Long Questions (Question 1)


Question 1:
The curve y = x3 – 6x2 + 9x + 3 passes through the point P (2, 5) and has two turning points, A (3, 3) and B.
Find 
(a) the gradient of the curve at P.
(b) the equation of the normal to the curve at P.
(c) the coordinates of and determine whether B is a maximum or the minimum point.

Solution:
(a)
y = x3 – 6x2 + 9x + 3
dy/dx = 3x2– 12x + 9
At point P (2, 5),
dy/dx = 3(2)2 – 12(2) + 9 = –3

Gradient of the curve at point P = –3.

(b)
Gradient of normal at point P = 1/3
Equation of the normal at P (2, 5):
yy1 = m (x – x1)
y – 5 = 1/3 (x – 2)
3y – 15 = x – 2
3y = x + 13

(c) 
At turning point, dy/dx = 0.
3x2 – 12x + 9 = 0
x2 – 4x + 3 = 0
(x – 1)( x – 3) = 0
x – 1 = 0  or x – 3 = 0
x = 1  x = 3 (Point A)

Thus at point B:
x = 1
y = (1)3– 6(1)2 + 9(1) + 3 = 7

Thus, coordinates of = (1, 7)

when  x = 1 ,   d 2 y d x 2 = 6 x 12 d 2 y d x 2 = 6 ( 1 ) 12 d 2 y d x 2 = 6 < 0 Since  d 2 y d x 2 < 0 ,   B  is a maximum point .

Short Questions (Question 22 – 25)


Question 23:
Given that y = 15 x + 24 x 3 ,

(a) Find the value of d y d x when x = 2,
(b) Express in terms of k, the approximate change in when x changes from 2 to
  2 + k, where k is a small change.

Solution:
(a)
y = 15 x + 24 x 3 y = 15 x + 24 x 3 d y d x = 15 72 x 4 d y d x = 15 72 x 4 When  x = 2 d y d x = 15 72 2 4 = 10.5

(b)
Approximate change in  y  to  x  in terms of  k , δ y δ x d y d x δ y = d y d x × δ x δ y = 10.5 × ( 2 + k 2 ) δ y = 10.5 k



Question 24:
If the radius of a circle increases from 4 cm to 4.01 cm, find the approximate increase in the area.

Solution:
Area of circle,  A = π r 2 d A d r = 2 π r Approximate increase in the area to radius, δ A δ r d A d r δ A = d A d r × δ r δ A = ( 2 π r ) × ( 4.01 4 ) δ A = [ 2 π ( 4 ) ] × ( 0.01 ) δ A = 0.08 π  cm 2



Question 25:
Given that y =3t+ 5t2 and x = 5t 1.
(a) Find d y d x  in terms of x,
(b) If increases from 5 to 5.01, find the small increase in t.

Solution:
y = 3 t + 5 t 2 d y d t = 3 + 10 t x = 5 t 1 d x d t = 5

(a)
d y d x = d y d t × d t d x d y d x = ( 3 + 10 t ) × 1 5 d y d x = 3 + 10 ( x + 1 5 ) 5 x = 5 t 1 t = x + 1 5 d y d x = 3 + 2 x + 2 5 d y d x = 5 + 2 x 5

(b)
Small increase in  t  to  x , δ t = d t d x × δ x δ t = 1 5 × ( 5.01 5 ) δ t = 0.002

Short Questions (Question 20 – 22)


Question 20:
The volume of water V cm3, in a container is given by   V = 1 5 h 3 + 7 h , where h cm is the height of the water in the container. Water is poured into the container at the rate of 15cm3s-1. Find the rate of change of the height of water in cms-1, at the instant when its height is 3cm. 

