Chapter 7 Applications Of Differentiation

M&S

Ch 7 Applications of Differentiation

S6/KC

7.1 Gradients


7.1.1 Slope  Slope / Gradient of AB =

f (x + h) − f (x ) x+h−x

B

f (x + h) − f (x ) = h



As h tends to zero, B becomes closer and closer to A and tends to the

A

tangent of A. ∴ Gradient / slope of tangent lim h →0






=

f ( x + h ) − f (x ) dy = h dx

Gradient / slope of tangent at A =

dy dx

A( x1 , y1 )

7.1.2 Tangent line  The gradient of the tangent is found previously. By using point-slope form, the equation of the tangent can be formulated.  Let P( x1 , y1 ) be a fixed point on the curve y = f ( x ) .

∴ Slope =

dy dx

x = x1

The equation of the tangent at P( x1 , y1 ) is

y − y1 dy = x − x1 dx


7.1.3 Normal line  a line passes through the contact point P( x1 , y1 ) and perpendicular to the tangent.  The equation of normal at P( x1 , y1 ) is

 y − y1  dy    x x − 1  dx 


x = x1

  = −1  x =x1 

Example 7.1.1  Find the equations of the tangent and normal to the curve y = ln x 3 + 4

(

)

at x = 1 P. 1/14

M&S 

Ch 7 Applications of Differentiation

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Analysis: To build an equation of a line, (1) Contact Point and (2) the slope of tangent are needed to know.



Soln: when x = 1, y = ln 5 -dy 3x 2 = 3 dx x + 4 dy 3 ∴ = dx x =1 5

(1) Contact point (1, ln 5)

--

(2) Slope =

The equation of the tangent is

y − ln 5 3 = x −1 5

i.e. y =

3 5

3 3 x − + ln 5 5 5

 y − ln 5  3  The equation of the normal is    = −1  x − 1  5  i.e. y =




−5 5 x + + ln 5 3 3

Example 7.1.2  Find the equation of tangent to the curve y = e −2 x which is parallel to the line L: 2x + y + 4 = 0



Analysis: “parallel” means they both have the same slope



Soln: Rearrange L, y = −2 x − 4

∴ The slope of tangent = -2 dy = − 2e − 2 x dx − 2 e −2 x = − 2 -- [Contact point (0, 1)] x = 0, ∴ y = e −2(0 ) = 1 The equation of tangent is y − 1 = −2( x ) i.e. y = −2 x + 1

7.2 Rate of Change
It can be applied in different areas, such as Economics (the marginal cost function), Physics (velocity and acceleration), and in multitude of other disciplines.
It describes the physical or geometric rate of change of a function. 

Average rate of change of a function f(t) with respect to t over the interval f (t1 ) − f (t 2 ) [t1, t2], t1 − t 2



Instantaneous rate of change of a function f ′(t ) = lim h→ 0



f (t + h ) − f (t ) , where t2=t + h. h

The derivative of a function is the rate of change of the dependent variable y with respect to the independent variable x at x = a. (instantaneous rate of change ) P. 2/14

M&S


Ch 7 Applications of Differentiation

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Example 7.2.1



The customer price index (CPI) if an economy is described by the function I (t ) = −0.2t 2 + 3t + 200 , 0 ≤ t ≤ 10 , where t is the number of years after 2000. At what rate was the CPI of the economy changing in (a) 2004 (b) 2009



dI = −0.4t + 3 dt



(a)






(b)

dI dt

t =4

dI dt

t =9

= −0.4(4) + 3 = 1.4 = −0.4(9) + 3 = −0.6

Example 7.2.2



In economics, the marginal cost is defined as the rate of change of cost C with respect to output x.



The marginal revenue is the derivative of the profit function. Given the cost function C = 260 ln x 2 + 100 , and the revenue function R = 3600 ln ( x + 50 ) ,

(

)

determine the marginal cost, the marginal revenue and the marginal profit at x = 50.

 

Soln: dC 260(2 x ) = dx x 2 + 100 ∴C ′(50) = 10 C ′( x ) =

dR 3600 = dx x + 50 ∴ R ′(50) = 36 R ′( x ) =

The profit function is P = R – C, ∴ P ′(50 ) = R ′(50 ) − C ′(50 ) . The marginal profit at x = 50 = 26.

