Calculus Tutorial-1 Differential Calculus

Introduction

Calculus is divided into two parts: Differential Calculus and IntegralCalculus.There is also a related subject-- Differential Equations.

Differential Calculus and Integral Calculus are like two opposing processes.Indifferential calculus, for a functionY = f(x), we take smaller and smaller intervals of x and take into account thecorresponding changes in y.In other words, what is the change in y when x is changed by an tiny amount. y = f (x) y1= f ( x + x)

where x [called 'delta x']is a small change in x.Then, differential coefficent is just this: d (y)/d(x) = (y1 - y) / x as tendsto zero. {I will explain this later.}

The integral calculus gives the total value of y (x) from say, two values of x, aninterval x1 to x2;As we will see ,this is the area under the curve of y versus x,from x1 to x2.

Differential Calculus What it really means to us?This field is actually a battle between chord and tangent to a curve.It helps us to calculate the instantaneous change instead of average change.

Take a simple example: y = f (x)Let us take y as the distance moved by a car or displacement. Take x as time tin minutes.

Let us rewrite the function as follows: d = f(t)

Suppose I draw a curve of d versus t , in a graph paper. ThenTake the table of values:d (miles) 0 0.5 1.2 2.4 3.5 4.9 6.2t (mins) 0 1 2 3 4 5 6

Suppose we want the average speed of the car, say at 4 mins. We take two points on either side of 4 mins. d1 = 1.2 miles at t1= 2 mins d2 = 6.2 miles at t2 = 6 mins.

The average speed is : speed = (d2 -d1) / (t2 -t1) = (6.2 - 1.2) / (6 -2) = 5 / 4 = 1.25 miles per minute.

We can also take the the two points closer to 4 mins. Speed = (4.9 - 2.4) / (5 -3) = 2.5/2 = 1.25 miles per minute.The same value.

But suppose we have data with finer divisions of minutes around 4 mins.

d (miles) 3.1 3.3 3.5 3.9 4.1 t (mins) 3.5 3.75 4.0 4.25 4.5

Now let us take avearge speed around 4 mins.

Speed = (4.1 - 3.1)/(4.5 - 3.5) = 1/1 = 1 mile per minute

or speed = (3.9 -3.3)/ (4.25 -3.75) =0.6 /0.5 = 1.2 miles/minute.

As we reduce the time interval, we get almost instantaneous change or speed.

Differential Calculus enables you to find this as the interval is reduced, thechange leading to instantaneous change.Thus we draw the tangent at the givenpoint on the graph of y versus x and take its slope.

Therefore, the average speed is found by drawing a chord, while instantaneousspeed found by drawing the tangent.[ Note: In all these discussions, we assume that the function f(x) is continuous.For various types of 'discontinuities', read your text book.]

The Limiting Process

The differential calculus is based on the concept of limits of a fucntion.Thelimiting process should be understood and the method of finding the limits of afunction should be mastered at this stage.

Take a simple function y = f(x)= 2x +3

This ,of course , is an equation for a straight line with slope =2 and intercept =3.Let us understand this equation in a different way....What happens to y as wedecrease the value of x ,say from x =1 to x =0.Let us construct a small table: x y 1.0 5 0.5 4

0.2 3.4

0.1 3.1

0.01 3.01

0.001 3.001When x is very close to zero, y tends towards a value of 3.0.This value of Y is called the Limit of the function as x tends to 0or Limit f(x) = 3 x -> 0

This ,of course , is a trivial example to illustrate this concept.

Let us take a more complex one.

Example 2 What is the Limit of f(x) = 1/ (x +1) as X tends to zero.

Again construct a table: X Y 1 1/2 =0.5

0.5 1/1.5 = 0.666

0.1 1/1.1 0.01 1/1.01We can see that as x tends to 0, y tends to 1So, Limit f (x) = 1 as X tends to 0 or X -> 0.

Example 3 Find the limit of f(x) = as x tends to -1 Construct a table starting with x = -2 X Y -2 4 - 1.5 2.25 - 3 + 4 = 3.25 -1 1 -2 +4 = 3Here again it is easy to find the limit.

Example 4 Find the limit of f(x) = 1/x as x tends to 0 Here the value of f (x) becomes very large as we approach x. The limit is ∞(infinity)

Example 5 Find the limit of f(x) = (2x + 3 ) / (5x - 2) as x tend to infinity.

