of the stone and then interpret the result.
Soln. (a) s(1) = 20 − 5 = 15 m;
s(2) = 20(2) − 5(4) = 20 m; s(3) = 15 m.
(b) When the stone hits the ground, s = 0. Thus t(20 − 5t) = 0, implying
t = 20/5 = 4 sec.
(c) v = ds
= 20 − 10t, implying v(1) = 20 − 10 = 10 m/s; v(2) = 20 − 20 = 0
dt
m/s; v(3) = 20 − 30 = −10 m/s.
(d) At maximum height, the stone stops, implying v = 0 = 20 − 10t, giving
88
t = 2 sec, which is the time taken to reach the max height. Thus max height
= 20(2) − 5(4) = 20 m.
(e) Acceleration is a = dv
= −10 m/s2 .
dt
On way up: the speed decreases by 10 m/s per sec.
On way down: the speed increases by 10 m/s per sec.
89
CHAPTER 7
The Real Number System.
Z = {· · · , −3; −2; −1; 0; 1; 2; 3; · · ·} is the set of all integers.
N = {1; 2; 3; · · ·} is the set of all natural numbers or positive integers.
A rational number is a number that can be expressed in the form p/q where
p and q are integers, with q 6= 0. The set of all rational numbers is denoted
by Q. A real number which is not a rational number is called an irrational
c
.
number. The set of all irrational numbers is denoted
by Q√
√
√
3
3
Examples of rational numbers: 3/2;
-8; √ 9; 0;√ 27; −1.
√
√ -5/6;
3
3;
4;
π; 27; 3 3.
Examples of irrational
numbers:
√
√
Numbers such as −4 and 4 −16 are called imaginarynumbers.
Sets: A set is a collection of clearly defined objects called the elements
of the set. If A and B are sets, the intersection of A and B is A ∩ B =
set of all elements belonging to both A and B. The union of A and B is
A ∪ B = set of all elements belonging to A or B. The difference of A and
B is A − B = set of all elements belonging to A and not B. The difference
of R − A is also denoted by Ac and is called the complement of A. Example:
Let A = {a; b; c; d; e; f ; g} and B = {a; c; g; h; j; k}. Then A ∩ B = {a; c; g};
A ∪ B = {a; b; c; d; e; f ; g; h; j; k}; A − B = {b; d; e; f ; }; B − A = {h; j; k; }.
Set-builder notation: The set {x ∈ R|x > 0} is expressed in set-builder notation and is read as: the set of all x in R such that x is greater than 0. Briefly
it means all real numbers that are positive. In future we will express sets in
set-builder notation.
Intervals
Intervals are sets of real numbers between certain given numbers or greater
than a given number or less than a given number. Some intervals are described below:
(1) [a, b] is a closed interval of all real numbers from a to b. It is also described
90
•
a
•
b
as follows: {x ∈ R|a ≤ x ≤ b}.
(2) (a, b) is an open interval of all real numbers between a to b. It is also
◦
a
◦
b
described as follows: {x ∈ R|a < x < b}.
(3) [a, b) is a left closed right open interval of all real numbers from a but less
•
a
◦
b
than b. It is also described as follows: {x ∈ R|a ≤ x < b}.
(4) (a, b] is a right closed left open interval of all real numbers greater than a up
91
◦
a
•
b
to b. It is also described as follows: {x ∈ R|a < x ≤ b}.
(5) [a, ∞) is a left closed right open interval of all real numbers greater than or
•
a
equal to a. It is also described as follows: {x ∈ R|x ≥ a}.
(6) (a, ∞) is an open interval of all real numbers greater than a. It is also de-
◦
a
scribed as follows: {x ∈ R|x > a}.
(7) (−∞, a] is a right closed left open interval of all real numbers less than or
92
•
a
equal to a. It is also described as follows: {x ∈ R|x ≤ a}.
(8) (−∞, a) is an open interval of all real numbers less than a. It is also de-
◦
a
scribed as follows: {x ∈ R|x < a}.
Examples : (1) Let A = [−1, 5) and B = (3, 6]. Find A ∩ B; A ∪ B; A − B;
B − A; Ac = R − A.
