Problems in Differential CalculusDr N K Srinivasan

IntroductionDifferential calculus is a branch of

mathematics ,full of many applications and

basic to many fields of sciences and‾¯

engineering. In recent years, differential

calculus is widely studied by students of

management fields.

But the subject is often made dull and

routine for beginning students. Though a

few interesting and useful applications may

be given in a regular course, the treatment

is usually "drill type" and students are

made to learn skills by repetitive

problems. Soon enough ,the student loses

interest and may even begin to feel that

the relevance of this math to his work

would be limited.

In this tutorial, you will find several

problems of differential calculus

introduced and completely solved ---to

explore several applications and methods of

solution. Therefore these examples would

help you to reinforce and review the work

done in high school or college courses and

also strengthen your background in this

field.

I have included only selected problems ,

drawn from many standard text books of

Calculus. I do not claim any new material

but only a presentation of relevant

applications and techniques .

Problems1 The position of an object thrown in air

,with free fall , is written as follows:

S(t) = - 4.9 t 2 + vt + h

where 4.9 is g/2 and the acceleration due to gravity ,g =9.8 m/s.s and v is the

initial velocity and h is the initial height.

Given that v = 120 m/s and h =3

find the velocity of the ball thrown up after 5 seconds and after 10 seconds.

Velocity is given by V = d S / dt

Recall that if y = kt.t , then dy/dt = 2 kt,

So, differentiating the equation with respect to time t , we get

V (t) = - 9.8 t + v = -9.8 t + 120

So, V (5) = - 9.8 x 5 + 120 = 71 m/s

V(10) = - 9.8 x 10 + 120 = 22 m/s

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2 Find the equation of a parabola

y = a x .x + b x + c

that passes through (0,1) and is tangent to the line

y = x -1 at (1,0).

We have to find the three 'unknowns' ,namely a, b,c --the coefficients in the

polynomial for parabola.

So,we have to set up three equations:

Since the parabola passes through (0,1), we get:

1 = a.0 + b.0 + c ----> c= 1

Since the parabola touches the tangent line at (1,0), the point (1,0) also lies on

the parabola.

So, we get: 0 = a.1 +b.1 + 1

or a + b = -1 ----> b = -1 -a

We need to find the value of 'a' now.

The given tangent line at (1,0) has a slope m=1

Since it is tangent to the parabola, the slope of the parabola at that point can be

found:

For this, let us differentiate the equation of the parabola: dy/dx:

dy / dx = 2ax + b

At the point (1,0) dy / dx = 2a + b

This is equal to m = 1

So , m = 1 = 2a + b

1 = 2a + (-1-a)

a = 2

Now we have all the unknowns: a = 2; b = -1-a = -3; c=1

So the equation for this parabola:

y = 2 x.x - 3x +

1

3. For the function f(x) = √ x , find the equation for the tangent passing through

(-4,0) but not on the graph.

The slope for the tangent is found from the derivative of the function:

y = sqrt(x)

dy / dx = 1/ ( 2 √ x)

For the tangent line, the slope with any arbitrary point would be : m = y - 0 / x

- (-4) = y / x+4

Now y = √ x,

so m = √x / (x+4)

Equating dy/dx with m , we get:

2x = x+4

or x = 4

If X = 4, then y= sqrt(4) = 2

So the point P (4,2) is also on the tangent line.

Now we have two points on the tangent line: (-4,0) and (4,2)

Let us find the equation of this tangent line using

y = mx + c

0 = m (-4) + c

2 = m (4) + c

Solving these two equations: c= 4m;

8m = 2

or m= 1/4 and c = 1

So we have the equation for the tangent line as follows:

y = (1/4) x + 1 or 4y -x

=1

4 Find a "linear approximation formula" for

y = f(x)= using the derivative near a specified point, say P(4,8)

The equation for linear approximation around x =3 is as follows:

f(x) = f(4) + f' (x) (x-4)

The first derivative f'(x) = (3/2)

At x=4 , f'(x) = 3

So, the linear approximation formula:

f(x) = 8 + 3 (x-4) = 3x - 4

Test: Let us test this approximation formula by taking x= 4.2.

f(x) = f( 4.2) = = 8.607 [The actual value]

As a linear approximation:

f(x) ~ 3x4.2 - 4 = 8.6

The error = 0.007 or 0.007/8.6 x 100 = 0.08%

So, the linear approximation is a very good and useful method for computing such

functions in the neighborhood of the specified point.

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5 For the circle: x.x + y.y = 25

at point (4,3) ,find the equations for tangent and normal lines.

y =

Then differentiate with respect to x:

dy/dx = -(1/2) (2x/ ) = - x/y

At point (4,3), we get dy/dx = - 4/3

The equation for the tangent line has the slope m = dy/dx at this point.

Therefore we write the equation of the tangent with the slope- point form of a

straight-line equation:

y-y' = m (x - x')

y - 3 = [- 4/3](x-4)

3y -9 = -4x +16

3y + 4x - 25 = 0

Normal equation:

Recall that if m is the slope of one line, then m' is the slope of a line

perpendicular to that line:

m. m' = -1

m' = -1/m

The slope of the normal line = m' = 3/4

Then the equation for the normal line:

y - 3 = [3/4] (x - 4 )

Simplifying, we get:

4y -12 = 3x - 12

or 4y - 3x = 0

Alternate method:

We can also find the derivative dy/dx using "Implicit differentiation":

x.x + y.y = 25

2x + 2y. dy/dx = 0

or dy/dx = - x/y

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6. Orthogonal Trajectories

For the given two graphs, find the points of intersection and show the two curves are

orthogonal trajectories---that is, their tangents are perpendicular to each other.

[Use the graphing utility to draw the curves and find their tangents at the points of

intersection.]

2x.x + y.y = 6 ------(1)

y.y = 4x -------(2)

The first graph is an ellipse. The second graph is a parabola.

From the second equation, y.y = 4x, substitute y.y in the first equation:

2 x.x + 4x -6 =0

or x.x + 2x -3 =0

Factoring , we get ( x+3)( x-1) =0

The two solutions are x= -3 and x = 1

X= -3 is not in this domain.

For x = 1, we get y.y = 4

So y has two values : y= 2 or y = -2

The two points of intersection are:

P (1,2) and Q (1,-2)

For the point P, the slope can be found for the two graphs:

y.y = 4x

so 2y dy/dx = 4

or dy/dx = 2/y = 2/2 = 1

For the second graph : 4x+ 2y dy/dx = 0

dy/dx= - 2x/y = -1

So the two tangents are perpendicular to each other.The product of the two slopes

m.m' = -1

Likewise, for point Q,

dy/dx = 2/-2 = -1 for equation (2) and

dy/dx = -2x/y = -2/-2=1 for equation (1)

Thus the tangents are prependicular to each other for the two curves at point Q also.

Hence the two graphs, the ellipse and the parabola are orthogonal

trajectories.

7 A population of 100 bacterias is introduced in a soup and their number grows with

time according to the growth equation as follows:

P(t) = 100 [ 1 + ( 4t /(50 + t.t)]

where t is in hours.

Find the growth rate of bacterias [number/hour] after 2 hours ,that is t = 1 , t= 2

hours and t = 7 hours.

We find the derivative of the given equation for the population:

Let y = 4t/(50 + t.t)]

Using the quotient rule:

dy/dt = [50+t.t)4 - 4t(2t)] / (50+tt).(50+tt)

Simplifying , we get:

dy/dt = [ (200 - 4 t.t )/ (50+t.t)(50+t.t)]

At t=2 , dy/dt = 184/(54.54) = 0.063

Therefore dP/dt = 100 dy/dt = 6.3

Or approximately 6 bacterias per hour.