7.3 Maxima and Minima (Extrema)
Increasing function A function f(x) is said to be increasing on the interval I if and only if for any point x1, x2 in I, where x1 < x2, then f(x1) < f(x2). [Larger the x, larger the f(x)] Decreasing function Similar, x1 < x2, then f(x1) > f(x2). [Larger the x, smaller the f(x)]




If we try to link up the relations between x1, x2, f(x1) and f(x2), as

f ( x 2 ) − f ( x1 ) x 2 − x1

it can be …. (1) the slope between two point (2) the rate of change over the interval I (3) for x1and x2 are closer and closer to each other, it becomes the slope of tangent at a point. P. 3/14

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Ch 7 Applications of Differentiation

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If f ( x0 ) > f ( x ) for all x in the domain of f(x), then

(x0 , f (x0 )) , then (x0 , f (x0 ))




maximum point of f(x) and f(x0) is called the absolute maximum value of f(x). If f ( x0 ) < f ( x ) for all x in the domain of f(x), then ( x0 , f ( x0 )) , then ( x0 , f ( x0 )) is called absolute




minimum point of f(x) and f(x0) is called the absolute minimum value of f(x). So for the absolute extrema, it can be located at the stationary points (/ turning points), or on the boundary of the interval. That means point x0 satisfies

dy dx

is called absolute

= 0 , that is called stationary point of x = x0

f’(x).




Definition of stationary point:

a point on a curve with zero gradients. i.e. the tangents of those points are parallel to the x-axis. For the points neither maximum nor minimum points are called points of inflexion.

7.3.1 Tests for maxima and minima
It is taking about the trend of the turn of the curve from either increasing to decreasing or vice-versa. Minimum point

Maximum point

dy >0 dx




dy <0 dx

dy <0 dx

dy >0 dx

First derivative test 



For all stationary points, it must satisfies

dy =0 dx



Maximum points: The value of

dy decreases from positive to negative. dx



Minimum points: The value of

dy increases from negative to positive. dx



Point of inflextion:

dy does not change sign as x increases through the point. dx

Example 7.3.1



Find the turning points of the curves y = x ln x and determine their natures.



dy 1 = (1) ln x + x  dx  x

For turning point,

[ product rule ]

dy =0 dx

i.e. ln x + 1 = 0 x = e-1

P. 4/14

M&S

Ch 7 Applications of Differentiation x < e-1

x = e-1

x > e-1

-ve

0

+ve

dy dx

nature




S6/KC

Minimum point

Second Derivative test 

Maximum point:

d2y <0 dx 2



Minimum point:

d2y >0 dx 2



The



Example 7.3.2

d2y d2y test is used when is easily obtained. (e.g. polynomial functions). dx 2 dx 2 Otherwise, the first derivative test is used.



Find the turning points of the curves y = x ln x and determine their natures.



dy = ln x + 1 dx d2y 1 = x dx 2

For turning point,

( )

dy =0, x = e-1 dx

∴ y = e −1 ln e −1 = −e −1

d2y dx 2

= x = e −1

(

∴ e −1 ,−e −1




)

1 > 0, e −1

is a minimum point.

Extra information:



Critical point: 

occurred when

dy dy = 0 or is undefined. dx dx

That means the turning points can be found at critical points.

7.4 Concavity
It is describing the direction in which the graph is curving / turning. 

Concave upward: bends up / convex downward d2y >0 [smile as it is positive!] dx 2 Concave downward: bends down / convex upward

 



d2y <0 dx 2

[Up +, down -]

P. 5/14

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Ch 7 Applications of Differentiation

Concave upward

Concave downward







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Relationship between f ′( x ) , f ′′( x) and the shape. sign of f ′( x) sign of f ′′( x) Characteristics of graph

＋

＋

Increasing, concave upward

－

＋

Decreasing, concave upward

＋

－

Increasing, concave downward

－

－

Decreasing, concave downward

shape

Points of inflexion: points across which the direction of concavity changes  Which means it satisfied (i)

d2y = 0 and around such a point, dx 2

(ii) the sign of y =



d2y changes. dx 2

Example 7.4.1  Find the intervals for y = xe − x , which is (i)

concave upward

(ii) concave downward (iii) Find any point of inflextion



dy d2y = e − x (1 − x ) , = e − x (x − 2) dx dx 2 d2y = 0, i.e. x = 2 dx 2 (i) The curve is concave upward when x > 2.