Soln.
A ∩ B = (3, 5)
◦
3
◦
5
A ∪ B = [−1, 6]
93
•
−1
•
6
A − B = [−1, 3]
•
−1
•
3
B − A = [5, 6]
•
5
•
6
Ac = R − A = (−∞, −1) ∪ [5, ∞).
Domain and Range of a function.
Let f :→ B be a function from A into B. Then A is called the domain of f .
f (A) = {f (a)|a ∈ A} is called the range of f . B is the co-domain of f . If a
domain of f is not specified, then take it to be largest set on which f can be
defined or exists.
√
Examples : (1) Let f (x) = x. Find the domain and the range of f .
94
Soln. dom(f ) = {x√∈ R|x ≥ 0} = [0, ∞) .
Range is R(f ) = { x|x ≥ 0} = [0, ∞).
√
(2) Let f (x) = x2 − 2. Find the domain and the range
√
√ of f .
2
R
|x
−
2
≥
0}
=
{x
∈
R
|x
≥
2
or
x
≤
−
2} =
Soln. dom(f
)
=
{x
∈
√
√
∞).
(−∞, − 2] ∪ [ 2,√
Range is R(f ) = { x2 − 2|x2 − 2 ≥ 0} = [0, ∞)
•√
− 2
√
•
2
√
. √
(3) Let f (x) = 2 − x2 . Find the domain and the range of f √
≥ 0} = {x√∈ R|x2 − √
2 ≤ 0} = [− 2, 2].
Soln. dom(f ) = {x√∈ R|2 − x2√
Range is R(f ) = { 2 − x2| − 2 ≤ x ≤ 2} = [0, 2]
√ 1
.
x2 −2
Find the domain and the range of f .
√
√
2
Soln. dom(f
)
=
{x
∈
R
|x
−
2
>
0}
=
{x
∈
R
|x
<
−
2
or
x
>
2} =
√
√
∪ ( 2, ∞).
(−∞, − 2) √
R(f ) = {1/( x2 − 2)|x2 − 2 > 0} = (0, ∞)
(4) Let f (x) =
(5) Let f (x) = x21−1 . Find the domain and the range of f .
Soln. dom(f ) = {x ∈ R|x2 − 1 6= 0} = R − {−1; 1} = (−∞, −1) ∪ (−1, 1) ∪
(1, ∞).
R(f ) = {1/(x2 − 1)|x 6= ±1} = R − (−1, 0) = (−∞, −1) ∪ (0, ∞)
(6) Let f (x) = log10(x + 1). Find the domain and the range of f .
Soln. dom(f ) = {x ∈ R|x + 1 > 0} = {x ∈ R|x > 1}(−1, ∞).
R(f ) = {log10 (x + 1)x > −1} = R = (−∞, ∞)
95
CHAPTER 8
Geometry
Congruent Triangles
Two triangles are congruent if they are equal in all respects. Thus three
sides and three angles of one triangle are equal to 3 sides and three angles
of the other triangle when the triangles are congruent. In symbols we write
4ABC ≡ 4ABC if the two triangles are congruent.
A
B
D
C
E
F
AB = DE, BC =
EF and AC = DF , thus the two tringles are congruent.
Proving two triangles are congruent: Show that (1) 3 sides = 3 sides; OR
(2) 2 sides and included angle = 2 sides and included angle ; OR
(3) 2 angles and a side = 2 angles and a side ; OR
(4) Right angle, hypotenuse and a side = Right angle, hypotenuse and a side;
A triangle with all its sides equal is an equilateral triangle. A triangle with
two of its sides equal is an isosceles triangle.
Now let 4ABC be such that AB = AC and draw AD ⊥ BC. Prove that
BD = CD and ∠ABD = ∠ACD.
Soln.
96
A
B
C
D
Now AB = AC (given); AD = AD (common); ∠ADB = ∠ACD = 90o .
Therefore 4ADB ≡ 4ADC (Rt angle, hyp, s). Thus BD = CD and
∠ABD = ∠ACD.