When t= 7, dy/dt= (200 - 196)/ 51.51 = 0.0015

dP/dt = 100 x 0.0015 = 0.1 --> almost zero.

For t= 10, dy/dt = [200 -400]/150.150 = -200/22500

= - 0.89

The growth rate decreases with time, reaches 0 at 7 min.

After that ,it is negative, which means that the population starts decreasing.

{ Note that the growth rate decreases with time since the second term has 50 + t.t in

the denominator and therefore the second term decreases with time.]

Note: Depending on the food available in the soup, the bacteria population will

increase and slowly reduce its growth rate.Then it reaches a constant value. This is

called "resource -limited " population. Find other examples.]

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8 An astronaut throws a stone upward ,standing on the moon. The height of the stone

as a function of time is given by:

h (t) = -2.7 t.t + 27 t + 6

Note that 27 ft/sec is the initial velocity of the throw and 6 ft, the height of the

astronaut.

1)Find the equation for the velocity and acceleration of the stone.

2)When does the stone reaches the maximum height on the moon?

3) What is the maximum height reached by the stone?

Velocity = v =dh/dt = -5.4 t + 27 [ feet/sec]

Acceleration = dv/dt = - 5.4 [ feet/sec.sec]

The stone reaches maximum height when V = 0

Therefore -5.4 t + 27 = 0

or t= 27/5.4 = 5 seconds

At 5 seconds, the height is :

h (5) = -2.7 (25) + 27(5) + 6 = 73 feet.

So the maximum height is 73 feet.

[Note that the acceleration due to gravity on the moon is nearly 1/6 of acceleration

due to gravity on Earth : g= 32 feet/sec.sec.]

----------------------------------------------------

Rate Problems

In these problems, we find an expression or function for rate of

change in one or more variables by differentiation. The rate of

change is often with respect to time. But you will come across

many examples where we measure the rate of change with other

variables like temperature or height and so on.

9 A circular oil spill develops on a nearby ocean.The

spill increases in radius at the rate of 100 meters per

hour. Find the rate of increase in area of the spill.

Area A = π r.r

Differentiating both sides with respect to time t,

dA/dt = π (2 r) dr/dt

Since dr/dt = 100 m/hour, we get

dA/dt = 6.28 r (100)= 628 .r

Note that the rate of increase in area is not constant

, but changes with the radius at any time!

If the instantaneous radius is 1000 meters,then

dA/dt = 6.28 x 1000 x 100 sq meter/hour.

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10 A hot air balloon rises vertically at the rate of 3 meters per second. It is

observed by a scientist located at 30 meters from the vertical.The scientist uses a

telescope to track the height of the balloon. What will be the rate of change of the

angle of elevation for observation , when the balloon is at a height of 30 meters.?

The problem is first modeled with a trigonometric

relation:

tan x = Opposite side/ adjacent side

where x is the angle of elevation.

The height of balloon is the "opposite side" or the vertical distance. The adjacent

side is the ground distance from the site of balloon launch to the observer which is

constant at 30 meters.

tan x = h(t)/30

where h(t) is the height, a function of time.

We differentiate both sides with respect to time 't':

d(tan x)/ dt = (1/30) d h/dt

If y = tan x,

dy/dx = sec x.sec x

Therefore, d(tan x)/ dt = d(tanx)/dx. dx/dt=

= (1/30) dh/dt

dh/dt= 3

dx/dt= (1/30) (3)

Since the vertical height is 30 meters and the ground distance is 30 meters, the

angle of elevation is 45 deg or cos x = 1/√ 2.

So, dx/dt = (1/10) (1/2) = 1/20 radians per second

The change is elevation angle is at the rate of (1/20) radians per second.

Note that as the balloon rises, the rate of change of elevation changes with time.

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11. The inside temperature of a refrigerator changes with time [ in hours] as

follows:

T = 700 { 1/ [t.t + 4t + 10]}

Find the rate of change of temperature T with time t at each of the times: at t=1

hour, t= 5 hours and t= 10 hours.

First we differentiate the expression:

y = 1/ (t.t + 4t + 10)

dy/ dt = -(2t +4) / (t.t + 4t + 10)ˆ2

dT/dt = 700 .dy/dt

We form the table for three specified times:

t dT/dt

1 -18.9

5 -3.22

10 - 0.71

Note that dT/dt is negative, meaning that the inner temperature is decreasing.

Further the rate of

change of temperature decreases with

time.

12 A ladder of length 25 feet is leaning against a wall. Let the height of ladder on

the wall be 'h'. The distance of the ladder from the wall on the ground be x.

The base of the ladder is being pulled away at 2 feet/sec. Find the rate at which the

height is decreasing when the ground distance x = 7 feet and x= 15 feet.

Using Pythagorian theorem,

h.h + x.x = s.s

The length of the ladder is constant ;so s.s is a constant.

h.h = s.s - x.x

Diffentiate both sides with respect to time 't':

2h dh/dt = - 2x dx/dt

or dh/dt = - (x/h) dx/dt

Given: dx/dt= 2

So dh/dt = -2 (x/h)

For x= 7 feet, h = sqrt(25.25-7.7) = 24

dh/dt = - 2 (7/24) = - 0.58 feet/sec

For x= 15 ft, h = sqrt(25.25-15.15) = 20

dh/dt = -2 (15/20)=-1.5

feet/sec.

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Using Mean Value Theorem of Diff Calculus

Mean value Theorem is considered as a fundamental theorem in differential calculus.

It states:

If a function f(x) is continuous in the closed interval [a,b] and differentiable in

the open interval (a,b) , then there exists a number c in (a,b) such that

f'(c) = [f(b) - f(a)]/ (b-a)

For the interval x=a to x=b,the line joining the f(a) and f(b) is called the Secant

Line.

The slope of this straight line ,call it 'm':

m = f(b) - f(a)/ (b-a)

The mean value theorem states that there is point between a and b ,call it 'c', at

which the slope to the graph is the same as that of the secant line.

Often we seek to find the point c .

Differentiate f(x) : then f'(c) = m

Equate f'(x) = m

Solving this equation gives x=c.

Then find the slope to the graph at x=c: f'(c).

We can find the equation of the secant line and the tangent line at x=c using y-y' =

m (x-x') :

y - f(b)= m (x-b) ---> the secant line

y-f(c) = m (x-c) ----> tangent line

12. For the function f(x) = sqrt(x) in the interval [1,9], find the equation for the

secant line and the tangent line using "Mean Value Theorem":

Secant slope = [f(9) - f(1)]/ (9-1) = (3-1)/8 = 1/4

The equation for the secant line ,taking the point (1,1)becomes:

y -1 = (1/4) (x-1)

Simplifying : 4y -x -3 =0

Using "MV T", slope of the tangent

= dy/dx= (1/2)(1/sqrt(x))

dy/dx = m = 1/4 = (1/2) (1/sqrt(x)

Solving we get x= 4

Therefore c = 4

At c=4, f(c)= 2

So the equation of the tangent line:

y - 2 = (1/4) (x-4)

Simplifying , we get 4y -x -4 =0

Note that the MVT offers a quick way of finding the slope for the graph using secant

line. If the interval is small, then this slope is close to the slope at the midpoint

of the interval.

Mean Value Theorem finds applications in numerical methods.

Note: There is another 'mean value theorem' in Integral calculus.Watch out!

----------------------------------------------------

13 A stone is dropped from a height of 500 meters. The equation for the position of

the stone is as follows:

S(t) = -4.9 t.t +500

Find the average velocity in the first three seconds.

