Brief History of Calculus

1. Brief History of Calculus • In the first half of the XVII century, mathematicians were interested in finding horizontal tangents to the graph of a function ( in order to find the maximum and minimum values of the function) and physicists were interested in finding the velocity and acceleration of a particle whose position function was given. • By 1666 Isaac Newton (English), introduced the 'method of fluxions' which allowed him to find the derivative of a function. The derivative can be interpreted as the slope of the tangent to the graph of y=f(x) and also as the instantaneous velocity of an object whose position is ruled by s=f(t). The derivative of a function f(x) is given by the limit of the quotient [f(x+h) – f(x)] / h when h0. • A few years later, Gottfried Leibniz (German), working independently introduced the idea of infinitesimals (a quantity whose square is zero) which allowed him to calculate the derivative without limit. He define the derivative of y=f(x) as the quotient dy/dx, where the differentials dy & dx are the respective values off(x+x) - f(x) and x when x  0.

2. Leibniz’s approach was taught until XIX century. It was the great mathematician Euler who did a gigantic work in applying Leibnitz’s method to both math & science. • But in late 1700 some mathematicians (like Berkeley, Cauchy, Weierstrass) began to criticize the lack of rigor of Calculus and in particular the Leibniz’s approach. In 1826 Cauchy taught an improved version of Newton’s approach in Paris. • “Leibniz’s differentials approach” didn’t pass the validation rigor test and the improved Newton’s approach was preferentially taught. • In 1965 A. Robinson (American), constructed an extension of real number and he called “Hipper-real numbers” that contains the real number and “the Leibniz’s infinitesimals elements”. This fact gave full validation to the Leibniz’s approach to Calculus. Calculus over the “Hipper-Real” is called Non-Standard Analysis and requires more theoretical knowledge of math. • In this project we begin with Newton’s approach and then we’ll introduce the differentials (Leibnitz’s way) as soon as possible and it will then make the work simpler. \

3. Assuming that x=f(t) is the function that relates the position x of a particle with time t, the derivative allows us to define instantaneous velocity at t=a as the limit of the average velocity from t =a to t=a+h when h  0. So, instantaneous velocity when t=a is defined as lim f(a+h) – f(a).h 0h Numerical approximation: Example1: Let f(t) = -16t2 +64t +312 be the function that relates the position (in feet) of a ball thrown vertically up versus with the time t in sec). Find instantaneous velocity of the ball when t=1 sec. Since the average velocity = f(1.1) – f (1) . 1.1 - 1 243.04 - 240. 0.1 = = 30.4 ft/sec. We guess that the instantaneous velocity is approximately 30.4 ft/sec Now, taking the values of t even closer to 1, we’ll get more accurate values of v(t). See the table below. We see (and it can be proved) that lim v(t) = 32 . t  1 It seems natural to say that the instantaneous velocity when t=1 is 32 ft/sec.

4. Example 2: The position of an object is given by s(t) = t2 +10 with s in feet and t in seconds. Find the speed of the object when t= 5sec Speed = lim s(5+h) – s(5) . h  0 h lim (5+h)2+10- (52+10)h  0 h = lim 25+10h+h2+10 – 25 -10 h  0 h lim 10h. h  0 h = = = 10 So speed when t=5 is v=10ft /sec Example 3: The position of an object is given by s(t) = at2 + bt+ c, with s in feet, t in seconds and a, b & c are constants. Find the speed of the object at any time t. s(t+h) – s(t). h a(t+h)2 +b(t+h)+c – (at2 +bt+c). h = at2 +2aht +h2+bt+bh+c- at2 –bt-c. h = . 2aht+h2+bh. . h = Speed at any time t = lim h(2at+b) / h . h  0 = lim 2at+b . h  0 = 2at +b