A line intersecting two parallel lines is a transversal.
L1
L2
1 2
3 4
5 6
7 8
Now let L1 k L2 and L a transversal drawn across them. Then alternate
angles are equal. Thus ∠3 = ∠6 and ∠4 = ∠5.
Corresponding angles are equal. Thus ∠1 = ∠5; ∠3 = ∠7; ∠2 = ∠6 and
∠4 = ∠8.
V ertically opposite angles are equal. Thus ∠2 = ∠3; ∠1 = ∠4; ∠5 = ∠8 and
∠6 = ∠7.
Prove that any triangle with two angles equal is isosceles.
97
P roof.
C
A
B
D
Suppose ∠A = ∠B. Draw CD ⊥ AB. Then in triangles ADC and BDC,
CD = CD (common); ∠A = ∠B and ∠ADC = ∠BDC = 90o . Therefore
4ACD ≡ 4BCD ( ∠, ∠, s). Thus CA = CD and hence 4ABC is isosceles.
NOTE: (1) The sum of the interior angles of a triangle is 180o .
(ii) the exterior angle of a triangle equals the sum of two interior opposite
angles of a triangle. See diagram:
C
1
A
2
3
4
B
P roof. ∠3 + ∠4 = 180o and ∠1 + ∠2 + ∠3 = 180o ; implying ∠3 = 180o −
(∠2 + ∠1) = 180o − ∠4. Thus ∠1 + ∠2 = ∠4.
Examples: (1) F M is a transversal cutting the parallel lines AB and CD in
O and T respectively. If ∠AOM = 3x − 20o and ∠BOF = 2x + 30o , calculate
x and ∠MT B.
Soln.
98
C
3x − 20 M
O
2x + 30
A
D
B
T
F
3x − 20 = 2x + 30o (vert. opp. L’s), implying x = 5)o and 2x + 30o = 130o .
Therefore ∠MT B = ∠MOD = 180o − 130o = 50o .
o
(2) In 4ABC, CD, with A on CDis drawn, and also AE, with B on AE,
so that AD = AE. If CB ⊥ DE, prove that 4ABC is isosceles.
Soln. Consider the diagram:
D
A
F
C
B
E
99
∠D = ∠E; ∠DF C = ∠EF B = 90o . Therefore ∠F BE = ∠ABC = ∠ACB.
Thus AB = AC, implying 4ABC is isosceles.
Theorem of Pythagors:
Let 4ABC be a right-angled triangle with ∠B = 90o . Then AB 2+BC 2 = AC 2
or a2 + c2 = b2 where a, c, b are the sides opposite ∠A, ∠C, ∠B respectively.
A
c
b
B
C
a
Converse of the Theorem of Pythagoras: Let 4ABC be such that AB 2 +
BC 2 = AC 2 or a2 + c2 = b2 where a, c, b are the sides opposite ∠A, ∠C, ∠B
respectively. Then ∠B = 90o . Thus 4ABC is right-angled.
QUADRILATERALS. A four-sided figure is called a quadrilateral.
Two lines are parallel if they maintain the same distance between them. Thus
parallel lines will never intersect.
A quadrilateral whose opposite sides are parallel is called a parallelogram. A
rhombus is a parallelogram whose sides are all equal. A rhombus whose interior angles are right angles is a square. A quadrilateral with a pair of opposite
sides that are parallel is called a trapezium.
100
A
B
D
C
AB k CD and the figure is a trapezium.
NOTE: (1) If a quad. has equal opposite sides, then it is a parallelogram:
(2) If a quad. has equal opposite angles, then it is a parallelogram:
P roof. Consider Quad ABCD with angles x, 1; 2; 3 and 4 as shown.
A
x
D
3
4
1
B
2
C
Then ∠1 + ∠2 = ∠3 + ∠4 · · · (1)
Also ∠1 = 180o − x − ∠3 and ∠4 = 180o − x − ∠2. Therefore ∠1 + ∠3 =
101
∠2 + ∠4 · · · (2). Thus (1) and (2) imply 2∠1 = 2∠4. So ∠1 = ∠4. Substitute
into (1), then ∠2 = ∠3. We have shown that alternate angles are equal. Hence
quad ABCD is a parallelogram.