Using the MVT, find the time at which the velocity is the same as the average

velocity.

Average velocity = [S(3) - S(0)]/3

= [-4.9 x 9 +500 - 500]/3

= -14.7 m/s

Velocity at any given time = V = dS/dt = -9.8 t

Using the MVT, we get V = - 9.8 t = - 14.7

Therefore t = 1.5 second.

At 1.5 sec, the instantaneous velocity is equal to the average

velocity.

Note: For a vehicle covering a distance of d miles in T hours, the average velocity

is v'= d/T.

MVT says that though instantaneous velocity may change, there is a time at which

this velocity equals the average velocity, if the vehicle is continuously running

from time t=0 to t.

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14 For the quadratic polynomial,

y = Axx + B x + C in the interval [a,b]

show that the point c using MVT is the midpoint of the interval.

For x = a , we get y1 = A aa + B a + C

For x = b , we get y2 = A bb + B b + C

The secant line slope becomes:

m= (y1-y2)/( a-b) = [ A (a.a - b.b) + B (a-b)]/(a-b)

= A ( a+b) + B

The slope of this polynomial dy/dx = 2Ax + B

We have to find x such that dy/dx = m at point c.

A(a + b) + B = 2 A c + B

or A(a+b) = 2 Ac

Therefore c = (a+b)/2

or c is the midpoint of the interval x=1 and x=b.

The slope at the midpoint of the parabola given by equation is equal to the secant

slope.

====================================================

Optimisation problems

1 A square sheet of metal of size 18 in is taken and four square corners of size x

are cut and removed. Then the sheet is folded into a square tray, with depth equal to

x. This is an open tray, without lid. What should be the value of x such that the

volume of the tray is a maximum?

Volume of the tray V = (18-2x)(18-2x)x

= [4x.x - 72 x + 324]x

To find the maximum of V with respect to x, find

d V/d x and set it to zero:

dV/ dx = (4x.x-72x+ 324) + x(8x - 72)

= 12 x.x -144x + 324 = 0

x.x - 12x + 27 = 0

(x-9)(x-3) = 0

So, x= 9 or x = 3

X=9 is not a meaningful solution for the size of sheet is only 18 inches.

So we take the solution as x= 3

Let us make sure that the solution gives a maximum value of V, by taking the second

derivative:

f"(v) = 24 x - 144 = 24 (3) - 144= 72 - 144= -72

So the value of x=3 gives a maximum value of volume since the second derivative is

negative.

For x = 3, the volume of the tray V =(18-6)(18-6)3=

=432 Cu in.

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2 Oftentimes we try to minimise the surface area of a prism or container to reduce

the cost of making the prism or container. At the same time we would like to find the

appropriate shape, for instance, for a cylinder the ratio of diameter to height or

base area to height and so on. Such problems are easily solved by applying the

differential calculus.

For an open box with square base of size x and height h, to have a volume of 144

cubic in, find the base and height such that the surface area is a minimum.

Volume of the box = V = x.x.h = 144

Surface area = A =x.x + 4 x.h

We wish to minimize the surface area, A.

Since there are two variables in the left side, namely x and h, we should first

eliminate one of the variables say,h.

Using the function for the volume, we get:

h = 144/(x.x)

So the modified function for surface area:

A = x.x + 4 x (144/x.x)

= x.x + 576/x

To find the min of A with respect to x, differentiate A with respect to x and set it

to zero:

dA/dx = 2x - 576/(x.x) = 0

Or x.x.x = 288

or x = 6.6 in

To check whether this lead to minimum, let us differentiate d A/dx again:

f"(x) = 2x +288/(x.x.x)

This is positive and hence x=6.6 in corresponds to the min surface area for the box.

Check: h = 144/(x.x) = 144/(6.6 x 6.6) = 3.3

V = 6.6 x 6.6 x 3.3 = 143.7

Note: For this problem we find that h = x/2 .

This is a general result that h = x/2 where x is the size of the base of the box.

Therefore for a open box with square base, the height would be half of base for

reducing the box material required.

Check with commercial boxes and write an essay on this!

-----------------------------------------------------

3. Redo the problem for a box with a lid:

h = 144/ (x.x)

A = 2xx + 4x(144/(x.x))

dA/dx = 4x - 576/(x.x) = 0

x.x.x = 144 Or x = 5.25

The h = 5.22

For a minimum of surface area, height is nearly equal to the base. This is indeed a

useful result ---make chunky boxes!

Volume v = 5.25 x 5.25 x 5.22 = 143.8

By this result we are 'proving' that a cube shaped box is an optimal design

!.

4 For a cylindrical can, find the radius and height that will minimise the sheet

metal if its volume is 256 cubic inches.

Given : V = π r.r.h = 256

To minimise: Surface area= A = 2.pi.r.r + 2.pi.r.h

= 2.pi r ( r + h)

We eliminate h from the equation for A:

h = 256/(pi.r.r)

So, A = 2. pi.r.r + 512/ r

Differrentiate A with respect to r and set it to zero:

dA/dr = 4 pi. r - 512/(r.r) =0

4.pi.r.r.r - 512=0

r=

= 3.4 in.

and h = 256/ (pix 3.4 x 3.4) = 7.05 in

Let us check the volume = V = pi.(3.4 x 3.4)7.05 = 255.9 cu in.

Note that h = d = 2r.

This can be shown as a general result for cylinder of any volume.

Exponential and Logarithmic Functions

Exponential function and logarithmic function are inverse of each other:

Let y = f(x) = eˆx

Switch x to y and y to x:

x = e ˆy

Now y = ln x

Therefore exponential and log functions are inverse of each other.

If y = exp(x)

then dy/dx = exp(x)

The function and the derivative are the same for exponential function.

1 Find the slope of the tangent to the function

y=f(x)=exp(3x) at the point (0,1)

Let u= 3x

Using Chain rule:

dy/dx = dy/du. du/dx = exp(3x).3

At x =0, dy/dx = 3

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2 Find the derivative of dy/dx :

x exp(y) -10 x + 3y = 0

x exp(y)dy/dx + exp(y) -10 + 3 dy/dx = 0

dy/dx[ xexp(y) + 3} = 10 - exp(y)

dy/dx = {10 - exp(y)]/(xexp(y) +

3)

3 Find the derivative of

f(x) = y =x.xexp(x)- 2x exp(x) + 2 exp(x)

Factoring out,

y = exp(x)[ x.x - 2x+2]

dy/dx = exp(x) { x.x - 2x + 2] + exp(x)(2x-2)

=exp(x) [x.x ]

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Hyperbolic Functions

sinh(x) = 1/2[exp(x) - exp(-x)]

cosh(x) = 1/2 [ exp(x) + exp(-x)]

tanh (x) = sinh(x)/cosh(x)

3 Find the derivative of sinh(x)

y= sinh(x)

dy/dx = (1/2)[ exp(x) + exp(-x)] = cosh(x)

If y = cosh(x) , dy/dz =

sinh(x)

3 The yield in millions of cubic feet per acre of lumbar with time (in years) is

given by :

Y = 7 exp(- 48/t)

When t approaches infinity, we get y = 7

This is the limiting value for y.

Find the rate at which y is varying when t = 20 years and when t= 60 years.

dY/dt = 7.exp(-48/t) - 48/t.t

For t = 20, we get : dy/dt = -336 exp(-2.4). (1/400)

= 0.84 exp(-2.4)=0.076 cubic feet per

year.

4 A bacteria for a terrible disease doubles every 30 minutes.If a soup contains 5000

of these bacterias,after t hours we will have

Find the rate of increase in population at 1 hour and 5 hours and interpret the

results.