5. GEOMETRIC INTERPRETATION OF THE DERIVATIVE Let’s assume that a particle is moving along a line and it’s position is given by the function s = f(t). The graph f is shown below. To visualize the (instantaneous) velocity when t=a we have to identify the expression [f(a+h)-f(a)] / h] on the graph. Let call A and B the points on the graph corresponding to t=a andt=a+h. See orange triangle. s So [f(a+h)-f(a)] / h] = AB slope = mAB. B f(a+h) a+h s= f(t) Taking an h1smaller than h we get point B1 (more close to A than B) . The Average Velocity is represented by mAB1 (the slope of AB1). See blue triangle. B1 Bn f(a+h) – f(a) = f A f(a) a h = x Taking the h smaller every time ( h 0), the secant’s line  tangent line at A and msecmtan t a+hn a+h1 So lim [ f(a+h)-f(a)] / h] = f ’(a) = mtan = slope of the tangent line .h 0to the curve at x = a

6. GLOSARY OF TERMS • f(x+ x)-f(x) or f(x) represents: . x x 1) the Rate of Change of f(x) = change of f(x) / change of x 2) Average Speed when x represents time. 3) The slope of the secant line passing though the points A= (x, (f(x)) & B=(x+ x, f(x+ x)). II) The derivative of f(x), f ’(x), is used for the value of lim f(x+ x) - f(x) = lim f x0 xx0x By “differentiating f(x)” we means “get the derivative of y=f(x)” . The notation for the derivative of y=f(x) is f ’(x) or df /dx or dy /dx. III) f ’(c) , the derivative of f(x) evaluating at x=c, represents the: 1) Instantaneous rate of change of f(x) at x=c 2) Instantaneous speed at x=c when x represents time. 3) Slope of the tangent line to the graph y=f(x) at x=c 4) Marginal value of f(x) in Business.

7. Example1: Find the derivative of f(x) = 2x+1 Since y=2x+1 is a line and the tangent line coincides with it, the slope of the tangent line is the slope of y =2x+1, e.g. m=2. So f ’(x)=2 Example 2: Find the value of x where the tangent line to the graph of f(x)= 5x2+x+1 has slope 0 Since f ’(x)= lim 5(x+h)2 +(x+h) +1– ( 5x2+x+1). . h  0 h = lim 10xh +5h2+h.. h  0 h = lim h(10x+5h+1). h  0 h = 10x+1 Som = 0 when 10x +1=0  x= - 0.1 Example 2: Find the points where the tangent line to f(x)= x3-3x2-9x+1 is parallel to the X axis. f ’(x)=3x2-6x-9 = mT= 0 But 3x2-6x-9 = 3(x+1)(x-3)=0  x= -1 or x= 3 Since f(-1)= 6 & f(3)= -26 The points are (-1,6) and (3,-26)

8. Theorem : f is differentiable on the interval (a,b).  f is continuous on the interval (a,b). Proof: Assume that f ’(c) exists for any c in (a,b). = lim f(c+h) - f(c) . h 0 h Then lim [ f(c+h)- f(c)]. h0 • h = f ’(c) • lim h. h 0 = f ’(c) • 0 = 0 So lim [ f(c+h) - f(c)] = 0. h0 , and from here we get lim f(c+h) = f(c) .. h0 So f is continuous at c for every c in (a,b). Example: Since the derivative of f(x)= 5x2+x+1 is f ’(x) = 10x+1, which exists for every real number x. So f(x)= 5x2+x+1 is continuous everywhere. Remark. The reverse of this theorem is not true. Counter example: We know that f(x) = |x| is continuous on R , but at x=0 it’s not differentiable since: lim l0+hl –l0lh 0 h +1 if h 0 –1 if h0 = lim lhl. h 0 h , which approaches to

9. Let f be a continuous function. A point P=(c,f(c)) is called: 1) a corner at P , when the graph of f has two different tangent lines at p (one for x c- & the other for x c+ ). 2) a cusp when the tangent line at (c,f(c)) is vertical Y We’ll usually say that f is smooth on (a,b) if f is differentiable on the interval (a,b). CORNER The graph on the right shows the two types of points where f’(x) doesn’t exist. X CUSP Example 1: Where is a parabola not smooth? The general equation of a parabola is f(x)= ax2+bx+c . Differentiating we get f ’(x) = 2ax+b; therefore f ’(x) exists for every real number x and it is smooth everywhere. In other words, parabolas have no corner points or cusps. All parabolas are smooth. Remark: The same result is valid for any polynomial functions.