Solution:

V = 1 5 h 3 + 7 h d V d h = 3 5 h 2 + 7 = 3 h 2 + 35 5 Given  d V d t = 15 h = 3 Rate of change of the height of water = d h d t d h d t = d h d V × d V d t Chain rule d h d t = 5 3 h 2 + 35 × 15 d h d t = 75 62  cms 1



Question 21:
A wire of length 88 cm is bent to form a circle. When the wire is heated, the length increases at the rate of 0.3 cms-1.
(a) Calculate the rate of change in the radius of the circle.
(b) Hence, calculate the radius of the circle after 5s.

Solution:
Length of circumference of a circle,  L = 2 π r d L d r = 2 π

(a)
Given  d L d t = 0.3 Rate of change in the radius of the circle = d r d t d r d t = d r d L × d L d t d r d t = 1 2 π × 0.3 d r d t = 0.0477  cms 1

(b)
2 π r = 88 r = 88 2 π = 44 π Hence, the radius of the circle after  5 s = 44 π + 5 ( 0.0477 ) = 14.24  cm



Question 22:

The diagram shows a conical container with diameter 0.8m and height 0.6m. Water is poured into the container at a constant rate of 0.02m3s-1. Calculate the rate of change of the height of the water level when the height of water level is 0.5m.

Solution:

Let,  h  = height of the water level r  = radius of the water surface V  = volume of the water r h = 0.4 0.6 Concept of similar triangles r h = 2 3 r = 2 3 h Volume of water,  V = 1 3 π r 2 h V = 1 3 π ( 2 3 h ) 2 h V = 4 27 π h 3 d V d h = ( 3 ) 4 27 π h 2 d V d h = 4 9 π h 2 The rate of change of the height of the water level when the height of water level is 0 .5  m = d h d t . d h d t = d h d V × d V d t Chain rule d h d t = 9 4 π h 2 × 0.02 Given  d V d t = 0.02 d h d t = 9 4 π ( 0.5 ) 2 × 0.02 d h d t = 0.0572   m s 1

Short Questions (Question 14 – 16)


Question 14:
Given y = x (6 – x), express y d 2 y d x 2 + x d y d x + 18 in terms of in the simplest form.

Hence, find the value of which satisfies the equation  y d 2 y d x 2 + x d y d x + 18 = 0

Solution:
y = x ( 6 x ) = 6 x x 2 d y d x = 6 2 x d 2 y d x 2 = 2 y d 2 y d x 2 + x d y d x + 18 = ( 6 x x 2 ) ( 2 ) + x ( 6 2 x ) + 18     = 12 x + 2 x 2 + 6 x 2 x 2 + 18     = 6 x + 18 y d 2 y d x 2 + x d y d x + 18 = 0    6 x + 18 = 0   x = 3



Question 15:
Find the coordinates of the point on the curve, y = (4x – 5)2 such that the gradient of the normal to the curve is 1 8 .

Solution:
y = (4x – 5)2
d y d x = 2(4x – 5).4 = 32x – 40

Given the gradient of the normal is 1/8, therefore the gradient of the tangent is –8.
d y d x = –8
32x – 40 = –8
32x = 32
x = 1
y = (4(1) – 5)2= 1

Hence, the coordinates of the point on the curve, y = (4x – 5)2 is (1, 1).



Question 16:
A curve has a gradient function of kx2 – 7x, where k is a constant. The tangent to the curve at the point (1, 4) is parallel to the straight line y + 2x–1 = 0. Find the value of k.

Solution:
Gradient function of kx2– 7x is parallel to the straight line y + 2x–1 = 0
d y d x = kx2– 7x

y + 2x –1 = 0, y = –2x + 1, gradient of the straight line = –2
Therefore kx2– 7x = –2

At the point (1, 4),
k(1)2 – 7(1) = –2
k – 7 = –2
k = 5

Short Questions (Question 11 – 13)


Question 11:
Given that the graph of function f ( x ) = h x 3 + k x 2  has a gradient function f ' ( x ) = 12 x 2 258 x 3 such that h and are constants. Find the values of and k.