Taking log both sides, we get

ln P = ln 5000 + 2t ln 2

Differentiating,

(1/P) dP/dt = 2 ln 2 = 0.69 x 2= 1.38

dP/dt = 1.38 P (t)

For t = 1, dP/dt = 1.38 (5000 ) 4 = 27600 per hour

For t = 5, dP/dt = 1.38 (5000)(8.8.16)=7065,600 per hour.

Since they double every 30 minutes, the rate also increases with increasing

population. Note that the rate itself depends on population at that time.

----------------------------------------

Logarithmic function: if y = ln x,

dy/dx = 1/x

4 Differentiate y = x ln x

dy/dx = ln x + x/x = 1 + ln x.

5 Differentiate y = ln (x.x + 2x + 3) and find the relative extrema {maxima or

minima]

dy/dx = [1/(x.x + 2x + 3)] ( 2x + 2)

Set dy/dx = 0 ---> 2x+2 =0

we get x= -1 , y= ln 2. This point is a relative

minimum.

Logsrithmic Differentiation

6 Find dy/dx by logarithmic differentiation:

y = x .sqrt(x.x - 1)

We take log on both sides:

ln y = ln x + ln (sqrt(x.x -1)

ln y = ln x + (1/2) ln (x.x-1)

(1/y)dy/dx= 1/x + (1/2) [1/(x.x-1)] 2x

= 1/x + x/(x.x-1) = (2x.x -1)/ [x(x.x-1)]

----------------------------------------------------

6 Find the derivative of y = (x-1)/[x tan x]

Taking log both sides, we get

ln y = ln (x-1) - ln [ x(tanx)]

= ln (x-1) -ln x - ln (tanx)

= ln (x-1) - ln x - ln (sin x) + ln (cos x)

Differentiate:

(1/y) dy/dx = 1/ (x-1) - 1/x - (1/sinx)cos x + (1/cosx).(-sin x)

Simplify

-------------------------------

7 Find the derivative for:

y = (sin x cos x)/ sqrt(x.x.x-4)

Use logarithmic differentiation for easy work:

Take log on both sides:

ln y = ln (sinx) + ln (cos x) - (1/2) ln ( x.x.x -4)

Differerntiate with respect to x:

(1/y) dy/dx = cos x/ sinx - sinx/cosx -(1/2) (3x.x) )/(xxx-4)

dy/dx = y [ cotx - sin x -(3/2) (x.x)/(x.x.x -4)

Extrema of functions

8 Find the relative extrema and points of inflection for the function:

y = x.x/2 - ln x

dy/dx = x - 1/x

Set dy/dx = 0 ---> (x.x-1)/x= 0

or x =+1 or x = -1

Let us find the second derivative: y" = 1 + (1/x.x)

For x =1, y" = 2--> a positive number ; so x=1 is a minimum.

[For ln function, x=-1 does not exist.]

x=1, y= 1/2 is the min

point.

9 Find the extrema for the function and also point of inflection if it exists:

y = f(x) = (1/√(2pi)) exp(- (x.x/2))

y' = (1/2pi) exp(-(x.x/2) .-x

Set y'= 0 --> x=0, y = 1/√(2pi)

At maximum occurs at x= 0.

y" = (1/sqrt(2pi) [ exp(-x.x/2)x.x + (-1)exp(-x.x/2)]

To find point of Inflection, set y" = 0

We get: x.x -1 =0 or x = +1 or x = -1

So the point of inflection occurs at

x= + (1/√2pi) or x = -1(√2pi)

y = 1/√(2.pi.e)

Note: This function y is the well known Normal distribution density function or

Gaussian , Bell shaped curve.See in your graphing utility. It is central to much of

probability

work!

Growth [Logistics] curve

Many growth processes are modeled with Sigmoidal or Logistics Growth curve where the

growth rate keeps increasing fast and after sometime, the growth rate decreases and

comes to zero; further growth stops.

The point of inflection [where f"(t) =0] gives the point in time after which the

growth rate starts decreasing. This happens at approximately half the value of steady

value of f(t).

The function used is:

y = f(t) = N / [1 + a exp(-bt)]

The steady value of y after a long time is N, that is when t goes to infinity.

The point of inflection occurs at t = t* when y = N/2.

1 Find the first derivative and the point of inflection for the Growth curve

[Gompertz relation] for N = 1 and a = 1.

y = f(t) = 1/[1+ exp(-bt)]

We have simplified the function by taking N =1 and a =1.

where

f'(t) = [-1 /x.x] (-b exp(-bt))= {1/xx}b exp(-bt)

f"(t) = .....................

= [-2b./x.xx].(dx/dt)exp(-bt) + {1/xx}(-bb)exp(-bt)

Setting f'(t) = 0, let the numerator be zero:

[2bb exp(-2bt) + x (-bb)exp(-bt) ]= 0

2exp(bt) - x =0

2exp(-bt) - 1 -exp(-bt) =0

exp(-bt) = 1

At the point of inflection,

t= t*, exp(-bt*) = 1

f(t*) = N /(1+ 1) = N/2

The point of inflection occurs when y = N/2

Note: Draw this curve using graphing utility with N = 1000 and a= 1, b = 0.1

Note the point of inflection.]

--------------------------------------------------

1 In a game preserve, the deer population grows according to the growth curve:

P(t) = 1000 /[1 + 10 exp(-0.17 t)]

where t is in months.

a) draw the graph and study its features, the point of inflection and the early

growth rate.

b) Find the rate of growth of deer population after 5 months.

c) find the point of inflection, the time when P"(t)

=0.

2 Asimpler growth model is the modified exponential form:

The sales of a new model of mircowave oven follows the equation:

S(t) = 1000(1-exp(-0.17t)

a) Find the time at which the sales reaches 500.

b) Find the rate of growth in sales after 2 years.

For S(t) = 500, 1- exp(-0.17t)= 0.5

exp(-0.17t)= 0.5

ln (0.5) = - 0.17t

t = 5.88 years

Rate of growth = P'(t) =1000 x 0.17 exp(-0.17 t)

For P'(2) = 0.17 exp(-0.34) = 121 ovens per year.

--------------------------------------------------------

Newton's method --to solve non-linear equations

This method, using first derivative, is useful in solving non-

linear equations or finding roots of equations--if it works.

Let us find the square root of 2.

The equation to solve: y = f(x) = x.x - 2 = 0

newton's method uses an approximate solution as the

seed to start the "iterative" process.Let the seed solution x(0) be 1.0.

Then x = x(0) - f(x(0)/ f'(x(0)

f'(x) = 2x

f'(x(0) = f'(1) = 2

f(x(0) = f(1) = -1

Then x = 1 - (-1)/2 = 1 + .5 = 1.5

This is the first solution. Using this ,we find the second better solution:

x(0) = 1.5 f(1.5) = 2.25-2 = 0.25

f'(1.5) = 2 x 1.5 = 3

So: x = 1.5 - 0.25/3 = 1.5 - 0.0833 = 1.4167

Let us "iterate" once again:

x(0) = 1.4167

f(1.4167) = 1.4167 x 1.4167 - 2 = 0.0070388

f'(1.4167) = 2 x 1.4167 = 2.8334

we get the next value:

x = 1.4167 - 0.00704/2.8334 = 1.4167 - 0.002485

x = 1.414215

The Calculator would give: sqrt(2) = 1.4142135

Our calculation by Newton's method after 3 iteration is accurate to 5 decimal places!

[Note: Most probably ,your calculator uses Newton's method it find the root; it can

do the 'algorithm' very fast!.]

-------------------------------------------------------- 2 Find the square root of 3

by Newton's

method.

