Calculus Tutorial 1 --differential Calculus

C alculus Tutorial-1 Differential C alculus Introduction Calculus is divided into two parts: Differential Calculus and Integral Calculus.There is also a related subject-- Differential Equations. Differential Calculus and Integral Calculus are like two opposing processes.In differential calculus, for a function Y = f(x), we take smaller and smaller intervals of x and take into account the corresponding 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) / to zero. {I will explain this later.}

x as

tends

The integral calculus gives the total value of y (x) from say, two values of x, an interval x1 to x2;As we will see ,this is the area under the curve of y versus x ,from x1 to x2. Differential C alculus 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 t in minutes. Let us rewrite the function as follows: d = f(t) Suppose I draw a curve of d versus t , in a graph paper. Then Take the table of values: d (miles) 0 0.5 1.2 2.4 3.5 4.9 6.2 t (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) t (mins)

3.1 3.3 3.5 3.9 4.1 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, the change leading to instantaneous change.Thus we draw the tangent at the given point on the graph of y versus x and take its slope. Therefore, the average speed is found by drawing a chord, while instantaneous speed 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.The limiting process should be understood and the method of finding the limits of a function 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 we decrease 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 3.001 When 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 0 or 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)

Again construct a table:

1

X

0.5

as X tends to zero.

Y 1/2 =0.5 1/1.5 = 0.666

0.1 1/1.1 0.01 1/1.01 We can see that as x tends to 0, y tends to 1 So, 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 = 3 Here 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 either higher 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.This will give left-hand limit.We can also approach as we decrease x from 2 to 1.This will give righ hand limit. Now if the function is continuous, the two limits must be the same. Think about this. Many times you can find the limit by factoring the expressions and simplifying the function. Example 7

Find the limit of f(x) =( + 5x + 6) / ((x+2) as x -> 2 f(x) = ( x + 3) ( x+2) / (x + 2) =x+3 Limit f(x) = 5 x -> 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 - 1 Limit 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 small value. 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 of y 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 differential coefficient 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 = 25 Limit of R as h tend to zero is then 25 only. DIfferential coefficient is just 25 , the constant in the function.Here the instananeous 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: ---------------(2)

dy/dx = Limit [f( x+h ) - f(x)] /h h -> 0

In words: dy/dx , the differential coefficient is the limit of f(x+h ) - f(x) divide d by h , as h tends to zero. Finding the differential coefficients for common functions There is another name for differential coefficients or dy/dx ----the derivative of 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 for several common functions . 1 Y = f (x) = k , a constant. changes here.!

Well, dy/dx = 0 in this case, because nothing

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

This was given earlier as y = 25 t. Let us go over this again] Applying equation (2) :

Subtract--->

f (x + h) = c ( x+h) f(x) = cx f (x+h) - f(x) = c( x+h) - c x = ch

f(x+h) - f (x) Limit -----------h tends to 0 h

= Limit of C = C

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 may learn 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-> Divide by h -->

f ( x+h) - f (x) = 2xh + 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 reached by a ball with initial velocity of 64 feet per second and intitial height of 6 feet from 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= 32 feet/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 be zero, at the top of the climb of the ball, what will be the maximum height reached 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 is given 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 rate of 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 h --> 0 h--> 0

+3 xh+

Therefore:

when y =

dy/ dx = 3

)=3

5 Generalise dy/dx for y = We noted that for y = and for y=

Therefore for y =

, ,

dy/dx = 2x dy/dx = 3

dy/dx = n

[ called the 'power rule']

This is an important formula: remember this as long as calculus work is to be done! We can use this formula even when n is negative or fractions. as we illustrate in the 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 4 find the derivative of 5 Find the derivative of

Ans:

--------------------------------------------------------------------------------------------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 x Subtract: f (x+h) - f(x) = six cos h + cos x sin h - sin x Divide by h : [ sinx ( cos h - 1) + cos x. sin h] / h Take the limit as h goes to zero: h-->0 Limit (cos h - 1) / h =0 Limit [ 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 sin x

cos x

rule!]

tan x

dy/dx cos x

- sinx sec2 x

[We derive this later,after the quotient

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 h^2+.... 0.}

is exp (x)

{since exp(h)/h =1/h+1+h + and its limit is 1 as h tends to

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.sinx Here u = x, v= sinx du/dx= 1 dv/dx = cosx dy/dx= x.cosx +sin x Example 12 Find the derivative of : y = x2 exp(x) du/dx=2x Here u = x2 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 be written as: y(t) =A e t sin t where A is a constant and t is the time .Find the derivative which gives the rate of decay of the amplitude. u (t) = e t du/dt = e t v(t) = sint

dv/dt = cost

dy/dt = A [e t sint + e t cost]= The Quotient Rule If y = f(x) = u (x) / v(x), then dy/dx = [v. du/dx - u.dv/dx ] / v 2 -------------------------------(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 = -sinx dy/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 x dy/dx = -cosec2 x [Try as was shown in the previious example] Example 16 Find the derivative of y= x / (x2 +1 ) Here u = x v= x2 +1

du/dx = 1 dv/dx = 2x

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

Example 17 Find the derivative of : y =

Applied problem 4 The 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/hour Applied 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 per year.

,Increase in population is:

= 750x1.16=870

The C hain 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 the given function into a simpler one. This rule is very powerful and enables us to differentiate any complicated function. Example 18 Find the derivate of y = sin (3x) Let u = 3x Then Y becomes: y = f(u) = sin u dy/du = cos u du/dx = 3 Therefore: 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 the expression} Example 19

Find the derivative.