T heorem : The diagonals of a parallelgram bisect each other.
A
B
D
C
T heorem : A quadrilateral is a parallelogram if the diagonals of a parallelgram bisect each other.
A median of a tringle ABC is a bisector of a side drawn through the vertex
of the triangle.
102
A
E
B
D
C
AD and BE are medians if BD =
DC and AE = EC.
Area of a Parallelogram: Consider parallelogram ABCD of height h and
∠ADC = θ.
A
B
h
D
θ
C
Then its area is A = base × height =
DC × h. But h/AD = sin θ, thus A = (DC)AD sin θ.
Examples : (1) ABCD is a quad. with AB = 52 mm, AD = 39 mm, BC =
60 mm, CD = 25 mm and ∠A = 90o . Prove that ∠BCD = 90o .
103
Soln. Consider the diagram of quad ABCD shown.
D
39
A
25
C
52
60
B
Triangle ABD is right-angled, thus BD2 =
522 + 392 = 4225 = 652 ; also BC 2 + CD2 = 602 + 252 = 4225 = BD2 , implying
triangle BCD is right-angled with ∠BCD = 90o .
(2) Triangle STM has ST = 150 mm, MT = 120 mm, SM = 90 mm. P is a
point on TM so that ∠T = ∠P ST . Prove that ∠M = 90o . Calculate P S.
Soln.
S
150
90
T
M
P
120
1202 + 902 = 22500 = 1502 implies T M 2 + SM 2 = T S 2; thus ∠M = 90o .
Also P S = T P since ∠T = ∠P ST , and P S 2 = SM 2 + P M 2 = 8100 + P M 2
implies (T M − P M)2 = 8100 + P M 2 . Therefore T M 2 − 2T M.P M = 8100,
so 14400 − 2(120)P M = 8100, giving P M = 105/4. Hence P S = T P =
104
120 − 105/4 = 375/4.
(3) In 4ABC, ∠A = 90o . AM is drawn with B on AM so that AB = BM
and AN is drawn with C on AN so that CN = AC. Prove that BN 2 +CM 2 =
5BC 2.
Soln.
M
B
N
C
BN 2 = AN 2 + AB 2 · · · (1)
CM 2 = (2AB)2 + AC 2 · · · (2)
BC 2 = AC 2 + AB 2 · · · (3)
Therefore BN 2 +CM 2 = AN 2 +AB 2+4AB 2 +AC 2 = 5AB 2 +(2AC 2)2 +AC 2 =
5(AB 2 + AC 2) = 5BC 2
A
CIRCLE GEOMETRY
Given a circle with center O; radii OA and OB, OD bisects AB. Prove that
OD ⊥ AB.
O
•
A
D
B
P roof.
In 4ODA and 4ODB, OA = OB(radii); OD = OD (common); AD = BD
since OD bisects AB. Thus 4ODA ≡ 4ODB (S,S,S). Hence ∠ODA =
105
∠ODB = 90o , implying OD ⊥ AB.
Conversely, given a circle with center O; radii OA and OD ⊥ AB. Prove
that OD bisects AB.
P roof.
O
•
A
D
B
In 4ODA and 4ODB, OA = OB(radii); OD = OD (common); ∠ODA =
∠ODB = 90o since OD ⊥ AB. Thus 4ODA ≡ 4ODB (right ∠,hyp.,S).
Hence AD = BD, implying OD bisects AB.
Examples : (1) AB is a chord of a circle and is 240 mm long. M is the
mid-point of AB. MD ⊥ AB and MD cuts the circle at D. Find the radius
of the circle if MD = 80mm.
Soln.
106
D
A
M
B
•
O
Since M is the mid-point of AB, OM ⊥ AB, thus O, M, D are collinear
(are on the same straight line). Thus AM = 120 = BM. radius = r = OD =
80+ OM implies r2 = OM 2 + AM 2 = (r − 80)2 +1202 = r2 − 160r +802 +1202 ,
giving r = 130 mm.