Newton's method is helpful in solving polynomials, provided you could make a quick

guess of the root for initial seed value.

3 Solve x.x.x +x - 1 =0 with three iterations.

y = f(x)= x.x.x +x -1 =0

If we take x =1 as type solution, f(x) 1+1-1 = -1, not close to zero.

Let us guess x = 0.5 Then f(x) = 0.125 + 0.5 -1= -0.375

x(0) = 0.5 f(.5) = -0.375

f'(x)= 3x.x +1 ---> f'(0.5) = 0.75 +1 = 1.75

First iteration: x = 0.5 - (-0.375)/1.75 = 0.7142

x(0)= 0.7142 f(0.7142) = 0.3643

f'(.7142) = 3 x 0.7142 +1=3.1426

Second iteration: x = 0.7142 - 0.3643/3.1426 =

x = 0.5983

Third iteration:

x(0) = 0.598 f(.598) = 0.2138 + 0.598 -1= -.1882

f'(0.598)= 1.794+1= 2.794

x = 0.5983 - (-0.1882)/2.794 = 0.6667

Check: f(0.6667) = 0.2958 + 0.6667 -1=-0.0374 -->close to

zero.

4 Solve the two equations simultaneously;

f(x) = 2x+1 and g(x) = sqrt(x + 4)

h(x) = f(x)-g(x) = 0 where f(x) = g(x)

h(x) = 2x+1 - √ (x+4) = 0

h'(x) = 2 +(1/2√(x+4))

Let us take x(0) = 1

x= 1-h(1)/h'(1) = 1 - 0.764/2.2237 = 0.6564

x(0)= 0.6564

h(0.656) = 1 + 1.3128- sqrt(4.6564) = 2.3128-2.1579=0.1549

We are getting close to the root!

h'(0.656) = 2 + (1/2 x 4.6564)= 2 + 1/9.328 = 2.1072

x = 0.6564 - (.1549/2.1072) = 0.6564 - 0.07351

x = 0.58289

Let us check : h(x) = h(0.5829) = 2.1658 - 2.1407=0.0251 ---very close to zero.!

--------------------------------------------------

Note: Newton's method is a fast numerical procedure to find solutions ,provided it

works! it will not work if f'(x) at the approximate root is zero or close to zero.

The second term will blow up!!.

Further , you must start with a value close to actual root as your seed

solution.

5 Find the root of the equation to three decimal places using Newton's method:

exp(-x) -x = 0

f(x) = exp(-x)- x

f'(x) = -exp(-x) - 1

let the initial root be x = 0.5

x = 0.5 + {exp(-0.5)-0.5)/[1+exp(-0.5)]

exp(-0.5) = 0.6065

x = 0.5 +.1065/1.6065 = 0.5 + 0.06629 = 0.56629

Second iteration

f(x) = exp(-.566) - 0.566 = 0.00179

f'(x) = -exp(-0.566) -1 = - 1.56779

x = 0.566 + 0.00179/1.56779 = 0.566 + 0.001142=0.56714

----------------------------------------------------

Historical Note:Newton's method is often called Newton-Ralphson method,because Joseph

Ralphson independently developed this method in 1690; though Newton did this work in

1671, Newton's book was published much later, only in 1736.

--------------------------------------------------------

Linear approximations for functions

If y = f(x), we find the linear approximation as follows at x= 1

P(x) = f(x) + f'(x)(x-1)= f(1) + f'(1)(x-1)

1 Find the linear approximation for y = 1/sqrt(1-x)

around x =0

y'= dy/dx = (-1/2)(1/(1-x).sqrt(1-x))

y'(0) = -1/2

f(x) = f(0) + (-1/2)(x-0)= 1- (x/2)

Check: Find y = 1/sqrt(1-0.3)= 1/sqrt(0.7)

Using the linear approx: y = 1-x/2 = 1-0.3/2 = 0.85

The actual value of sqrt(0.7) =0.8366

The error is 0.014 or 0.014/0.836 x 100 = 1.67 %

----------------------------------------------------

2 Find the linear approximation for y = sin x

near x =0 [x is in radians]

y= sin x

y' = cos x y'(0) = 1

y = sin(0) +y'(0)(x-0)= y'(0) (x)

y = 1.x

sin x = x

Check: Calculate sin 5 deg using linear approximation and check with calculator

value.

5 deg = (5/180)(3.1416) = 0.0873 rad

sin (.0873) = 0.0873

The calculator gives: sin (5 deg) = 0.087156

The error is 0.0002

---------------------------------------------------

Extrema [maxima or minima ] problems some more problems

1 Find the relative maxima or minima for

y = x.x.x - 3x.x +3

Find the point of inflection.

Differentiating y with respect to x and set to zero:

dy/dx = 3x.x -6x =0

or x(x-2) =0

The roots are : x=0 and x = 2

y" = 6x-6

For x =0, y" is negative --> so the point P(0,1) is a relative maximum

For x = 2, y"= 12-6 = 6 is positive--> so the point Q(1,1) is a relative minimum.

For point of inflection, set y" = 0

6x -6 = 0 or x=1, y=1.

[Note: draw the graph with graphing utility and study the figure.--------------------

--------------------

------------------------------------------------------

2 Find the extrema for :

y = x.sqrt(4-x)

dy/dx = y' = sqrt(4-x) + x (-1/2)(1/sqrt(4-x))

= sqrt(4-x) [1 - x/(2(4-x)}

Setting dy/dx = 0 and simplifying , we get

(4 -x) - (x/2) =0

x= 8/3

*Show that the second derivative y" is negative for x = 8/3.

Hence x=8/3, y= (8/3)sqrt(4/3) or (8/3, 16/(3.√3)) is a maximum point.

To check whether this point is a maximum, let us take x= 2 and x = 3 .

f(2) =2sqrt(2)=2.828

f(3)=3 f(8/3) = 16/(3(1.732)) = 16/5.196=3.07

So, x=8/3 is a relative maximum point.

-------------------------------------------------------

3 Find the closes points on the graph of y = -x.x +4, a parabola, to the point (0,1)

on the y axis.

Using the distance formula:

d = sqrt( (x-0)(x-0) + (y-1)(y-1)]

=sqrt(x.x + (y-1)(y-1)]

We can mimimize the function which is d.d:

f(x) = x.x + y.y - 2y + 1

Substitute for y: y = -x.x + 4

We get:

Z= f(x) = x.x + y.y - 2x.x -8+1

= -x.x + (4-x.x)(4-x.x) -7

= -x.x + x.x.x.x -8x.x +16 -7

= x.x.x.x - 9x.x +9

Find z' and set it to zero:

z' = 4x.x.x -18 x =0

= x(4x.x-18)=0

The roots are : x= 0 and x = + sqrt(9/2) and x = -sqrt(9/2).

The closest points are +[sqrt(9/2), 1/2] and (sqrt(-

9/2),1/2]

4 Find any relative minima or maxima for the function:

y= f(x) = x.x.x.x/4 - 8x

f'(x) =x.x.x -8 =0

x = 2--relative minimum.

-------------------------------------

Parametric Equations

Consider the functional relationship between y and x:

y = f(x)

Let t be a variable or 'parameter' common to both x and y.

Then we can find relationship like this:

x = f(t)

y = g(t)

Then we can find the derivative dy/dx as follows:

Using the chain rule:

dy/dx = (dy/dt)(dt/dx)

dt/dx = 1/(dx/dt)

1 Find dy/dx if

x = √ t

and y = t.t/4

dx/dt = (1/2)[1/sqrt(t)]

dy/dt = t/2

So, dy/dx = (t/2)(2 sqrt(t) = t.sqrt(t)

We can check the result by writing:

y = t.t/4 = x.x.x.x/4

dy/dx = x.x.x =

t.sqrt(t)

Note: The common 'parameters' in real applications are time,temperature, angle,height

and so on.