Let us rewrite the function as follows: f(x) = ( 2 + 3/x) / ( 5 - 2/x)As x tends to infinity, f(x) tends to ( 2+0) / ( 5 -0)

Limit f(x) = 2/5 as x tends to ∞ .

Example 6 Find the limit of f(x) = sin x / x as x tned to 0. The limit is 1. Note: Read the text book.

Left-hand limit and right -hand limit

For finding the limit , you can find values of a given function f(x) from eitherhigher values or lower values of x.

Take y = f(x) = 1/ x +2 as x tends to 1.

We can appraoch the function at x = 1 , as we increase x from say, 0 to 1.Thiswill give left-hand limit.We can also approach as we decrease x from 2 to 1.Thiswill give righ hand limit. Now if the function is continuous, the two limits must bethe same. Think about this.

Many times you can find the limit by factoring the expressions and simplifying thefunction.

Example 7 Find the limit of f(x) =( + 5x + 6) / ((x+2) as x -> 2 f(x) = ( x + 3) ( x+2) / (x + 2) = x + 3Limit f(x) = 5x -> 2

Example 8 Find the limit of f(x) = ( - 1) / (x + 1) as X tends to 1 f (x) = ( x+1) (x-1) / (x+1) f(x) = x - 1Limit f(x) = 0 as x tend to 1.

The differential coefficient

Let us take a function y= f (x)Let its value be y1 at x = x1 y1 = f (x1)Let its value be y2 at x = x2 y2 = f (x2)You have done problems with rates before.Now what is the rate of change of y as we change x? Take the ratio: R =( y2 - y1) / ( x2 -x1) ---------------------------(1)Simple!Now let us take x1 and x2 very close values. x2 = x1 + h where h is very smallvalue. We can take 'h' as small as we please. Then R = ( y2 - y1)/ ( x2 - x1)

The denominator x2 - x1 = h and y2=f(x2)=f(x1+h)

So, R becomes : R =( y2 - y1) / h or R = ( f (x1+h) -f (x1) ) / h

The process of making h smaller and smaller is useful.We can find the change ofy for very small changes in x.How we do this? --- by the process of limits.

Find the limit of R as h goes to zero.

Limit R = Limit ( y2 - y1)/h as h tends to zero.

h -> 0

This limit is called the differential coefficient of f (x).If f(x) is the distance travelled and x equals time t , the this differentialcoefficient becomes the INSTANTANEOUS SPEED .

Some examples:

A car's distance from the starting point follows the equation : y = f(t) = 25 t

Find the differential coefficient and its instantaneous speed.

y2 = 25 t2 y 1 = 25 t1

R = (y2 - y1) / ( t2 - t1)

let t2 - t1 =h

R = 25 ( t2 - t1)/ ( t2 - t1)

R = 25 h /h = 25Limit of R as h tend to zero is then 25 only.DIfferential coefficient is just 25 , the constant in the function.Here theinstananeous speed is 25, same as constant speed or average speed.

We denote the differential coefficient as dy/dx dy/dx =Limit f(x2) - f(x1) / (x2 - x1) h -> 0 Writing x1 = x, x2 = x1 +h = x+h We can simplify writing hereafter: dy/dx = Limit [f( x+h ) - f(x)] /h ---------------(2) h -> 0

In words: dy/dx , the differential coefficient is the limit of f(x+h ) - f(x) dividedby h , as h tends to zero.

Finding the differential coefficients for common functions

There is another name for differential coefficients or dy/dx ----the derivativeof a function.From now on we use this term which is simpler.

Using the equation #2 above, let us find out the derivatives or dy/dx forseveral common functions .

1 Y = f (x) = k , a constant. Well, dy/dx = 0 in this case, because nothingchanges here.!

2 y = f (x) = c x --> Here c is a constant, this function is just a straight linepassing through the origin.!

