9.4 First Derivatives of the Quotient of Two Polynomials

9.4 Find the Derivatives of a Quotient using Quotient Rule

Method 1

The Quotient Rule

Example:




Method 2 (Differentiate Directly)



Example:
Given that  y = x 2 2 x + 1 ,  find  d y d x

Solution:

y = x 2 2 x + 1 d y d x = ( 2 x + 1 ) ( 2 x ) x 2 ( 2 ) ( 2 x + 1 ) 2  = 4 x 2 + 2 x 2 x 2 ( 2 x + 1 ) 2 = 2 x 2 + 2 x ( 2 x + 1 ) 2



Practice 1:
Given that  y = 4 x 3 ( 5 x + 1 ) 3 ,  find  d y d x

Solution:
y = 4 x 3 ( 5 x + 1 ) 3 d y d x = ( 5 x + 1 ) 3 ( 12 x 2 ) 4 x 3 .3 ( 5 x + 1 ) 2 .5 [ ( 5 x + 1 ) 3 ] 2  = ( 5 x + 1 ) 3 ( 12 x 2 ) 60 x 3 ( 5 x + 1 ) 2 ( 5 x + 1 ) 6  = ( 12 x 2 ) ( 5 x + 1 ) 2 [ ( 5 x + 1 ) 5 x ] ( 5 x + 1 ) 6  = ( 12 x 2 ) ( 5 x + 1 ) 2 ( 1 ) ( 5 x + 1 ) 6  = 12 x 2 ( 5 x + 1 ) 4

Long Questions (Question 4)

Question 4:


In the diagram above, AXB is an arc of a circle centre O and radius 10 cm with ∠AOB = 0.82 radian. AYB is an arc of a circle centre P and radius 5 cm with ∠APB = θ. Calculate:

  1. the length of the chord AB,
  2. the value of θ in radians,
  3. the difference in length between the arcs AYB and AXB.

Solution:

(a)
  1 2 AB=sin0.41×10( Change the calculator to Rad mode ) 1 2 AB=3.99 The length of chord AB=3.99×2=7.98 cm.

(b)
  Let 1 2 θ=α, θ=2α sinα= 3.99 5 α=0.924 rad θ=0.924×2=1.848 radian.

(c)
Using s =
Arcs AXB = 10 × 0.82 = 8.2 cm
Arcs AYB = 5 × 1.848 = 9.24 cm

Difference in length between the arcs AYB and AXB
= 9.24 – 8.2
= 1.04 cm

 

 

 

Long Questions (Question 3)


Question 3:
Diagram below shows a sector QPR with centre P and sector POQ, with centre O.

It is given that OP = 17 cm and PQ = 8.8 cm.
[Use π = 3.142]
Calculate
(a) angle OPQ, in radians,
(b) the perimeter, in cm, of sector QPR,
(c) the area, in cm2, of the shaded region.

Solution:
( a ) OPQ=OQP x+x+30=180    2x=150   x=75 OPQ= 75×3.142 180    =1.3092 radians


( b ) Length of arc QR=rθ    =8.8×1.3092    =11.52 cm Perimeter of sector QPR =11.52+8.8+8.8 =29.12 cm


( c ) 30 o = 30×3.142 180 =0.5237 rad Area of segment PQ = 1 2 r 2 ( θsinθ ) = 1 2 × 17 2 ×( 0.5237sin30 ) = 1 2 ×289×( 0.52370.5 ) =3.4247  cm 2 Area of sector QPR = 1 2 r 2 θ = 1 2 × 8.8 2 ×1.3092 =50.692  cm 2 Area of shaded region =3.4247+50.692 =54.1167  cm 2

Long Questions (Question 2)


Question 2:
Diagram below shows a semicircle PTQ, with centre O and quadrant of a circle RST, with centre R.

[Use π = 3.142]
Calculate
(a) the value of θ, in radians,
(b) the perimeter, in cm, of the whole diagram,
(c) the area, in cm2, of the shaded region.

Solution:
( a ) sinROT= 2.5 5  ROT= 30 o θ= 180 o 30 o = 150 o   =150× π 180   =2.618 rad


( b ) Length of arc PT=rθ    =5×2.618    =13.09 cm Length of arc ST= π 2 ×2.5   =3.9275 cm O R 2 + 2.5 2 = 5 2   O R 2 = 5 2 2.5 2 OR=4.330 Perimeter=13.09+3.9275+2.5+4.330+5    =28.8475 cm


( c ) Area of shaded region =Area of quadrant RSTArea of quadrant RQT Area of quadrant RQT =Area of OQTArea of OTR = 1 2 ( 5 ) 2 ×( 30× π 180 ) 1 2 ( 4.33 )( 2.5 ) =1.1333  cm 2 Area of shaded region =Area of quadrant RSTArea of quadrant RQT = 1 2 ( 2.5 ) 2 ×( 90× π 180 )1.1333 =3.7661  cm 2

Long Questions (Question 1)


Question 1:
Diagram shows a circle, centre O and radius 8 cm inscribed in a sector SPT of a circle at centre P.  The straight lines, SP and TP, are tangents to the circle at point Q and point R, respectively.

[Use p= 3.142]
Calculate
(a) the length, in cm, of the arc ST,
(b) the area, in cm2, of the shaded region.