10. Ex 2: Find the value of x that f(x) = (2x-6) 2/3 is not smooth. f ’(x) = 4/3(2x-6)-1/3 , exists for every x except at x=3. f ’(3) + ∞ or -∞ when x3. But f is continuous at x=3, this means that the tangent line at (3,0) is a vertical line. So f has a cusp at x=3 . Y Since f(x) is never negative, the shape of the graph of f like the figure on the right. y= (2x-6) 2/3 X 3

11. DIFFERENTIALS Let f(x) be a differentiable function on an interval (a,b). Then lim f x0x lim f - f ’(x) x x0x – f ’(x) = 0 = y=f(x) Y Meaning that [f - f ’(x) x] 0when x0 Q See figure on the right f = f(x+ x) – f(x) (f is represented by the blue color curve. ) f To help us to identify the expression f ’(x) x we draw the tangent line in red color to the graph of f at P and check the red right triangle. df P dx . || x X Now we’ll rename x as dxand we’ll call df the other leg of the red right triangle (which measures the difference from the tangent line). x x+x f(x+dx)-f(x) df . dx dx On the red triangle we have: f ’(x) = m tan at P = = which implies that df = f ’(x)dx . (Note that this is an independent selection of x ).

12. In summary, when using df = f ’(x) dx you can find derivatives without limits. y=f(x) Y Let’s review that df = f ’(x) dx one more time and that m=dy/dx Tangent y=mx+b The graph on the right shows the tangent to y=f(x) at the point P(x,y) and dx, df & f at x+x as well. f df P dx The difference f – dy is represented in green color X x x+x On the graph see how f – dy 0 as x  0. Remark: This could explain why Leibniz used dy/dx notation for the derivative instead of the f ’(x) notation.

13. THE DIFFERENTIAL TO FIND DIFFERENTIATION RULES. First we’ll prove that (dx)2 = 0 (as mentioned before) . Let f(x)= x2 and find df in two ways: 1) f ’(x)= lim (x+x)2 – x2 . x0x lim x2 +2x x + (x)2 – x2x0 x lim x0 (2x+x) = 2x = = So df = f ’(x) dx = 2xdx 2) Using differential definition = 2xdx +(dx)2 df =(x+dx)2 - x2 = x2+ 2xdx +(dx)2 - x2 From 1) & 2) we get: 2x dx = 2xdx+(dx)2 . So (dx)2 = 0. REMARK: Using df = f(x+ dx) – f(x) plus the fact that dx2 =0, or better (dx) n = 0 for n ≥ 2 , we can perform the differentiation without using limit process. Example 1: Find f ’(x) for f(x)= x3 df = (x+ dx)3 – x3 0 0 df = x3 + 3x2dx + 3x(dx)2 +(dx)3 – x3 = 3x2dx d x3 .dx So dividing both sides by dx , we get = 3x2

14. Example 2. Differentiate f(x) = 5x3 – 2x2 –10 df(x) = 5(x+ dx )3 – 2(x+ dx)2 –10 –(5x3 – 2x2 –10) = 5(x3 +3x2 dx +3xdx2+ dx3 )–2(x2+ 2xdx +dx2 ) -10 –(5x3 – 2x2 –10) Replacing dx2 = 0 and collecting … d f(x) =5x3+15x2dx - 4x dx - 2x2 - 10– 5x3+2x2+10 Factoring out dx … df(x) = (15x2 +4x)dx Dividing by dx … f ’(x)= 15x2 +4x Example 3: Find dy/dx if y=xn , for any positive whole number n 0 Since (x+dx)n - xn = xn +nxn-1dx + (sum of terms containing (dx)n with n≥2) - xn So d((xn)= nxn-1dx and dividing by dx we get dy/dx = nxn-1 Example 4. Find d( x) / dx Let’s call y =  x . Squaring both side we get y2 = x So d(y2) = dx 2ydy = dx dy = dx / (2x )  dy/dx = ½ x