[ This was given earlier as y = 25 t. Let us go over this again]

Applying equation (2) : f (x + h) = c ( x+h) f(x) = cxSubtract---> f (x+h) - f(x) = c( x+h) - c x = ch

f(x+h) - f (x) Limit ------------ = Limit of C = C h tends to 0 h

So, dy/dx = c, which is the slope of this line.-------------------------------------------------------------------------------------------------------------

We need another important result now:

Suppose we have two functions : y1 = f(x) and y2= g(x) Let us call : y = y1 + y2 = f(x) + g(x)

What is dy/dx ? It is simply this: dy/dx = dy1/dx + dy2/dx

{ In advanced math, we call differentiation dy/dx as a linear operator. You maylearn Operational calculus later!}

Suppose we have a function : y = k f(x) = ky1 where k is a constant. Then dy/dx = k dy1/dx

Let me illustrate these results with examples:

Example1 Find dy/dx or derivative of y = 4x dy/dx = 4

Example 2 Find dy/dx when y = 4x + 6x dy/dx = 4 + 6= 10

Example 3 Find the derivative dy/dx for the straight line : y = 3x + 4 dy/dx = 3 + 0 = 3 since 4 is a constant.

--------------------------------------------------------------------------------------------------------Let us now proceed to find dy/dx for other functions:

3 y = f(x) =

Let us go through the steps as for the previous function:

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

Subtract-> f ( x+h) - f (x) = 2xh + Divide by h --> f(x+h) - f (x) -------------- = 2x + h h

Limit ( f(x+h) - f(x) ) /h = 2x h ---> 0

Therefore: dy/dx = 2x when y = f(x) =

Some problems:

Example 4 Find dy/dx or the derivative of Y = 2 dy/dx= (2) . 2x = 4x

Example 5 Find dy/dx of y = 2 + 3x + 4

dy/dx = 4x + 3------------------------------------------------------------------------------------------------------Applied Problem 1: Brian throws a ball up with some force.The height reachedby a ball with initial velocity of 64 feet per second and intitial height of 6 feetfrom Brian's hand is given by the equation: [ fall due to gravity] h (t) = - 16 + 64 t + 6 where t is time in seconds.Find d h / dt or the velocity during upward motion and descent of the ball.[ Here - 16 represents -g/2 where g is the acceleration due to gravity: g= 32feet/sec.sec. See your physics text book for this equation.]

d (h) / d t = (-16) 2t + 64 = - 32 t + 64

Applied problem 2 From the previous problem , when the velocity will bezero, at the top of the climb of the ball, what will be the maximum heightreached by the ball.?

dh/dt is the instantaneous velocity of the ball. At maximum height , dh/dt = 0 Therefore, -32t + 64 = 0 t = 2 seconds

The height reached in 2 second is given by the original equation: h ( 2) = - 16. (2) + 64 (2) + 6 = -32 +128 +6= 102 feet.[the maximum height]

Applied Problem 3 An astronaut throws a rock on the moon into the air. The height of the rock isgiven by the equation: where s is in feet and t in seconds.Find the velocity and acceleration of the rock.Find the maximum height ,by setting ds/dt = 0 and finding the height at that time.-----------------------------------------------------------------------------------------------------------

Example 6 The area of a circular field is given by : A = As the radius r increases, the cost of making the field increases. What is the rateof increase in area when r = 200 feet.?

A ( r) = d A / dr = ( 2 r )

The rate of increase in area = 6.28 x r = 6.28 x 200 square feet/feet

4 y = f(x) = Find the derivative.

Let us go through the steps to find dy/dx:

= + 3 h + 3 x +

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

Limit ( f(x+h) - f (x) )/h = Limit ( 3 + 3 x h + ) = 3h --> 0 h--> 0

Therefore: dy/ dx = 3 when y =

5 Generalise dy/dx for y =

We noted that for y = , dy/dx = 2x and for y= , dy/dx = 3

Therefore for y = dy/dx = n [ called the 'power rule']

This is an important formula: remember this as long as calculus work is to bedone!We can use this formula even when n is negative or fractions. as we illustrate inthe following examples.