Solution:
f ( x ) = h x 3 + k x 2 = h x 3 + k x 2 f ' ( x ) = 3 h x 2 2 k x 3 f ' ( x ) = 3 h x 2 2 k x 3 But it is given that  f ' ( x ) = 12 x 2 258 x 3 Hence, by comparison,  3 h = 12  and   2 k = 258 h = 4    k = 129



Question 12:
Given that  y = x 2 x + 3 , show that  d y d x = x 2 + 6 x ( x + 3 ) 2 Find  d 2 y d x 2  in the simplest form .

Solution:
y = x 2 x + 3 d y d x = ( x + 3 ) ( 2 x ) x 2 .1 ( x + 3 ) 2 = 2 x 2 + 6 x x 2 ( x + 3 ) 2 d y d x = x 2 + 6 x ( x + 3 ) 2  (shown) d 2 y d x 2 = ( x + 3 ) 2 ( 2 x + 6 ) ( x 2 + 6 x ) .2 ( x + 3 ) ( x + 3 ) 4 d 2 y d x 2 = ( x + 3 ) [ ( x + 3 ) ( 2 x + 6 ) 2 ( x 2 + 6 x ) ] ( x + 3 ) 4 d 2 y d x 2 = [ 2 x 2 + 6 x + 6 x + 18 2 x 2 12 x ] ( x + 3 ) 3 d 2 y d x 2 = 18 ( x + 3 ) 3



Question 13:
If  y = x 2 + 4 x , show that  x 2 d 2 y d x 2 2 x d y d x + 2 y = 0.

Solution:
y = x 2 + 4 x d y d x = 2 x + 4 d 2 y d x 2 = 2 x 2 d 2 y d x 2 2 x d y d x + 2 y = x 2 ( 2 ) 2 x ( 2 x + 4 ) + 2 ( x 2 + 4 x ) = 2 x 2 4 x 2 8 x + 2 x 2 + 8 x = 0  (Shown)

Short Questions (Question 5 – 7)


Question 5:
Given that f (x) = 3x2(4x 1)7, find f’(x). 

Solution:
f (x) = 3x2(4x 1)7
f’(x) = 3x2. 7(4x 1)6. 8x + (4x 1)7. 6x
f’(x) = 168x3 (4x 1)6 + 6x (4x 1)7
f’(x) = 6x (4x 1)6 [28x2+ (4x 1)]
f’(x) = 6x (4x 1)6 (32x 1)



Question 6:
Given that y = (1 + 4x)3(3x 1)4, find d y d x

Solution:
y = (1 + 4x)3(3x2 – 1)4
d y d x
= (1 + 4x)3. 4(3x2 – 1)3.6x + (3x2 – 1)4. 3(1 + 4x)2.4
= 24x (1 + 4x)3(3x2 – 1)3 + 12 (3x2 – 1)4(1 + 4x)2
= 12 (1 + 4x)2(3x2 – 1)3 [2x (1 + 4x) + (3x2 – 1)]
= 12 (1 + 4x)2(3x2 – 1)3 [2x + 8x2 + 3x2 – 1]
= 12 (1 + 4x)2(3x2 – 1)3 [11x2 + 2x  – 1]



Question 7:
Given that  f ( x ) = 3 x 4 x 2 1   ,  find  f ' ( x ) . .

Solution:
f ( x ) = 3 x 4 x 2 1 = 3 x ( 4 x 2 1 ) 1 2 f ' ( x ) = 3 x . 1 2 ( 4 x 2 1 ) 1 2 .8 x + ( 4 x 2 1 ) 1 2 .3 f ' ( x ) = 12 x 2 ( 4 x 2 1 ) 1 2 + 3 ( 4 x 2 1 ) 1 2 f ' ( x ) = 3 ( 4 x 2 1 ) 1 2 [ 4 x 2 + ( 4 x 2 1 ) ] f ' ( x ) = 3 ( 8 x 2 1 ) ( 4 x 2 1 )

Short Questions (Question 1 – 4)


Question 1:
Differentiate the expression 2x (4x2 + 2x – 5) with respect to x.