2 Find dy/dx if x= 2 sin t and y= 3 cos t

dx/dt = 2 cos t dy/dt = -3 sin t

dy/dx = -3sint/(2cost) = - (3/2) tan

t

3 For a sphere, volume v= (4/3) pi. r.r.r where r is the radius.

Surface area = A = 4.pi.r.r

Find dA/dv

dA/dr = 8 .pi.r

dV/dr = 4.pi.r.r

So, dA/dv = (dA/dr)(dr/dV) = (8pi r)/(4 pi r.r) = 2/r

Note that as r decreases, this rate increases to higher values.

Note:For a heated sphere, heat content in calories is proportional to volume, while

the rate of cooling is proportional to surface area. Thus dA/dv is related to cooling

rates of spheres of different diameters. Small spheres or balls cool faster than

bigger

ones.

4 For the graph of y = f(x), using the parametric equations x=2t and y = t.t- 1, find

dy/dx at t=2 and find the equation for the tangent at that point.

x= 2t --> dx/dt = 2

y = t.t-1 --> dy/dt = 2t

dy/dx= 2t/2 = t

At t= 2, dy/dx = 2

At t=2, x= 4, y=3

So the tangent equation: y - 3 = 2(x-4)

Simplifying: y - 2x = -5

------------------------------------------------

5 The electrical resistance of a wire varies with temperature as follows"

R = 0.01 + 0.102 T - 0.003 T.T

The temperature of a furnace varies with time t as follows:

T = 100 + 4 sqrt(t) where t is in minutes.

Find how resistance varies with time:

dR/dt = dR/dT. dT/dt

dR/dT = 0.102 - 0.006 T

dT/dt = 2/ sqrt(t)

So, dR/dt = (0.102 - 0.006T)2/sqrt(t)

-----------------------------------------------

Differentials

Consider two points P(x,y) and Q(x',y') very close on the graph of y = f(x)

Let us call: y-y'= δy and x-x' =δ x

[ delta y and delta x ,which means small y and small x}

delta-y and delta- x are called 'differentials'.

Let (x',y') be the given point or chosen point.

Take the slope dy/dx = f'(x) at the given point : f'(x').

Since the changes are small, we approximate:

or

This equation, though approximate, enables us to estimate the delta y for a given

delta x.

This can be used to estimate small changes in functions when x changes a bit and also

for error estimation in measurements.

1 Macdonald , a farmer, measures the area of his circular field by measuring its

radius with a measuring tape which has an error of 1 in. If the radius is 5000

inches, find the error Macdonald has to include in the area measurement.

We should begin with the basic equation always:

area A = pi.r.r where r is the radius which the farmer measures.

Differentiate A with respect to r:

dA/dr = 2.pi.r

the small change or error in r : δr = 1 in

Let δ A = (dA/dr) δr

dA/dr at r= 5000 becomes dA/dr = 2.pi.5000= 31400

So the error in area = δA = 314000. 1 = 31400 sq in.

------------------------------------------------------2 2 Find the sqrt(17) .

Using differentials, take the chosen point as x=16 , y= 4 .

Start with the basic equation: y = f(x) = sqrt(x)

dy/dx = (1/2)1/sqrt(x)

dy/dx at the chosen point (16,4) becomes

dy/dx = (1/2)(1/4) = 1/8

so delta(y) = (1/8) delta x.

Since we wish to compute sqrt(17), delta x= 1

So: delta y = 1/8 .1

deltay = 0.125

So sqrt(17) = sqrt(16) + delta y

sqrt(17) = 4 + 0.125 = 4.125

Check with your calculator: sqrt(17) =

4.1231

We can extend this method of differentials to find relative errors or % errors in

measurements.

3 An electrical engineer measures the electrical resistance of wires. The electrical

resistance R depends on the length of the wire and its area (pi r.r) where r is the

radius.

R = k l/A where k is a constant called "resistivity".

Assume that the error in measuring length is small and can be neglected. But there

could be considerable error in measuring the radius. The engineer uses a screw gauge

to measure the radius and the error is +/- 0.01 mm.

Find the relative error in measuring the resistance for a wire of 1000 mm length and

3mm radius.

We use logarithmic differentiation for problems of this kind.

Taking log both sides:

ln R = ln k + ln l -ln (pi.r.r)

ln R = ln k + ln l - ln pi - 2 ln r

Taking differentials, d(ln R) = (1/R) dR

dR/R = -2 dr/r

The relative error in radius= dr/r

Replace dr by delta(r)

relative error in radius = δr/r = 0.01/3

Then relative error in resistance = δR/R = -2(0.01/3)= -0.0066

Note that the relative error in resistance varies with the radius of the wire.

To find the % error , multiply relative error by 100.

This the % error in resistance = 0.0066 x 100 =0.66%

-------------------------------------------------

4 Using the differentials, find the error in the range calculation of a projectile

when the angle of elevation is changed from 10 deg to 11 deg.

Range R = (v.v/32) (sin 2x)

where x is the angle of elevation of a gun, in radians. V is the initial velocity =

2200 ft per sec.

Find the derivative with respect to the angle x:

dR/ dx = (v.v/32) cos 2x.(2)= (v.v/16) cos 2x.

Taking differentials:

Convert dx into radians: 1 deg = 0.0174

δ R = [(2200)(2200)/16 ][. 0.936 x 0.0174] = 4927 feet.

------------------------------------------------------

5 Compute sin 46 deg using differential :

Given: sin 45 = cos 45 =0.7071

1 deg = 0.01745 radians

y = f(x) = sin x

f'(x) = cos x

f'(45) = 0.7071

= 0.7071 (0.01745)= 0.01237

sin 46 = sin 45 + δ y = 0.7071 + 0.01237 = 0.7195

The calculator gives:0.71934

The relative error = 0.0002/0.7193 = 0.000278 or

0.02%

6 The diameter of a sphere is measured to be 18 cms. If the error in measurement is

0.05 cm, find the error propagated to surface area and the % error involved.

A = 4 pi. r.r

delta A = 4.pi. 2r.δr

= 8 . 3.14.9 (0.05)= 3.6 pi

A = 4 x 81 pi=324 pi

% error in A = (3.6/324) x 100 = 1.11%

--------------------------------------------

Some Business Applications

1 A company found that the total cost of making mobile carts follows the equation:

C = 8x.x - 96 x +200

where x is the number of carts produced and cost is in thousands of dollars. Find the

number of units to be produced for a minimum of total cost.

[This is an equation of a parabola with a min at the apex.]

Differentiate C with respect to x:

dC / dx =f'(x)= 16 x - 96

Set dC/dx = 0 and solve for x:

16 x - 96 = 0

or x = 6

To check whether this result leads to minimum cost, differentiate again and find the

second derivative for

x = 6:

f"(x) = 16 which is positive.So x=6 represents a min point.

2 Find the number of units of mobile carts to be produced for a minimum value of

average unit cost?

The expression for the average cost C'(x) as function of the units produced is:

C'(x) = total cost/ x = (8x.x - 96 x + 200)/x

C'(x) = 8x - 96 + 200/x

[Note that the average cost formula is usually of the form: y = a x + b/x +

constant]

To find the units for min of C'(x), differentiate with respect to x and then set to

zero:

d C'(x)/ d x = 8 - 200/(x.x) = 0

x.x = 200/8= 25

So, x = 5

If number of units produced is 5 carts [per batch] the average cost will be at the

minimum.