15. BASIC FORMULAS TO DIFFERENTIATE A FUNCTION Let u & v be functions of x, then 1) d(u ± v) = du ± d v Proof: d(u± v) = [u(x+ dx)± v(x+ dx)] – [u(x)± v(x) ] =[ u(x+ dx) – u(x) ]± [ v(x+ dx) -v(x) ] = d(u)± d(v) 2) d( ku ) = k du where k is a constant Proof: d(ku) = (ku)(x+dx) – (ku)(x)= k[u(x+dx)] – k[u(x)] = k[u(x+dx) – u(x)] = kdu 3) d(u v) = (du)v + u(dv) Proof: d(u·v) = [ u(x+dx)· v(x+ dx) ] – u(x)·v(x) = [u(x) + u’(x) dx ]·[v(x) + v’(x) dx ] – u(x)· v(x) Expanding … = u(x)v(x)+ u(x)v’(x) dx + u’(x) dx v(x)+ u’(x)dx v’(x) dx – u(x)v(x) = u(x)v’(x) dx + u’(x) dx v(x) + u’(x) v’(x) dx2 0 So, d(u v) = du·v + u·dv

16. 1 u - du. u2 4) d( ) = 1 u Proof: Notice that d ( u · ) = d(1) = 0 1 u 1 u 1 u du . u 1 u But d(u · ) = du·( ) + u·d( ) = 0  + u·d( ) = 0 1 u 1 u - du. u2 Isolating d( ) we get d( ) = u v (du) v – u (dv). v2 5) d( ) = (du)v–u(dv). v2 u v 1 v 1 v -dv . v2 du ( ) + u ( ) = Proof: d( ) = d( u· ) = FORMULAS TO FIND DERIVATIVES OF BASIC FUNCTIONS Let u & v function of x. Dividing by dx above formulas we get. 1) d (u  v)d ud v. dx dx dx 2) d (k·u)d u. . dx dx =  = k, k is a constant u v dudv dx dx d ( ) v - u 3) d (u·v)dudv. dx dx dx = v + u 4) ______ ___________ = dx v2 dxndx Besides was proved that = nxn-1for any integer n.

17. Ex 1. Find [√(x) - 2x3 +1]’ = [√x]’- [2x3 ]’+[1]’ = ½(x-1/2) – 2 ·3x2 + 0 = 1/(2√x ) – 6x2 Ex 2. For y= (5x2– 3x + 6) / (x2– 5) , find y’(3) y’ = (10x – 3)( x2-5) – (5x2 – 3x+6) (2x) . (x2– 5)2 y’(3) = 27(4)-42(6) = -9 . 16 Ex 3: Find dy/dx when x=1, for y = (x2+3x-1)(5x3-x2-2x+5) y’ = (x2+3x-1)’(5x3-x2-2x+5)+ (x2+3x-1)(5x3-x2-2x+5)’ = (2x+3) (5x3-x2-2x+5) + (x2+3x-1)(15x2-2x-2) So, y’(1) = (5)(7)+(3)(11)= 35+33 = 68 For more exercises click here

18. NON CONTINUITY v/s CONTINUITY v/s DIFFERENTIABILITY A) A function f is discontinuous at x=c when lim f(x) f(c) . x c Two different situations can occur. 1) there exists an interval containing x=c , where the function is continuous except only at x=c. Graphically this happens if: a) f(x) has a jump. See figures below b) A point is missing or has been relocated. See figure below. OR Finite jump Infinite jump Point missing or relocated Y Y Y Y X X X X c c c c 2) It f doesn’t exist on the interval as defined in part A) then the there exists an interval containing x=c where the function is discontinuous everywhere. See next example.