Example 7 Find dy/dx for y =

dy/dx = 9

Example 8 Find dy/dx for y = 1/x y = 1/x = x-1

Here n = -1 dy/dx = (-1) x -1-1

dy/dx = -1 / x2

Example 9 Find dy/dx for y = 1/ x2

Here n = -2 dy/dx = (-2) x -2-1

dy/dx = -2 / x 3

Example 10 Find dy/dx when y =

y = x 1/2

n = 1/2

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

Practice Problems 1 Find the derivative for Ans: 2 Find the derivative of Ans:

3 Find the derivative of Ans: 4 find the derivative of 5 Find the derivative of

-----------------------------------------------------------------------------------------------------Derivatives for Trig functions

6 Find the derivative for y = sin x

Let us go over the steps: f ( x+h ) = sin ( x +h) = sin x .cos (h) + cosx sin ( h ) f (x) = sin xSubtract: f (x+h) - f(x) = six cos h + cos x sin h - sin xDivide by h : [ sinx ( cos h - 1) + cos x. sin h] / hTake the limit as h goes to zero: h-->0

Limit (cos h - 1) / h =0Limit [ cos x. sinh /h ] = cos x (since Limit [sin h / h ] = 1 as h -> 0.)h -> 0

Therefore d [sinx ] / dx = cos x

7 We do not derive the derivatives for other trig functions, but list them here.

y dy/dx

sin x cos x cos x - sinx

tan x sec2 x [We derive this later,after the quotientrule!] cot x

7 Find the derivative for exponential function y = f (x) = exp (x)

f ( x+h ) - f ( x) = exp(x) .exp (h) - exp (x)[f( x + h ) - f(x)]/h = exp( x) exp(h) /h- exp (x)/h

The Limit of this expression as h ->0 is exp (x) {since exp(h)/h =1/h+1+h +h^2+.... and its limit is 1 as h tends to0.}

efore for y= f(x)= exp (x) , dy/dx = exp (x)

Note that the function and its derivative are the same for exponential function!Why is this? Think for a moment.

8 For y = ln x Limit ln(1+h/x) as h tends to 0 is 1/x dy/dx= 1/x

The Product Rule

Suppose y is a product of two functions of x: y = u(x).v(x)Then dy/dx = u(x).dv/dx + v du/dx --------------------------(3)

Example 11 Find the derivative for y= x.sinxHere u = x, v= sinxdu/dx= 1 dv/dx = cosx dy/dx= x.cosx +sin x

Example 12 Find the derivative of : y = x2 exp(x)Here u = x2 du/dx=2x

v= exp ( x) dv/dx=exp(x) dy/dx = x2 exp(x) + 2x.exp(x) = xexp(x) [ x + 2]

Example 13 Find the derivative of y = Here u = and du/dx = 2x and dv/dx = 3

dy/dx = =

Check :

Applied problem 3 The vibrational amplitude of a suspension spring in an automobile can bewritten as: y(t) =A et sin t where A is a constant and t is the time .Find thederivative which gives the rate of decay of the amplitude. u (t) = et du/dt = et

v(t) = sint dv/dt = cost

dy/dt = A [et sint + et cost]=

The Quotient Rule

If y = f(x) = u (x) / v(x), then dy/dx = [v. du/dx - u.dv/dx ] / v2 -------------------------------(4)

[This rule can be derived from the product rule.Try doing this.!]

Example 14 Find the derivative of y = tan x

y = tan x = sin x/ cos x

Here u(x) = sin x v(x) = cos x

du/dx = cos x dv/dx = -sinxdy/dx = d (tanx)/dx = [cos x .cos x - sin x (- sin x)] / cos2 x dy/dx = 1/ cos2 x = sec2 x

This result was given earlier under Trig functions.

Example 15 Find the derivative of y = cot xdy/dx = -cosec2 x [Try as was shown in the previious example]

Example 16 Find the derivative of y= x / (x2 +1 )

Here u = x du/dx = 1v= x2 +1 dv/dx = 2x

dy/dx = [(x.x+1)x - x.2x ]/ (x2 +1)2

Example 17 Find the derivative of : y =

Applied problem 4The bacteria population in a broth [culture] varies with time as follows: P(t) = 100 ( 1 + 4 t / (50 + t2 )) where t is time in hours

Find the rate of growth at t = 2 hours. P(t) = 100 +400 t / (50 + t2 )u = t v= 50 + t 2

du/dt = 1 dv/dt = 2t

d P / dt = 400 [ (50 + t2 ) - t.2t] / ( 50 + t2 )2

To find dp/dt at t=2, substitute t =2 in the above equation: dP/dt = 400 [ 46 / 54.54] = 6.3 bacterias/hourApplied Problem 5 The population of a city increases from 25000 in the year 1990 exponentially: where t is in years since 1990.Find the rate of population growth and the value after 5 years.