(ii) The curve is concave downward when x < 2. (iii) The point of inflexion is (2, 2e-2).

7.5 Asymptotes
It is a line which the curve approaches it nearer and nearer but never touches.


Vertical asymptotes:

vertical line, x = a, of the function f (x) iff (iff and only if) lim f ( x ) = ∞ (or − ∞ ) or lim− f ( x ) = ∞ (or − ∞ )

x→a +

x→ a

[a+ and a- means slightly greater than and less than a respectively.] P. 6/14

M&S


Ch 7 Applications of Differentiation

Horizontal asymptotes: horizontal line y = b of the function f (x) iff lim f ( x ) = b x→∞




S6/KC

or lim f ( x ) = b x → −∞

Example 7.5.1



Find the asymptotes for the graph of y = ex + 1



lim e x + 1 = ∞ , ∴ x → ∞ , the graph does not settle down. x→∞

(

)

(

)

lim e x + 1 = 1 , ∴ x → ∞ , it tends to y = 1

x → −∞

Q y → ∞ only when x → ∞ , the graph has no vertical

asymptote.

7.6 Problems on Maximization and Minimization
it involves the practical problems, which need to transfer the word problem to the mathematical model.


Procedure to due with the application problems






1.

Try to link up all the relationship by modeling / formulating.

2.

Locate the domain or range of the variable is.

3.

Find the turning point by

4.

Nature determining by performing the First or Second Derivative tests.

5.

Find the minimum or maximum value of f (x).

dy = 0. dx

Example 7.6.1



Two ships A, B are initially 100km due North and 80km due East from a buoy O respectively. At 0:00 midnight of 1st July, 2008, both ships start to sail, with the velocity 2km/h due South and 1km/h due East respectively.

2km/h

Find the time when they have minimum distance between each other.



Let the time traveled be t hour after staring

100km

Let the distance between each other be s s 2 = (100 − 2t ) + (80 + 2t ) 2

2

(Pyth. Th.)

80km

Differentiate both sides with respect to t 2s

1km/h

ds = 2(100 − 2t )(− 2 ) + 2(80 + 2t )(2 ) dt

ds 1 = (8t − 40 ) dt s For turning point, Checking,

ds dt

ds =0 ⇒t =5 dt

< 0, t =5−

ds dt

>0 t =5+

∴ When it is 5:00a.m. of 1st July 2008, they have minimum distance between each other. P. 7/14

M&S

Ch 7 Applications of Differentiation

S6/KC

7.7 Approximation
If y = f (x) is a differentiable function of x and ∆x =dx is a small increment (difference / changes) of x.
It is different from rate of change of y. [Reminder: rate of change is the derivative of a function could be observed as some of the physical quantities.]




By Differentiation  When x is increased by a small increment ∆x , it will produce a corresponding increment ∆y = f ( x + ∆x ) − f ( x ) From the First Principle of Differentiation, f ′( x ) = lim



 f ( x + ∆x ) ≈ f ( x ) + f ′( x )∆x = f ( x ) + dy If ∆x is sufficiently small, then ∆y ≈ f ′( x )∆x tends to dy = f ′( x )dx



So there are two meaning of

∆x → 0




dy dx

1.

Derivatives of y = f (x)

2.

Quotient of the differentials dy and dx. (Similar to the chain rule)

To Compute the approximate relative error and percentage error  As error is changing from time to time, it is more meaningful to compare the error at different situation. ∆Q  Relative error in Q = Q 

Percentage error in Q = relative error in Q × 100% Applied in the differential of y = f (x).  Actual Error: ∆y ≈ f ′( x )∆x ∆y f ′( x )∆x  Relative Error: = y f (x ) 




∆y ∆x



Percentage Relative Error:

f ′( x )dx ∆y × 100% = × 100% y f (x )

Example 7.7.1

 

Use differential to approximate Analysis:

  

3

123 .

43 = 64, 53 = 125 123 is closer to 125 than 64.

Let y = 3 x

1 dy = 2 dx 3x 3 When x = 125, dx = 123 – 125 = -2 [original value - x] dy = f ′( x )dx

P. 8/14

M&S

Ch 7 Applications of Differentiation =

1 3(125)

2 3

S6/KC

(− 2 ) = −0.0267

f ( x + ∆x ) ≈ f ( x ) + dy f (125 − 2) ≈ f (125) − 0.0267 3




123 ≈ 4.9733

Example 7.7.2



An oil spill is increasing such that the surface covered by the spill is always circular.