Let du/dx = 2x dy/dx = dy/du . du/dx

Example 20

y = sin (cos ( x))

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

Find dy/dx

y = sin u 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 x y = 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 x dy/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 Problems Find the derivative for 1 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 the derivative at a given point (x,y). That value is the tangent to the original curve at that point.! Example 24 Consider a ball thrown in the air. where t is the time (seconds) Let the displacement and s in feet The derivative is the velocity: Evaluate the derivative at t= 1 second and t =2 seconds For t = 1. ds/dt = velocity = -32+80 = 48 feet/second For 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 the maximum 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 the tangent 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/m The equation for the normal : y -y1 = m' (x -x1) Let us see some examples. Example 25 +3 at x= 2

Find the equation for the tangent line to the parabola y = 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 that the 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 tangent to 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=1 The slope of the parabola at (1,0) : dy/dx=slope = 2ax + b =1 At 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 discussed later.} 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 = 3r Find the rate of change of volume dV/dt if dr/dt = 2 in/min when a liquid flows for 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 the change 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. Since the cost of making the can is proportional to the surface area, find the rate of increase in cost if the cost per sq in is 2 Cents.

ds/dr = For r = 6 in, ds/dr cost = s x 2 cents 4 A cylindrical silo of diameter 60 in is being filled by pouring grains at the rate of 200cuin/sec. Find the rate of increase in height of the filled grains at any given time. Volume of a cylinder = Since r is a constant,

Implicit Differentiation Example 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 = -2x Equating 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, find its 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 or horizontal distance and the height or altitude, both varying with time. ground distance 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 Using Chain Rule:

dx/dt = dg/dt dy/dx = dy/dt . dt/dx

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

)=

if dx/dt is not 0.

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 = 2 dx/dt = 2 dy/dt= 2t dy/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) dx Now 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 the tangent to the curve, we make a small right triangle with in x axis and axis.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.

in y

Example 38 The area of a circular field is given by . John wants to increase the area by increasing the radius by 1%.What will be the increase in area 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 Problem The target at a distance x has a diameter D meters in a field telescope used in army tanks . The diameter is given by D = x where theta is the angular measure, in radians.If the error in theta measurement is 1 milliradians, find the error 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 differentiation You want to find the derivatives for complex expressions---here is a really,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 rule too, but this illustrates the steps. Example 40 Find the derivative for Taking logarithm of the function both sides:

Simplify! Practice Problems 1 Find the derivative for

(1/y)dy/dx = 4x/(x^2+3) + 2/sin(2x) dy/dx= y [ 4x/(2x^2+3)+2/sin(2x)

2 Find the derivative for Relative changes Finding the realtive changes in variables connected by an expression or equation is easy with log differentiation and differentials, given earlier. Example 41 The surface area of a sphere is When a spherical baloon has a radius of 2 feet,if the relative change in r is 2%, what is the relative 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 and inversely proportional to its cross-sectional area. What would be the relative change 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 increase the resistance by 2%. The estimation of relative changes is helpful in many analyses of small changes in variables and for error estimates. Practice Problem 1 An aircraft ball bearing has balls of diameter 10 mm. If the radius increases due 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 to Isaac Newton, and the name of one Raphson is also tagged on. Hence 'NewtonRaphson 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) =0 Start with an approximate solution or root of the equation: x= xo= 45 This is called an initial root. We can improve on this by using the derivative: dy/dx at x= xo.

Rewriting this: Let

x - xo = (y - yo)/f'(xo) 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 the above 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 the actual value.! In general we write:

xn+1 = Xn - f(Xn)/f'(Xn)

------Iterative N-R

equation

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=42.767 Iterate : 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 of Newton-Raphson method, compared to other 'brute' force method like bisection or 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 term f(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 the root we are looking for. We have used N-R method which is a popular method among numerical methods for solving algebraic equations.Try the following practice problems: Practice Problem 1 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 x=1.73177

Ans:

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 , mathematician,had a simple iterative formula. To find

,a greek

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

This is Heron's formula! Let us apply this to find x1=1/2(2+5/2)=2.25

and start with xo=2.

x2= 1/2(2.25+5/2.25)= 2.2361 x3=1/2(2.2361+5/2.2361)=2.23606 The calculator gives :2.236068 {Note: Most probably your calculator chip uses the same algorithm for square roots as Heron's formula!} 2 Isaac Newton ,historians say, first used this method in 1669 to solve this equation: 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 =0 in the year 1225. The answer was: x= 1.368 808 107.Check this answer by the N-R method. Applied Problems 1 Find the break -even point between the use of PC's and main-frame computer for workload of W files.The cost equations are: PC's : c1= kw where k=1.2 Main-frame computer: c2 = 100+log (w+1) At break -even point c1=c2 Therefore: 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 varies downstream from an effluent point as follows: C = 10 -15 [exp(-0.1x)-exp(0.5x)] where x is the distance in miles.Find the distance 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 the basic 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 and applied 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 the problems. My email ID : [email protected] Leave your feedback/comments there. We shall see more tutorials on Calculus later.The applications of calculus are enormous and fascinating. Who invented or developed Calculus? Sir Isaac Newton (1642-1727),of course.Well, Leibnitz (1646-1716) was also close on the heels of Newton in developing these.These two mathematicians fought a bitter verbal battle in their claims.Newton applied the calculus for a variety of problems and also developed numerical methods . There are plenty of bulky calculus texts which take you into all kinds of by-lanes and diversions without focussing on the basic concepts.If you can read the books by the following authors,try along with class-room text: 1 Silvanus Thompson 2 Tom Apostol 3 G B Thomas 4 George Simmons 5 Richard Silverman 6 F B Hildebrand 7 Serge Lang 8 Richard Courant For numerical methods: 1 Chapra and Canale 2 Baron and Salvadori 3 Richard Hamming

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