When t= 5 ,Increase in population is: = 750x1.16=870per year.

The Chain Rule

If the given function y = f(x) can be split as follows: y= f( u ) and u = g(x),then, dy/dx = [dy/du].[du/dx] ------------------------- (5)

Consider the function 'u' as an intermediary function, that helps you to split thegiven function into a simpler one.

This rule is very powerful and enables us to differentiate any complicatedfunction.

Example 18 Find the derivate of y = sin (3x) Let u = 3xThen Y becomes: y = f(u) = sin u dy/du = cos u du/dx = 3Therefore: dy/dx = (dy/du)(du/dx)= cos u .3 = 3.cos(3x){ Note: Don't forget to substitute for u in the final expression and simplify theexpression}

Example 19 Find the derivative.

Let

du/dx = 2x

dy/dx = dy/du . du/dx

Example 20 y = sin (cos ( x)) Find dy/dx

Let u = cos ( x) y = sin u du/dx = - sin x dy/du = cos u

dy/dx = cos u . (- sin x) = - cos ( cos x).sin x

Example 21 y = ln ( sin x)Let u = sin x du/dx = cos xy = ln u dy/du = 1/u

dy/dx = 1/u . cos x = cos x/ sin x = cot x

Example 22 Y= Find the derivative. Let u = sin x

dy/du = 2u

du/dx = cos xdy/dx = 2u.cos x = 2 sinx .cosx = sin (2x)

The chain rule can be extended to three variables too: dy/dx = (dy/dv) (dv/du) ( du/dx)

Example 23 Find the derivative: Let Now carefully subtitute for the variables:

Practice ProblemsFind the derivative for1 y= cos(sin x)2 3 4 5 6 7 8 9 10 Evaluating a derivative.

After finding the derivative as an expression, you can find the value of thederivative at a given point (x,y). That value is the tangent to the original curveat that point.!

Example 24 Consider a ball thrown in the air. Let the displacement where t is the time (seconds) and s infeetThe derivative is the velocity:

Evaluate the derivative at t= 1 second and t =2 secondsFor t = 1. ds/dt = velocity = -32+80 = 48 feet/secondFor t = 2 , ds/dt = velocity = -64 + 80 = 16 feet per second.ds/st = 0 when t= 80/32 = 2.5 seconds ;the velocity is zero;the ball is at themaximum height.

Example 24. Evaluate the derivative for y = at x= 15 degrees. See example 22 dy/dx=sin(2x) = sin(30)Example 25 Evaluate the slope for at x= 1, x= 2 and x=3

dy/dx = 2.2.x = 4x dy/dx = 4 for x=1 dy/dx= 8 for x=2 dy/dx= 12 for x = 3

A practical application of finding the derivative is to find the equation for thetangent and the normal at a given point on a curve.

Finding tangent and normal to a curve

Note that the derivative is the slope for the curve.Choose a point P (x1, y1) on the curve.Recall the point-slope form equation of a straight line: y - y1 = m (x-x1) where m is the slope.Replace m by dy/dx.------> y - y1 =[ dy/dx] (x-x1)Normal has the slope : m' = -1/mThe equation for the normal : y -y1 = m' (x -x1)

Let us see some examples.

Example 25 Find the equation for the tangent line to the parabola y = x2

+3 at x= 2

dy/dx = 2x At x =2, y= 4+3 = 7 At x= 2, dy/dx = 4

The equation for the tangent: y - y1 = slope [x - x1] y - 7 = 4 [x-2]Simplifying: y - 7 = 4x - 8 y = 4x -1[Note: Use a graphing calculator,draw the parabola and the tangent;check thatthe tangent touches the parabola at a point P(2,7)

Example 26: Find the tangent line for the curve y =

through the point (1,-9) At x= 1, y- (-9) = -6(x-1) y+9 = -6x+1 y=-6x-8

Example 27 The parabola y = passes through (0,1) and is tangentto the line y = x -1 at (1,0).Find the equation of the parabola, that is ,find a,b,c.