(2) ST is a diameter of a circle with center O and is perpendicular to chord
M N and cuts MN at P . If P S = 4P T , calculate MN in terms of the radius
of the circle.
Soln.
S
O
•
M
P
N
T
Let r = radius of the circle. Then OP 2 + MP 2 = r2 ; OP = 2r − r − P T =
107
r − P T implies r2 − 2r.P T + P T 2 + MP 2 = r2 . Thus MP 2 = 2r.P T − P T 2 =
2r(2r/5) − 4r2 /25 = 16r2 /25. Therefore MP = 4r/5; hence MN = 2(4r/5) =
8r/5.
(3) P T is a diameter of the circle with center O and perp to chord AB and
cuts AB at Q. If P Q = 9QT , calculate AB in terms of the radius of the circle.
Soln.
P
O
•
A
Q
B
T
Let r = radius of the circle. Now QA2 = r2 − OQ2 ; OQ = 2r − P O − QT ;
P Q = 2r − QT = 9QT implies 2r = 10QT , thus QT = r/5. Therefore OQ = r − r/5 = 4r/5. So QA2 = r2 − 16r2 /25 = 9r2 /25. Hence
AB = 2AQ = 2(3r/5) = 6r/5.
(4) AB is a chord of a circle with center M, and AC with B on AC is drawn
so that BC = 2(AB). Prove that MC 2 = MB 2 + (3/2)BC 2 .
Soln.
108
A
D
B
C
•
M
Now MC 2 = MD2 + DC 2 = MB 2 − DB 2 + (DB + BC)2 = MB 2 +
2DB.BC + BC 2. DB = (1/2)AB = (1/2)(1/2)BC = (1/4)BC; thus MC 2 =
M B 2 + (2/4)BC 2 + BC 2 = MB 2 + (3/2)BC 2 , as required.
T heorem. The angle at the center of a circle is twice the angle on the circle
if both angles are subtended by the same arc.
P roof.
A
3 4
O
•
1 2
B
C
D
We must show that ∠BOC = 2∠BAC. Now draw AD through O. ∠1 =
2∠3 and ∠2 = 2∠4. Thus ∠1 + ∠2 = 2(∠3 + ∠4), i.e. ∠BOC = 2∠BAC.
109
Corollary : An angle on a semicircle is 90o .
P roof.
A
x
O
•
B
C
Angle at the centre is 180o = 2x where x is an angle on the circle subtended
by the same arc as the angle at the center. Thus x = 180o /2 = 90o .
Examples: (1) 4ABC has AB = AC. A circle on AB as diameter cuts
BC at D. Prove that BD = CD.
Soln.
A
O
•
B
D
C
In triangles ABD and ACD, ∠ABC = ∠ACD since AB = AC; ∠ADB =
∠ADC = 90o (angle on semicircle and BDC a straight line); AB = AC (given).
Thus 4ADB ≡ 4ACD (∠, ∠, S). Therefore BD = CD.
110
(2) ABCD is a quad. inscribed in a circle. BC is a diameter of a circle.
AB = CD and BA and CD intersect at O. (a) Prove that AO = OD. (b) If
CD = 20 mm and ∠DBC = 30o , calculate AC to the nearest mm.
Soln.
O
D
C
30
A
30
B
(a) In triangles BAC and CDB, ∠BAC = ∠CDB = 90o (angle on semicircle), BC = BC (common), AB = DC (given); thus 4BAC ≡ 4CDB
(rt angle, hyp., S). Therefore ∠ABC = ∠DCB, thus OB = OC, implying
OA + AB = OD + DC. Hence OA = OD.
(b)√Using triangle BDC, we have DC/DB =√tan 30o ; thus DB = 20/ tan 30o =
20 3 mm. But AC = DB, hence AC = 20 3 mm.
Cyclic Quadrilaterals
A quad having its vertices on a circle is called a cyclic quadrilateral.
T heorem : The sum of opposite angles of a cyclic quad is 180o .
111
B
A
1
•
2
C
D
P roof.