{Note that the total cost is a minimum at 6 units, while the average cost is a min at

5 units!]

----------------------------------------------------

3 The 'Profit Equation' for profit as a function of number of pizzas sold in a Pizza

restaurant is as follows:

P = 2 x - (x.x)/20000 - 1000

for x pizzas, in the range of [0,40000]

Find the number of Pizzas for maximum profit.

dP/dx = 2 -x/10000 = 0

So, x= 20000.

-----------------------------------------------------

4 The inventory cost for a retailer for ordering and storing x units of bicycles is

:

C (x) = 2x + 320000/x

Find the min order quantity for this item.

Differentiate C(x) with respect to x and set it to zero:

d C/ dx = 2 -320000/(x.x) =0

or x.x = 320000/2= 160000

or x = 400

400 units can be ordered and

stored.

5 Profit = Revenue - Cost

Both revenue and cost are expressed as a function of the quantity produced ,x.

For a company making bikes,

P (x) = R(x) - C(x) (in millions of $)

R(x) = 20 x and C(x) = x.x.x -15 x.x + 38 x

Therefore,

P(x) = R(x) - C(x) = -x.x.x + 15 x.x - 18 x

Differentating P(x) with respect to x, we get:

dP/dx = -3x.x + 15 x - 18 =0

or 3 x.x - 15 x + 18 = 0

x.x - 5x + 6 = 0

or (x-3)(x-2)=0

or x= 3 0r x= 2

Now taking the second derivative, we get:

P''(x) = -6x + 15

For x = 3, P" (3) = -3 this means x=3 gives max profit.

For x = 2 , P'(2) = 3; this means x=2 gives min profit.

So, we choose the solution x = 3 for the max profit.

For x= 3, the max profit = P(3) =-27+ 135 - 54 = 54 (million $)

For x = 2, P(2) = -8 + 60 -36 = 16

--------------------------------------------------------

6 The demand function for cell phones is :

p = 90 - x

where p is the price at which it is sold and x is the price.

[In general, with increase in price, the demand decreases.]

The revenue becomes : R(x) = px = x(90-x)

The cost function for the same item for x numbers:

C (x) = 100 + 30 x

Find the price for maximum profit.

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

P(x) = x(90-x) - (100 + 30x)

= -x.x + 90x - 30 x -100

= -x.x + 60 x -100

[Note that this function is a parabola turned upside down with a maximum because the

function has -x.x term.]

To find the value of x for maximum profit,

set dP/dx = 0

dP/dx = -2x + 60 = 0

or x = 30

For x = 30, P(30) = -900 +1800 -100 = 800 $

To show that x =30 corresponds to the maximum of P, the second derivative is taken:

P"(x) = - 2 which is negative and so we have a maximum.

----------------------------------------------------

7 For the above problem, find the marginal revenue . Given the cost function,find

marginal cost and average cost.

At what level of x, the average cost is a minimum?

Revenue function: R(x) = 90x - x.x

Marginal Revenue: dR/dx = 90-2x

The given Cost function: C(x) =100 + 30x+ 0.25 x.x

Marginal cost = dC/dx = 30 + 0.5 x

Average cost function : c (x) = C/x = 100/x + 30 + 0.25x

The min of average cost: d c(x)/dx =0

or: -100/x.x +0.25 =0

or: x.x = 400 ---> x= 20

Min Average cost= c(20)= 100/20 + 30 + 0.25x20 = 5+30+5 =40

Marginal cost = 30 + 0.5 x20 = 40

Note that at x= 20, the min average cost = marginal cost.

This is a general

result!

8 Prove that the average cost is a minimum at the value of x where the average cost

equals the marginal cost.

The cost function be C(x)

Marginal cost = dC/dx

Now average cost= C' =C(x)/x

At min, dC'/dx =0 , using quotient rule,

x dC/dx-C(x) =0

or dC/dx = C(x)/x

which is: marginal cost = average

cost

9 A company estimates that it can sell N (x) items by spending on advertisement x

(million $) according to the equation:

N(x) = 500 -450 x + 60 xx - 2 x.x.x

Find the point of inflection such that the rate of change of sales with advertising

is positive [increasing] below this point and it is negative or decreasing beyond

this point.

Let us find dN/dx =N' and N'', the second derivative.

N'(x) = -450 + 120 x - 6xx

N''(x) = 120 - 12 x

Setting N''(x) = 0, we get

x =10

Therefore x=10 is the point fo inflection. This point is also called : "The point of

diminishing returns" by economists and business experts.

This point corresponds to the maximum rate of change :

N'(10) = -450 + 1200 -600 = 150 units/million $ of advertising.

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Quadratic Approximation

Quadratic approximation for a function y = f(x) involves the first and second

derivative.

Given x= a and y= f(a),we develop the approximation for values near x = a.

The approximations are:

P(x) = f(a) + f'(a) (x-a) [linear]

P (x) = f(a) + f'(a)(x-a) + (1/2) f" (a)(x-a)(x-a)

[Quadratic]

1 Find linear and quadratic [second order] approximation for y = sin x (where x is

in radians.)around x = 10 degree = 0.1744 rad

f(x) = sin (.1744) = 0.1736

f'(x) = cos x ---> f'(0.1744) = 0.9848

f"(x) = -sinx --------> f"(0.1744) =-0.1736

So we get linear approximation:

P(x) = f(10) + 0.9848 (x-0.1744)

sin x = 0.1736 + 0.9848( x -0.1744) (x in radians)

sin (0.2093) = 0.1736 + 0.9848(0.2093-0.1744)

sin (12deg)= sin (0.2093) = 0.1736 + 0.03436=0.20796

Quadratic approximation:

P(x) = - (1/2)0.1736 (x-0.1744)(x-0.1744)

Find sin 12deg using this approximation:

12 deg = 0.2093 rad

f(x) = 0.20796 - (1/2) (0.2093-0.1744)(0.209 - 0.1744)

= 0.20796 - (1/2)(.0349 x 0.0349)

= 0.20796 - 0.000609 =

So sin (12 deg) = 0.207351

The calculator gives : sin 12 = 0.2079117

The error is : 0.0006 = 0.0006/0.208 x 100 = 0.29%

------------------------------------------------------

2 Find linear and quadratic approximation for f(x) = exp(x) near x = 0. Calculate

exp(.2)

f(0) = 1

f'(x) = exp(x) f'(0) = 1

f"(x) = exp(x) f'(0) = 1

So we get:

P(x) = 1 + 1(x-0) + (1/2)1 (x.x)

So the approximation is : y = exp(x) = 1 + x + x.x/2

y(.2) = exp(.2) = 1 + 0.2 + 0.02= 1.22

The calculator gives: exp(.2) = 1.22140

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Taylor series expansion for functions

A function y = f(x) can be expanded into a power series and a polynomial using

derivatives .

Let the function be expanded around or 'in the neighborhood of ' x= a.

Find the following functions first:

f(x) at a ----> f(a)

f'(x) ---> f'(a)

f" (x) -----> f"(a)

f''' (x) -------> f'''(a )

and so on for higher order derivatives.

Taylor series which is an "infinite series" is written as follows:

f(x) = f(a) + f'(a)(x-a) + f"(a)/2 (x-a)(x-a) + f'''(a)/3! [ (x-a)(x-a)(x-

a)]+......to infinity.

You will note that the first two terms of this series are the same as the linear and

quadratic approximation given earlier.

If the series is expanded around x=0 or a=0, then we get a simpler series:

f(x) = f(0) + f'(0)x + f''(0)/2 [x.x] + f'''(0)/3! [x.x.x] + f''''(0)/4! [x.x.x.x]

......

This series is called MacLaurin Series.