The point (0,1) on the parabola : 1 = 0+0+c c=1The slope of the parabola at (1,0) : dy/dx=slope = 2ax + b =1At x =1, slope = 2a+b =1 ---------(1) Further, at x =1, y=0 0 = a+b +1 a+b = -1 -------------(2)Solving, we get a= 2, b = -3

Example 28 For the ellipse find the slope of its tangent at P ( )

{Hint: Find dy/dx using implicit differentiation discussedlater.} Ans: 1/2

Practice Problems

1 Find the derivatives of the following functions: a. y = sec x = 1/cosx

b. y = cosec x= 1/sinx

c

d [ Evaluate the derivative at (2,2)]

e

f Find the derivative at x =

g

h

i

j

Applied Problems

1 The volume of a funnel in conical shape is The value of h = 3rFind the rate of change of volume dV/dt if dr/dt = 2 in/min when a liquid flowsfor r = 6 inches.

cuin/min

2 The inventory cost C for a company is given by the equation:

C = 108 / Q + 6.3 Q [$millions]

where q is the order size. Find the cost at q = 350 and q = 351 and find thechange in C for change in q by one unit.Find the derivative dC/dq at Q =350 and compare with the rate found earlier.

dc/dQ =

dC /dQ at Q = 350 is = 6.3 nearly

3 The surface area of a cylindrical can with lid is: For a can with h = 2r, find the rate of increase in surface area ar r= 6 in. Sincethe cost of making the can is proportional to the surface area, find the rate ofincrease in cost if the cost per sq in is 2 Cents.

ds/dr = For r = 6 in, ds/dr cost = s x 2 cents4 A cylindrical silo of diameter 60 in is being filled by pouring grains at the rateof 200cuin/sec. Find the rate of increase in height of the filled grains at anygiven time. Volume of a cylinder =

Since r is a constant,

Implicit DifferentiationExample 29: Consider an expression like this:Equation for a circle: Here the radius is 3 units.How do you find the dy/dx now?Rewrite the equation with y terms on the left and x terms on the right side: Let Then using the chain rule: d g/dx= dg/dy . dy/dx = 2y.dy/dx Let df/dx = -2xEquating the two : we get: 2ydy/dx = - 2x

dy/dx = - x/y

Example 30 Find dy/dx if

Use product rule:

Keep dy/dx on the left side: dy/dx= -y/x

Example 31 Find dy/dx for sin x + 2 cos 2y =1 2cos2y = 1-sinx 4(-sin(2y)) dy/dx=-cosx dy/dx= cos x /4sin2y

Example 32 For tan (x + y ) = x,find dy/dx Ans: 0

Example 33 For xy = 4, find dy/dx at (-4,-1)

Example 34 . A well-known expression is called 'Witch of Agnesi" : Find its slope at point (2,1) Ans: -1/2

{ Note: For these problems, use the graphing calculator to draw the curve, findits slope and tangent and check with the work.}

Parametric Equations Often times, it is easy to express x and y as functions of another variable called'parameter.'

For instance, an aircraft path in space can be expressed as ground distance orhorizontal distance and the height or altitude, both varying with time. grounddistance x = x(t) and height = h (t) 't' is called the parameter.In these cases, we can find dy/dx as follows:

y = f(t) x = g(t)

dy/dt = df/dt dx/dt = dg/dt

Using Chain Rule: dy/dx = dy/dt . dt/dx

= (dy/dt) / (dx/dt) if dx/dt is not 0.

Example 35 A curve is given by and Find the slope at the point x=2,y=3 t=4

and

Now, ) =

Slope at (2,3) : t=4; dy/dx= 8

Example 36 For and

,

Find the slope

These parametric equations give an ellipse:

Example 37 For and

,

find the dy/dx and slope at t = 2dx/dt = 2 dy/dt= 2tdy/dx = t = 2

Differentials--splitting dy/dx

From differential coefficents, we introduce differentials: If Y = f(x) [We use the notation: dy/dx=f'(x) ]Now we do a clever manipulation: dy = f'(x) dxNow replace dy by a small change in y , [delta y]Replace dx by a small change in x, Note that by doing this we are not really using the limit process, but at thetangent to the curve, we make a small right triangle with in x axis and in yaxis.The tangent is the hypotenuse. Then : This is an approximation, but quite useful in applications.The value of must be evalauted at a given point.