Let ABCD be a cyclic quad on a circle of center O. See diagram. First
we show that ∠A + ∠C = 180o : Now ∠2 = 2∠A and ∠1 = 2∠C. This implies 360o = ∠1 + ∠2 = 2(∠A + ∠C) , thus ∠A + ∠C = 180o . Similarly
∠B + ∠D = 180o .
Conversely: If the sum of opposite angles of a quad is 180o , then the quad
is a cyclic quad.
P roof. Homework.
112
1
8
6
D 5
O
4
3
2
7 C
Corollary: The exterior angle of a cyclic quad equals the interior opposite
angle.
P roof. Consider the diagram of a cyclic quad shown below:
D
A
C
B
E
113
We show that ∠CBE = ∠D. Now ∠CBE + ∠ABC = 180o (sum of angles
on a straight line); and ∠D + ABC = 180o (sum of opp. angles of a cyclic
quad); implying ∠CBE = ∠D.
Converse: If the exterior angle of a quad equals the interior opposite angle,
then the quad is a cyclic quad.
T heorem : Consider quad ABCD with the angles 1 to 8 as indicated. Then
ABCD is a cyclic quad iff ∠3 = ∠6 OR ∠2 = ∠7 OR ∠1 = ∠4 OR ∠5 = ∠8.
2
1
O
8
D 7
3
4
5
6 C
Omit proof.
Examples: (1) ABCD is a cyclic quad. AF with B on AF and DF with
C on DF are drawn cutting F so that ∠F = 90o . If ∠BAC = 17o and
∠CAD = 25o , prove that ∠BDA = 31o .
Soln. See diagram.
114
F
90
B
17
A 25
C
D
∠BCF = 42o (ext angle of cyclic quad = int. opp. angle); therefore
∠F BC = 180o − 90o − 42o = 48o . Therefore ∠ADC = ∠F BC = 48o . Thus
∠BDA = 48o − 17o = 31o .
(2) ABCD is a cyclic quad with AB = BC and AC and BD intersecting at
O. Prove that (a) ∠BCO = ∠ODC, (b) BD bisects ∠ADC.
Soln. Consider the diagram below:
115
A
6
D 5
1
8
O
4
3
B
2
7 C
(a) Since AB = BC, we have ∠1 = ∠2. Since ABCD is a cyclic quad, we
have ∠1 = ∠5, implying ∠5 = ∠2, i.e. ∠BCO = ∠ODC .
(b) From (a), ∠5 = ∠2, but ∠6 = ∠2 (ABCD cyclic quad). Therefore
∠5 = ∠6. Hence BD bisects ∠ADC.
(3) In triangle ABC, AB = AC. A circle through B and C cuts AB at D
and AC at E. BE and DC intersect at O. (a) Prove that OB = OC; (b)
4DBC ≡ 4ECB; (c) AD = AE.
116
A
E
D
O
1
B 2
3
4 C
Soln. Consider the following diagram:
(a) ∠1 + ∠2 = ∠3 + ∠4 (since AB = AC); ∠1 = ∠3 (since DECB is a cyclic
quad.); therefore ∠4 = ∠2, implying OB = OC.
(b) In triangles BDC and ECB, ∠4 = ∠2 by (a), BC = BC (common),
∠DBC = ∠1 + ∠2 = ∠3 + ∠4 = ∠ECB (since AB = AC, given). Therefore
4DBC ≡ 4ECB, (∠, ∠, S).
(c) Since 4DBC ≡ 4ECB, we have DB = EC; therefore AD = AB − DB =
AC − EC = AE.
SIMILAR TRIANGLES
Triangles ABC and DEF are similar and we write 4ABC///4DEF iff they
are equi-angular; i.e. ∠A = ∠D, ∠B = ∠E and ∠C = ∠F . Consequently, for
similar triangles ABC and DEF , their corresponding sides are in proportion,
AB
AC
i.e. DE
= DF
= BC
.
EF
Examples. (1) ∠BAC of 4ABC is 90o and AD ⊥ BC. Prove that (a)
4ABD///4CAD///4CBA; (b) AD2 = BD.CD; (c) AB 2 = CB.BD; (d)
CA2 = CD.CB .