If the area

of the oil spill is increased at the rate of π cm2s-1 when the radius is 2 m (a) Find the rate at which the radius r of the surface is changing with respect to time when the radius is 2 m. (b) Find the relative rate of change of the surface area of the oil spill. (a) If the surface covered by the spill is always circular, Let A = πr 2 dA = 2πr dr



From the question, dA =π dt dA dA dr = ⋅ =π dt dr dt

π dr dr 1 =π ⋅ = = dt dA 2π (2 ) 4 ∴



when r = 2m

the rate = 0.25 cm / s

(b) Relative rate of change =

dA 2πr (dr ) 2dr = = A r πr 2

** EXTRA NOTES ** 1.

Procedure for sketching a graph.
Wholly or partly (i)

find the first derivative and determine the turning points.

(ii) find the second derivative and determine (1) concavity and (2) the points of inflexion. (iii) find x-intercept by putting y = 0, y-intercept by putting x = 0. (iv) find the horizontal or vertical asymptotes. (v) symmetries of f (x) (if necessary) i.

Symmetrical about…. 1.

x-axis

 

when the function is even powers of x If given y = f (x), then make x in terms of y, x = f-1(y) = g (y) ,

[inverse function if exists] g(y) = -g(y) P. 9/14

M&S 2.

3.

2.

Ch 7 Applications of Differentiation

S6/KC

y-axis



when the function is odd powers of x

 

f (x) = f (- x)

it is called even function

the origin



the equation remains unchanged

 

f (- x) = - f (x)

it is called odd function.

Polynomial functions
If it is in the form y = P ( x ) = a n x n + a n −1 x n −1 + ... + a1 x + a 0




Have no horizontal and vertical asymptotes.





Number of x-intercept < degree of the function. Example



A curve y = x(x - 2)(ax + b) touches the x-axis at the point where x = 2, and the line y = 2x at the origin. (a) Find the values of a and b. (b) Then, sketch the curve.



(a) y = x(x - 2)(ax + b) = ax3 + (b – 2a)x2 -2bx dy = 3ax 2 + 2(b − 2a )x − 2b dx

Sub

Sub

--

(1)

2a + b = 0

--

(2)

dy = 2 and x = 0 into (1) dx

[Since it touches the line y = 2x, and

dy = 0 and x = 2 into (1) dx dy = 2] dx

b = -1

So a =



(b)

y=

1 2

1 2 x(x − 2 ) 2

For turning points, from (1), x=

2 or 3

3 2 x − 4x + 2 = 0 2

x=2

d2y = 3x − 4 dx 2

P. 10/14

M&S

Ch 7 Applications of Differentiation

d2y dx 2

= −2 < 0 , x=

2 3

d2y dx 2

=2>0

--

S6/KC

(4)

x=2

 2 16   ,  is the maximum point and (2,0) is the minimum point.  3 27 

--

*

--

**

--

***

d2y 3 = 3x − 4 = 0 , x = 2 4 dx

From (4), it shows the sign of

d2y 3 changes near x = 2 4 dx

 3 75   ,  is the point of inflexion.  4 128  When x = 0, y-intercept = 0, i.e. x-intercept = 0.

3.

Rational Function
For the vertical asymptotes, P(x ) If f ( x ) = , where Q( x ) = 0 and P(x) and Q(x) are polynomial functions. i. Q(x ) If Q (k ) = 0 and P (k ) ≠ 0 ,

ii.

then line x = k is a vertical asymptote rule for rational functions.




For the horizontal asymptotes, P ( x ) a n x n + a n −1 x n −1 + ... + a1 x + a = i. If f ( x ) = Q( x ) bm x m + bm −1 x m −1 + ... + b1 x + b 1.

2. 3.

If n > m, f (x) becomes infinitely large as x → ∞ i.e. there is no horizontal asymptote a If n = m, lim f ( x ) = n [as the numerator and the denominator are divided by xn.] x→∞ bm If n < m, lim f ( x ) = 0 , y = 0 is a horizontal asymptote. x→∞

Given y =




ax + b a c , where ≠ , there are followings properties. cx + d c d

i.

x-intercept =

ii.

x=

iii.