Example 38 The area of a circular field is given by . John wants toincrease the area by increasing the radius by 1%.What will be the increase inarea when r = 200 feet? At r= 200 feet, Change in area The change in r: = 2 feet ---> 1 %

Example 39 Find the square root of 16.2 16= 16+0.2 = x+ δx dy/dx (at x= 16) = 0.125

Take {The calculator gives a value of 4.02492 ---pretty close}The diffferentials are useful in estimating errors in measurements.

Applied ProblemThe target at a distance x has a diameter D meters in a field telescope used inarmy tanks . The diameter is given by D = x where theta is the angularmeasure, in radians.If the error in theta measurement is 1 milliradians, find theerror in the diameter for a distance of 2000 meters.Recall the arc distance S = r x theta when theta is in radians. Error in diameter of the target=

Logarithmic differentiationYou want to find the derivatives for complex expressions---here is areally,really easy way ---using logarithms.First a few things to recall: ln (AB) = ln A + lnB ln (A/B) = ln A - ln B If z = ln y, then dz/dx = dz/dy . dy/dx Let us see some simple examples first.Example 40 Find the derivative of

Take logarithm both sides: Differentiate both sides with respect to x: Plug in for y and simplify:

This ,of course, is a simple problem; you could have done with product ruletoo, but this illustrates the steps.

Example 40 Find the derivative for Taking logarithm of the function both sides: (1/y)dy/dx = 4x/(x^2+3) + 2/sin(2x) dy/dx= y [ 4x/(2x^2+3)+2/sin(2x)Simplify!Practice Problems1 Find the derivative for

2 Find the derivative for

Relative changes Finding the realtive changes in variables connected by an expression orequation is easy with log differentiation and differentials, given earlier.Example 41 The surface area of a sphere is When a sphericalbaloon has a radius of 2 feet,if the relative change in r is 2%, what is therelative change in the surface area. Taking log both sides: Differentiating: Converting to differentials: Therefore: A change in radius by 2%,results in a 4% change in surface area.

Example 41 The electrical resistance of a wire is proportional to its length andinversely proportional to its cross-sectional area. What would be the relativechange in resistance for change in the radius of the wire and its length?

R = k L/ where k is a constant,called resistivity. Taking log both sides: Taking differentials: This means that a 1% reduction or error in the radius of the wire will increasethe resistance by 2%.The estimation of relative changes is helpful in many analyses of small changesin variables and for error estimates.

Practice Problem1 An aircraft ball bearing has balls of diameter 10 mm. If the radius increasesdue to frictional heating by 0.02 mm, find the change in the volume of the ball.volume:

Solving algebraic equations-Newton's method You can consider this as an application of differential calculus.It is attributed toIsaac Newton, and the name of one Raphson is also tagged on. Hence 'Newton-Raphson method'. Suppose you want to find the square root of a number ,say 1824.Write as follows:

Now we try to solve the equation : y=f(x) =0Start with an approximate solution or root of the equation: x= xo= 45This is called an initial root.We can improve on this by using the derivative: dy/dx at x= xo.

Rewriting this: x - xo = (y - yo)/f'(xo) Let x =x1 = xo - f(xo)/f'(xo) since y=0

This value of x ,called x1 ,is the improved root,closer to the actual value.

Iteration: We can continue by repeating the procedure: plug back x1 in theabove equation for xo and get another value x2: x2= x1 - f(x1)/f'(x1)This method is an iterative procedure. Hopefully , the successive values of the root ,xo,x1,x2,x3 ---will converge to theactual value.!

In general we write: xn+1 = Xn -f(Xn)/f'(Xn) ------Iterative N-Requation

Example 42: Find the in two iterative steps using Newton-Raphson[N-R method] starting with xo=45;

Using the N-R equation: x1= xo - f(xo)/f'(xo)= 45 -201/90=45-2.2333=42.767Iterate : f(x1)=42.767^2-1824=5.0163 f'(x1) = 2(42.767)=85.534 x2 = x1- f(x1)/f'(x1) = 42.767-5.0163/85.534 =42.767 - 0.0586=42.7084 The answer we get is: Your calculator will give: 42.70831 The error is 0.0001 or 0.00002%A phenomenal accuracy by applying twice the NR equation .This is the power ofNewton-Raphson method, compared to other 'brute' force method like bisectionor secant method.!