Soln. Consider the following diagram:
117
A
1 2
B
3 4
D
5
6
C
(a) In triangles ABD and CAD we have ∠3 = ∠4 = 90o ; ∠1 + ∠2 = 90o =
∠6 + ∠2, implying ∠1 = ∠6. Therefore ∠5 = ∠2; hence 4ABD///4CAD.
In triangles CAD and CBA we have ∠4 = ∠BAC; ∠6 = ∠6 (common); therefore ∠5 = ∠2, implying 4CAD///4CBA.
(b) By similarity of triangles ABD and CAD, we have AD
= BD
, implying
CD
AD
2
AD = BD.CD.
= BD
, implying
(c) By similarity of triangles ABD and CBA, we have AB
CB
BA
AB 2 = CB.BD.
(d) Homework. Use similarity of triangles CAD and CBA.
(2) Line segments AB and CD intersect at O with AD k CB. Prove that
AO
(a) 4AOD///4BOC; (b) BO
= OD
= DA
. (c) If AO = 30 mm, BO = 50
OC
CB
mm, CD = 100 mm, calculate DO.
Soln. Consider the following diagram:
A
3
5
D
1 O
2
C
6
4
B
(a) In triangles AOD and BOC we have ∠1 = ∠2 (vert. opp. angles);
∠3 = ∠4 and ∠5 = ∠6 (alt. angles, AD k BC). Thus 4AOD///4BOC.
118
AO
(b) BO
= OD
= DA
from similarity in (a).
OC
CB
(c) Using (b) and the diagram, we have 35 =
5(OD), giving OD = 300/8 = 37, 5 mm.
OD
,
100−OD
implying 300 − 3(OD) =
Example: Prove that a line segment joining the mid-points of two sides of
a triangle is parallel to the third side and its length is half the length of the
third side.
P roof. Consider the following diagram:
A
3
D1
B
5
E
6
2
F
4C
Let triangle ABC be such that D is the mid-point of AB and E is the midpoint of AC. Draw DE and then extend DE to DF such that CF k AB. Then
4ADE ≡ 4CF E (two angles and a side). Therefore CF = AD = BD and
DE = F E, implying BDF C is a parallelogram (pair of opp. sides equal and
parallel). Hence DF k BC, implying DE k BC. Also DF = BC = 2(DE),
implying DE = (1/2)BC.
CHAPTER 9
COSINE and SINE FORMULAE (RULES)
Sine Formula: For Triangle ABC, we have sina A = sinb B = sinc C where a, b and
c are , respectively, the sides opposite the angles A, B and C. See diagram.
119
A
b
c
h
C
a
D
B
We prove the formula for the case all interior angles of the triangle are acute.
P roof. Draw AD ⊥ BC. Then sin C = AD
and sin B = AD
, implying
b
c
b sin C = c sin B = AD, hence sinc C = sinb B . Similarly sina A = sinc C = sinb B .
We leave the case of one angle being obtuse as an exercise.
AREA of a Triangle
Consider triangle ABC as sketched below with height h.
A
b
c
h
C
a
B
Its area is Area = (1/2)ah. But h/b = sin C, so h = b sin C, implying Area
= (1/2)ab sin C.
Similarly Area = (1/2)bc sin A or Area = (1/2)ac sin B.
COSINE RULE
For triangle ABC, c2 = a2 + b2 − 2ab cos C.
We prove the rule for the case one angle of the triangle is obtuse.
P roof. Consider the diagram below:
120
A
c
b
D
C
B
a
Draw AD ⊥ DB with C on DB. Then c2 = (DC + a)2 + AD2 since triangle
ADB is right-angled. Thus c2 = DC 2 + 2a.DC + AD2 + a2 = b2 + 2a.DC + a2.
Now DC/b = − cos C since angle C is obtuse. Thus DC = b cos C, implying
c2 = a2 + b2 + 2a(−b) cos C = a2 + b2 − 2ab cos C.
Similarly a2 = c2 + b2 − 2cb cos A and b2 = c2 + a2 − 2ca cos B.