−b −d , y-intercept = a c

−d a is the vertical asymptotes, y = is the horizontal asymptotes. c c dy ad − bc = dx (cx + d )2





If ad – bc > 0,

dy > 0 for all x. dx

If ad – bc < 0,

dy < 0 for all x. dx

In either case, there is no turning points.

P. 11/14

M&S 4.

Ch 7 Applications of Differentiation

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Partial Fraction
Procedure to perform partial fraction i.

Factorize the denominator completely

ii.

Show all the linear case 1. (x - 1) (x + 2) 2. (x – 1)3 3. (x - 1) (x + 2)2 4. (x2 + 2) (x -1)

iii.

 

(x – 1) and (x + 2)

 

(x – 1), (x + 2) and (x + 2)2

(x – 1), (x – 1)2 and (x – 1)3 (x2 + 2) and (x -1)

For the decomposited rational function, Numerator Q( x) = ∑ An x n while Denominator R( x ) = ∑ Bn +1 x n +1

 iv.

5.

The maximum degree of the numerator should be less than the denominator by 1.

Find the coefficient An by 1.

Substitution, by any number, better to substitute the roots of the denominator.

2.

Compare Coefficients

3.

Long Division (not preferable, as it is troublesome.)

Other application related to Rate of Change


Physics i. If s is the displacement traveled by a body in time t, ii.

the velocity v is defined by the rate of change of displacement

iii.

the acceleration a is defined by the rate of change of the velocity.




v=

ds dv d 2 s dv , a= = 2 =v dt dt dt ds

Cost i.

Define the notation C(x)

--

total cost for producing x items

R(x)

--

total revenue for selling x items

P(x)

--

total profit obtained from selling x items

P(x) = R(x) - C(x)

ii.

Marginal cost --

 iii.

The average cost, C ( x ) =

Marginal revenue --

 iv.

the rate of change of C with respect to x,

dC dx

C (x ) x

the rate of change of R with respect to x,

dR dx

1  R ′( x ) = p1 +  , it attains maximum when the demand has unit elasticity.  E

Marginal profit

--

the rate of change of P with respect to x,

dP dx P. 12/14

M&S v.

Ch 7 Applications of Differentiation

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Maximum profit attains when marginal revenue is equal to marginal cost.




Demand function i. x = x(p), relating the price of an item and the quantity x that will be sold




Elasticity of demand i. A measure of how a change in the price of a product will affect the quantity of demand. px ′( p ) Elasticity of demand E ( p ) = x( p )

x ′( p ) is negative in general, so E ( p ) is used for convenience. ii.

Elastic -- small change in price results in a great change in quantity demanded, E ( p ) > 1

iii.

Inelastic – a small change in price result a small change in quantity demanded, E ( p ) < 1

iv.

Unit elastic – any changes of price will result an equal percentage of quantity demanded,

E( p) = 1


Rate of change f ′(x ) f (x )

i.

Relative rate of change of f (x) =

ii.

Percentage rate of change of f (x) =

f ′( x ) × 100% f (x )

6. Extra Exercise (1)The demand equation for a given commodity is given by x( p ) = −30 p + 90 , 0 ≤ p ≤ 2 where p is the price of a commodity and x is the number of commodities sold. (a)Determine the elasticity of the demand at p = 1. Interpret the result. px ′( p ) p = x( p ) p =1 p − 3

= −0.5 p =1

It is an inelastic demand. The result shows that for 1% increases on the price, 0.5% decreases on the quantity demand. (b)Determine the range of p such that the elasticity of demand is elastic. Demand is elastic if i.e.

p < −1 p−3

or

p > 1. p −3 p >1 p−3

[sign changes as 2 > p > 0 from the question, it is negative] p>3–p

⇒ p>

or

p < p – 3 (for all p)

3 2 P. 13/14

M&S (2)The function y = (a)

Ch 7 Applications of Differentiation

S6/KC

x+a , where a, b are constants and b > 0, has relative extrema at x = 1 and x = -1. x2 + b

Determine the values of a and b. [a = 0, b = 1]

(b)Find the relative maximum and minimum points and the points of inflexion of the graph. −1  1  [max. 1,  , min  − 1,  , points of inflexion: 2   2 

 3 ,  3,   4  

 − 3  , (0, 0)]  − 3,   4  

(c)Show that y = 0 is the only asymptotes of the graph of f (x). (d)Sketch the graph.

P. 14/14