: There is a caveat to this. NR method works only when: 1. f'(xo) is not close to zero. [If f'(x0) is close to zero,the termf(xo)'f'(xo) will go to a large number and take you away from the root.] 2. The initial choice of xo is important.It should be close to theroot we are looking for. We have used N-R method which is a popular method among numericalmethods for solving algebraic equations.Try the following practice problems:

Practice Problem1 Find using N-R method with two iterations:start with x0= 1.5

2 Find with N _R method,two interations with xo=12

3 Solve f(x) = x -exp(x) =0 with N-R method with initial root Xo=1.5 Ans:x=1.73177

4 Solve f(x0) = exp(x)-3x=0 [One of the real roots lies between 0.4 and 0.9 Ans:0.6185

1. To find the square root of a number , ,a greekmathematician,had a simple iterative formula. To find

We can see that this formula can be derived from N-R method:

This is Heron's formula!Let us apply this to find and start with xo=2. x1=1/2(2+5/2)=2.25 x2= 1/2(2.25+5/2.25)= 2.2361

x3=1/2(2.2361+5/2.2361)=2.23606The calculator gives :2.236068{Note: Most probably your calculator chip uses the same algorithm for squareroots as Heron's formula!}

2 Isaac Newton ,historians say, first used this method in 1669 to solve thisequation: Starting with x=2, we get x1= 2 - f(2)/f'(2)= 2- (-1)/10=2.1 x2= 2.1 - f(2.1)/f'(2.1)= 2.1 -0.61/11.23= 2.0457

3 Before Newton, Leonardo of Pisa, solved the equation: f(x) = x^3 +2x^2+10x -20 =0in the year 1225. The answer was: x= 1.368 808 107.Check this answer by theN-R method.

Applied Problems

1 Find the break -even point between the use of PC's and main-frame computerfor workload of W files.The cost equations are: PC's : c1= kw where k=1.2Main-frame computer: c2 = 100+log (w+1)At break -even point c1=c2Therefore: f(w)= kw - log(w+1)-100=0

Solve for w . Take the initial root as wo= 50 Ans: w=84.945

2 An environmental group finds that the oxygen concentration C variesdownstream from an effluent point as follows: C = 10 -15 [exp(-0.1x)-exp(0.5x)] where x is the distance in miles.Find thedistance at which the concentration C=4 by N-R method. Start with x= 10 miles. Ans: x=8.872 miles

Summary and comments We have come to the end of "Elementary Differential Calculus"I have given the basic concepts and the methods:

limits,

definition of dy/dx,

derived the rules for a few simple functions ,

additional rules such as : product rule,quotient rule and chain rule.

implicit differentiation

parametric equations

differentials and approximations

logarithmic differentiation.Newton-Raphson method for finding the roots of algebraic equations

Each section could be one or more chapters in your text book.But focus on thebasic concepts and methods of doing things.Most text books have too much verbiage and also distractions and diversions.Concentrate on essential concepts and techniques in the first round of study.In the second round, you can drill yourself with lot of practice problems andapplied problems.I have given some practice problems and applied problems to get you started.Use graphing calculators to draw the cuves and find the slopes for some of theproblems.My email ID : nksrinivasan@hotmail.com Leave your feedback/comments there.We shall see more tutorials on Calculus later.The applications of calculus areenormous and fascinating.Who invented or developed Calculus? Sir Isaac Newton (1642-1727),ofcourse.Well, Leibnitz (1646-1716) was also close on the heels ofNewton in developing these.These two mathematicians fought a bitterverbal battle in their claims.Newton applied the calculus for avariety of problems and also developed numerical methods .

There are plenty of bulky calculus texts which take you into all kinds of by-lanes and diversionswithout focussing on the basic concepts.If you can read the books by the following authors,try along withclass-room text:1 Silvanus Thompson 2 Tom Apostol 3 G B Thomas 4 George Simmons 5 Richard Silverman 6 F BHildebrand 7 Serge Lang8 Richard CourantFor numerical methods: 1 Chapra and Canale 2 Baron and Salvadori 3 Richard Hamming

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