The case when angle C is acute is similarly done.
Examples
(1) In triangle ABC, find ∠B, AB and BC given that ∠A = 23o ; ∠C = 47o
and AC = 123 mm.
Soln. Consider the following diagram:
A
123
C
23o
c
47o
a
B
123
= sin
,
∠B = 180o − 47o − 23o = 110o . From the sine rule we have sinAB
47o
B
sin 47o ×123
AB
BC
implying AB = sin 110o = 95, 73 mm. sin 47o = sin 23o , implying BC =
sin 23o ×95,73
= 51, 14 mm; or use the cosine rule: a2 = b2 + c2 − 2bc cos A =
sin 47o
(123)2 + (95, 73)2 − 2(123)(95, 73) cos 23o , hence a = 51, 14 mm.
121
(2) Given triangle ABC with ∠B = 25o ; AB = 7, 2 cm; AC = 5, 7 cm, find
∠A, ∠C and side BC.
Soln. Consider the following diagram:
A
5, 7
7.2
25o
C
B
a
25
C
25
By the sine rule, sin
= sin
, implying sin C = 7,2×sin
, whence ∠C = 32, 3o
7,2
5,7
5,7
or ∠C = 180o − 32, 3o = 147, 7o .
Case ∠C = 32, 3o
122,7o
∠A = 180o − 32, 3o − 25o = 122, 7o , implying sinBC
= sin5,725 by the sine rule.
o
×5,7
= 11, 35 cm.
Therefore BC = sin 122,7
sin 25
o
Case ∠C = 147, 7
7,3o
∠A = 180o − 147, 7o − 25o = 7, 3o , implying sinBC
= sin5,725 by the sine rule.
7,3o ×5,7
= 1, 71 cm.
Therefore BC = sin sin
25
(3) In triangle ABC, a = 9 m, b = 7 m and c = 8 m. Calculate the angles
of triangle ABC.
Soln. Consider the following diagram:
C
a=9
b=7
A
c=8
B
122
a2 = b2 + c2 − 2bc cos A = 49 + 64 − 2(56) cos A, implying cos A = 49+64−81
,
2(56)
sin 73,4o
sin B
o
whence ∠A = 73, 4 . Now using the sine rule, we have 9
= 7 , implying
7×sin 73,4
o
o
sin B =
, whence ∠B = 48, 2 . Therefore ∠C = 180 − 73, 4o − 48, 2o =
9
58, 4o .
(4) In triangle ABC, c = 8 cm, b = 5 cm and ∠A = 63o . Find the third side
and two other angles.
Soln. Consider the diagram below:
A
63o
c=8
b=5
C
B
By the cosine rule, a2 = b2 + c2 − 2bc cos 63o = 25 + 64 − 80 cos 63o , whence
63o
= sin5 B , implying sin B =
a = 7, 26 cm. Now by the sine rule, we have sin
7,26
o
5×sin 63
, whence ∠B = 37, 85o . Thus ∠C = 180o − 63o − 37, 85o = 79, 15o .
7,26
(5) In triangle ABC, AB = 7, 3 m, BC = 8, 4 m and ∠ABC = 131o .
Calculate AC.
Soln. Consider the following diagram:
A
7, 3
131o
B
C
8, 4
123
By the cosine rule, AC 2 = b2 = a2 + c2 − 2ac cos B = (8, 4)2 + (7, 3)2 −
2(8, 4)(7, 3) cos 131o , whence AC = 14, 29 m.
(6) In triangle ABC, AB = 5 m, AC = 7 m and ∠B = 50o . Calculate the
area of 4ABC.
Soln. Consider the diagram below:
A
5
7
o
B 50
C
Area = (1/2)bc sin A, so we need to compute ∠A.
50o
=
implying sin C = 5×sin
, whence ∠C = 33, 2o .
By the sine rule,
7
o
o
Therefore ∠A = 180 −50 −33, 2 = 96, 8 . Hence Area = (1/2)(7)(5) sin 96.8o =
17, 38 m2.
sin C
5
o
sin 50o
,
7
o
124