Solution:
2x (4x2 + 2x – 5) = 8x3 + 4x2– 10x
d d x (8x3 + 4x2 – 10x)
= 24x + 8x –10 



Question 2:
Given that  y = x 3 + 2 x 2 + 1 3 x ,  find  d y d x . .

Solution:
y = x 3 + 2 x 2 + 1 3 x y = x 3 3 x + 2 x 2 3 x + 1 3 x y = x 2 3 + 2 x 3 + 1 3 x 1 d y d x = 2 x 3 + 2 3 1 3 x 2 d y d x = 2 x 3 + 2 3 1 3 x 2



Question 3:
Given that  y = 3 5 x + 1 ,  find  d y d x

Solution:
y = 3 5 x + 1 = 3 ( 5 x + 1 ) 1 2 d y d x = 1 2 .3 ( 5 x + 1 ) 3 2 ( 5 ) d y d x = 15 2 [ ( 5 x + 1 ) 3 ] 1 2 d y d x = 15 2 ( 5 x + 1 ) 3



Question 4:
Given that  y = 3 5 u 5 , where u = 4+ 1. Find d y d x in terms of x.

Solution:
y = 3 5 u 5 ,   u = 4 x + 1 y = 3 5 ( 4 x + 1 ) 5 d y d x = 5. 3 5 ( 4 x + 1 ) 4 .4 d y d x = 12 ( 4 x + 1 ) 4

9.9 Small Changes and Approximations

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9.9 Small Changes and Approximations


If  δ x  is very small,  δ y δ x  will be a good approximation of  d y d x , ,

This is very useful information in determining an approximation of the change in one variable given the small change in the second variable. 


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Example:
Given that y = 3x2 + 2x – 4. Use differentiation to find the small change in y when x increases from 2 to 2.02.

Solution:
y = 3 x 2 + 2 x 4 d y d x = 6 x + 2

The small change in is denoted by δy while the small change in the second quantity that can be seen in the question is the x and is denoted by δx.

δy δx dy dx δy= dy dx ×δx δy=( 6x+2 )×( 2.022 )δx=new xoriginal x δy=[ 6( 2 )+2 ]×0.02  Substitute x with the original value of x, i.e2. δy=0.28

Related Rates of Change


(A) Related Rates of Change


1. If two variables and y are connected by the equation y = f(x)


Notes:
If x changes at the rate of 5 cms -1   d x d t = 5
Decreases/leaks/reduces Þ  NEGATIVES values!!!


Example 1 (Rate of change of y and x)
Two variables, and y are related by the equation   y = 4 x + 3 x . Given that y increases at a constant rate of 2 units per second, find the rate of change of x when x = 3.

Solution:
y = 4 x + 3 x = 4 x + 3 x 1 d y d x = 4 3 x 2 = 4 3 x 2 d y d t = d y d x × d x d t 2 = ( 4 3 x 2 ) × d x d t when  x = 3 2 = ( 4 3 3 2 ) × d x d t 2 = 11 3 × d x d t d x d t = 6 11  unit  s 1


(B) Rates of Change of Volume, Area, Radius, Height and Length




(C) Rate of Change of Any Combination of Two Variables



9.7 Second-Order Differentiation, Turning Points, Maximum and Minimum Points

(A) Second-Order Differentiation

1. When a function y = x3 + x2 – 3x + 6 is differentiated with respect to x, the derivative  d y d x = 3 x 2 + 2 x 3

2. The second function   d y d x can be differentiated again with respect to x. This is called the second derivative of y with respect to and can be written as d 2 y d x 2 .

3. Take note that   d 2 y d x 2 ( d y d x ) 2 .

For example,
If y = 4x3 – 7x2 + 5x – 1,

The first derivative   d y d x = 12 x 2 14 x + 5

The second derivative    d 2 y d x 2 = 24 x 14

(B) Turning Points, Maximum and Minimum Points



(a) At Turning Points A and B,




(b) At Maximum Point A



(c) At Minimum Point B,