Solution:
(a)
For triangle  O P Q sin 30 = 8 O P O P = 8 sin 30 = 16  cm Radius of sector  S P T = 16 + 8 = 24  cm S P T = 60 × 3.142 180 = 1.047  radian Length of arc  S T = 24 × 1.047 = 25.14  cm


(b)
For triangle  O P Q : tan 30 = 8 Q P P Q = 8 tan 30 = 13.86  cm Q O R = 2 ( 60 ) = 120 Reflex angle  Q O R = 360 120 = 240 240 = 3.142 180 × 240 = 4.189  radian Area of shaded region = (   Area of  sector  S P T ) ( Area of major    sector  O Q R ) ( Area of triangle  O P Q  and  O P R ) = 1 2 ( 24 ) 2 ( 1.047 ) 1 2 ( 8 ) 2 ( 4.189 ) 2 ( 1 2 × 8 × 13.86 ) = 301.54 134.05 110.88 = 56.61  cm 2

Short Questions (Question 3 & 4)


Question 3:
Diagram below shows a circle with centre O.
The length of the minor arc is 16 cm and the angle of the major sector AOB is 290o.
Using  π = 3.142, find
(a) the value of  θ, in radians. (Give your answer correct to four significant figures)
(b) the length, in cm, of the radius of the circle.

Solution:
(a) 
Angle of the minor sector AOB
= 360o 290o
= 70o
= 70o × 3.142 180
= 1.222 radians

(b) 
Using s =
r × 1.222 = 16
radius, r = 13.09 cm


Question 4:
Diagram below shows sector OPQ with centre and sector PXY with centre P.
Given that OQ = 8 cm, PY = 3 cm ,  ∠ XPY = 1.2 radians and the length of arc PQ = 6cm ,
calculate
( a)  the value of θ , in  radian ,
( b)  the area, in cm2 , of the shaded region .

Solution:
(a) s = θ
 6 = 8 θ
 θ = 0.75 rad

(b) 
Area of the shaded region
= Area of sector OPQ – Area of sector PXY
= 1 2 ( 8 ) 2 ( 0.75 ) 1 2 ( 3 ) 2 ( 1.2 )
= 24 – 5.4
= 18.6 cm2

Short Questions (Question 1 & 2)


Question 1:


The figure shows the sector OCB of radius 13 cm at the centre O. The length of the arc CB = 5.2 cm. Find
(a) the angle in radians,
(b) the perimeter of the shaded region.

Solution:
(a)
s = r θ 5.2 = 13 ( C O B ) C O B = 0.4  radian

(b)
cos C O B = O A O C cos 0.4 = O A 13  (change calculator to Rad mode) O A = 11.97  cm A B = 13 11.97 = 1.03  cm C A = 13 2 11.97 2 C A = 5.07  cm

Perimeter of the shaded region = 5.07 + 1.03 + 5.2 = 11.3 cm.



Question 2:


The figure shows the sector AOB of a circle, centre O and radius 5 cm. The length of the arc AB is 6 cm. Find the area of:
(a) the sector AOB,
(b) the shaded region.

Solution:
(a) Arc AB = 6cm
  s = θ
  6 = 5 θ
θ = 6/5 rad

Area of sector  A O B = 1 2 r 2 θ = 1 2 ( 5 ) 2 ( 6 5 ) = 15 c m 2

(b)
Area of shaded region = 1 2 r 2 ( θ sin θ )   ( change calculator to Rad mode ) = 1 2 ( 5 ) 2 ( 6 5 sin 6 5 ) = 3.35  cm 2

8.3 Area of a Sector of a Circle

8.3 Area of a Sector of a Circle
(A) Area of a Sector of a Circle

1. If a circle divided into two sectors of different sizes, the smaller sector is known as the minor sector while the larger sector is known as the major sector.



2. If AOB is the area of a sector of a circle, of radius r, that subtends an angle θ radians, at the centre O, then




Example 1:
In the above diagram, find the area of the sector OAB.

Solution:
Area of the sector OAB
= 1 2 r 2 θ = 1 2 ( 10 ) 2 ( 0.354 ) = 17.7  cm 2


(B) To Calculate the Area of a Segment of a Circle




Example 2:


The above diagram shows a sector of a circle, with centre O and a radius 6 cm. The length of the arc AB is 8 cm. Find
(i) AOB
(ii) the area of the shaded segment.


Solution:
(i) Length of the arc AB = 8 cm
rθ = 8
6θ = 8
θ = 1.333 radians
AOB = 1.333 radians

(ii)
the area of the shaded segment
= 1 2 r 2 ( θ sin θ ) = 1 2 ( 6 ) 2 ( 1.333 sin 1.333 r ) = 1 2 ( 36 ) ( 1.333 0.972 ) = 6.498  cm 2

8.2 Length of an Arc of a Circle

(A) Formulae for Length and Area of a Circle



r = radius, A= area, s = arc length, q = angle, l = length of chord



(B) Length of an Arc of a Circle




Example 1:
An arc, AB, of a circle of radius 5 cm subtends an angle of 1.5 radians at the centre.  Find the length of the arc AB.

Solution: 
s = rθ
Length of the arc AB = (5)(1.5) = 7.5 cm



Example 2:
An arc, PQ, of a circle of radius 12 cm subtends an angle of 30° at the centre.  Find the length of the arc PQ. 

Solution: 
Length of the arc PQ
= 12 × 30 × π 180 = 6.283  cm



Example 3:

In the above diagram, find
(i) length of the minor arc AB
(ii) length of the major arc APB

Solution: 
(i) length of the minor arc AB = rθ
= (7)(0.354)
= 2.478 cm

(ii) Since 360o = 2π radians, the reflex angle AOB
= (2π – 0.354) radians.
Length of the major arc APB
= 7 × (2π – 0.354)
= 7 × [(2)(3.1416) – 0.354]
= 7 × 5.9292
= 41.5044 cm