In most cases of approximations, we use only the three terms [using third derivative]

as given in the examples .-------------------------------------------------------

1. Find the Taylor series expansion or Taylor polynomial for y= f(x) = sin x around

x=0 or a=0

f(0) = sin (0) = 0

f'(x) = cos x ---> f'(0) = 1

f''(x) = - sin x ----> f''(0) = 0

f'''(x) = - cosx ------> f'''(0) = -1

f''''(x) = sin x ------> f''''(0) = 0

f '''''(x) = cosx ------> f'''''(0) = 1

So Taylor series becomes:

f(x) = 0 + x +0 - x.x.x/3! +0 + x.x.x.x.x/5! +........

f(x) = x - x.x.x/6 + x.x.x.x.x/120

Find the value of sin 10 degrees using the above Taylor polynomial and compare with

table /calculator values.

10 deg = 0.1744 radians

sin (0.1744) = 0.1744 - +

= 0.1744 - 0.000884 = 0.173516

The last term is not included.

The calculator gives: sin (10)=0.1736481

The error is 0.0001

--------------------------------------------------

2 Expand y= f(x)= cos x at x= 0, using the three terms in Taylor polynomials .

f(x) = 1 - x.x/2 + x.x.x.x/24

---------------------------------------

3 Find the Taylor polynomial for y= f(x)= exp(x) at x= 0. Find the approximate value

for exp(0.1) with four terms.

f(x)= exp(x) f(0)= 1

f'(x) = exp(x) f'(0) = 1

f''(x) = exp(x) f''(0) = 1

and so on .

So we get:

exp(x) = 1 + x + x.x/2! + x.x.x/3! + x.x.x.x/4! +....

= 1 + x + x.x/2 + x.x.x/6 + x.x.x.x/24 ......

For exp(.1), we get:

exp(0.1) = 1 + 0.1 + 0.01/2 + 0.001/6 +0.0001/24......

= 1.1 + 0.005 + 0.0001666 + 0.0000041= 1.1051707

The calculator gives: exp(0.1) = 1.105171

--------------------------------------------------

Note: The student should read about the reminder theorem for Taylor series from the

text

book.

4 Find the Taylor polynomial for y= f(x)= ln x for a= 1.

f(x) = ln x f(1) = 0

f'(x) = 1/x f'(1) = 1

f''(x) = -1/(x.x) f''(1) = -1

f'''(x) = 2/(x.x.x) f'''(1) = 2

f''''(x) = -6/x.x.x.x f''''(1) = -6

The Taylor Polynomial :

ln x = (x-1) -(x-1)(x-1)/2 + (1/3)(x-1)(x-1)(x-1)

- (1/4) (x-1)(x-1)(x-1)(x-1).

5 Expand y= f(x)= ln (1+x) around x=0 and find

ln(1.1) using the four terms.

As before , we get

ln (1+x) =0 + x -x.x/2 + x.x.x/3 -x.x.x.x/4

ln (1.1) = 0.1 - 0.01/2 + 0.001/3 - 0.0001/4

=0.095308

The calculator gives:

0.0953101

6 Expand in Maclaurin Series.

Using the expansion for exp(x):

Replace x by x.x/2:

-------------------------------------------------

7 Find the Taylor series polynomial for f(x)=sec x around a=0.

f(x) = sec x f(0) = 1

f'(x) = sec x. tan x f'(0) = sec (0)tan (0) = 0

f"(x) = sec x . secx.sec x + tan x.sec x. tan x

tan(0) =0

So f" (0) = 1

f"'(x) = 3 sec x sec x .sec x tan x + sec x 2 tanx sec x f"'(0) = 0

f""(x) = 3 secxsecxsecxsecxsecx +2 secxsecxsecxsecx+......

f""(0) = 5

f(x) = f(0) + f"(0)x.x/2 +5 f""(0)x.x.x.x/4!

= 1 + x.x/2 +

5x.x.x.x/4!

8 Find the Taylor polynomial for y = tan x around x = 0.

f(x) = tan x f(0) = 0

f'(x)= secx.sec x f'(x) = 1

f"(x) = 2 sec x. secx tan x f''(0) = 0

f"'(x) = 2 secx.secx.secx.secx + 2 tan x d(sec x)/dx = 2

tan x = x + x.x.x/3 + ......

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Additional Problems

1 A study showed that the relationship between systolic blood pressure and weight for

nearly 5000 normal children followed the equation:

p(x) = 17.5 (1 + ln x)

where x is the weight in lbs and p is the pressure in mm of mercury.

Find the rate of change in pressure with weight at 40 lb level and 90 lb level.

dp/dx = 17.5 (1/x)

At 40 lbs, dp/dx = 17.5/40 = .4375

At 90 lbs level , dp/dx = 17.5/90

=0.1944

2 A drug was added to a bacteria colony to increase its population. The population

after t minutes, is as follows:

P(t) = 1000 + 30 t.t -t.t.t

for t [0,20].

i) What is the maximum rate of growth?

ii) Find the inflection point after which the rate of growth starts decreasing.

dP/dt = P'=60 t -3 t.t

Point of inflection: P" = 60 -6t =0

t =10 minutes.

The growth rate starts decreasing after 10 minutes.

The maximum growth rate at 10 min. = 60 x10 - 3 x100 = 300 bacteria per minute.

Check the growth rate for t= 8 minutes and for t=12 mins.

For t= 8: dP/dt = 60 - 48 = 12

For t= 12: dp/dt = 60 - 72 = -12

so, the growth rate is maximum at the inflection point and starts decreasing after

that time.

--------------------------------------------------------

3 A drug is ingested in the right arm of a patient. The concentration of the drug in

the left arm is given by the equation , with respect to time, as follows:

C(t) = 0.14 t/ (t.t+1)

where t is in minutes.

Find any local maxima or minima and the point of inflection.

Let (t.t +1) = d

dC/dt = [0.14 (t.t +1) - 0.14t (2t)]/ d.d

Setting dC/dt = 0 , we get

0.14 t.t - 0.28 t.t + 0.14 =0

-0.14 t.t +0.14 =0

or t.t = 1 or t=1 [ t=-1 has no meaning.]

For t< 1, say t=0.8, dc/dt = [-0.14 (.64) +0.14]/(1.64)(1.64)

dC/dt = 0.36 x0.14 /2.69 = 0.187

Let us find dC/dt for t = 1.1

dC/dt = [-0.14 x 1.21 + 0.14]/ (2.21)(2.21)

= - 0.0294/4.88 = -0.0060

So , the rate is increasing or positive for t<1 and the rate is negative or

decreasing for t>1.

Hence t= 1 is the point of inflection, where the rate reaches a maximum.

-------------------------------------------------

4 The salvage value S for an aircraft is found to be:

S(t) = 300000 exp(-0.1 t)

where t is in years.

Find the rate of depreciation after 5 years and after 10 years.

S'(t) = 300,000 exp(-0.1t)(-0.1)

= -30000 exp(-0.1t)

For t = 5, s'(t) = -30000 (.6065) = 18195 $ per year

For t= 10 , s'(t) = -30000(0.3678)= 11034 $ per year.

Note: The deprecation rate is higher in the early years and keeps decreasing with

each passing

year.

5 In the learning theory, we use the concept of learning curve. If a person learns N

items in time t hours,

N(t) = 20 sqrt(t)

Find the rate of learning at 1 hour and 4 hours.

dN/dt = 20 (1/2)(1/sqrt(t))= 10 /sqrt(t)

For t =1, we get dN/dt = 10 items per hour

For t = 4, dN/dt = 10/2 = 5 items per hour.

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----- The End--------