19. 1 if x is rational 0 if x is irrational Example: Let f(x) = This function is discontinuous everywhere. Its graph cannot be drawn. B) A continuous function f is not differentiable at x=c , if and only if, lim f(c+x) – f( c) does not exist. x0 x At x=c the graph of f can either: 2) has a corner ( there are two tangent lines instead of just one tangent line at x=c.) 1) has a cusp ( the tangent line at x=c is vertical) OR See graph below See graph below Y Y X X c c

20. CHAIN RULE Theorem: Let y = f(u) and u= g(x) be differentiable functions (Notice that y = fg(x) = f(g(x)) depends only onx).Then y u x dy dx dy du du dx = (See figure on the right). dy/du Using Newton’s notation: (f(g(x))’= f ’(g(x))·g’(x) dy/dx Proof: By definition dy = d(f(u))= f ’(u)·du du/dx But du =u’(x)dx Replacing du … dy = f ’(g(x)) u’(x)dx Dividing by dx we get dy . dx = f ’(g(x)) g’(x) Ex 1: Find dy /dx if y = √3x2 –5 y= √u u = 3x2– 5 x dy/du =1/(2√u) Let u = 3x2 – 5 , so y = √u dy/dx Then dy/du = 1/(2√u). & du/dx = 6x Using the chain Rule dy/dx =[ 1/(2√u )]6x du/dx= 6x But u = 3x2– 5 ,we get dy. dx .3x .√(3x2 – 5) =

21. Remark: (f(g(x))’= f ’(g(x)) g’(x) says that to get the derivative of “nested functions” you multiply the derivative of each one starting from left to right and so on). (beware that sin2(3x-1) means [sin(3x-1)]2 Ex 1. Find y ’ for y = (5x2 –17x+3)5 = 5(5x2 –17x+3 )4• (10x-17) Ex 2: Find y’(1) for y = (3x2-2)3( 5x3-x-3)4 y ’= 3(3x2 -2)2 (3x2-2)’( 5x3-x-3)4 + (3x2-2)3 4( 5x3-x-3)3 ( 5x3-x-3)’ y ’= 3(3x2 -2)2 (6x) ( 5x3-x-3)4 + (3x2-2)3 4( 5x3-x-3)3 (15x2-1) y ’(1)= 3(3-2)2 (6) (5-1-3)4 + (3-2)3 4 (5-1-3)3 (15-1) = 74 . 2 x - 1√5x2+4 dy . dx Ex 3: For y = , find when x=1 . y ’ = (2x–1)’ √5x2 +4 – (2x-1) ((5x2 +4)1/2 )’. √(5x2 +4)2 y ’ = (2)√5x2+4 – (2x-1) ((½(5x2+4)-1/2(10x). (5x2 +4) 13. 27 2√9 -(2-1) [½(9)-1/2](10). 9 y ’(1) = =

22. APPROXIMATING f by df error .x Since when x0 both error = f - df  0 and  0 , we can say that for small values of x, df is a good approximation of f. error .f(x) f - df.f(x) We’ll call the relative error of approximating f - df = Ex 1: Use the differential to approximate √(27) and estimate the relative error percent Let f(x) = √x and x=25 (since 25 is the perfect square of 5 and is near to 27), then x=2 . Substituting x=25 and x=2 in the approximating formula f(x+x)- f(x)  f ’(x) x we get f(x+x)  f(x)+ f’(x) x √(25+2)√25 + 1/(2 √25) (2) = 5 + 1/5 = 5.2 . So √(27)  5.2 Since (5.2 -5)/5.2= 0.038 then % error = 3.8 %

23. Ex 2: Given f(x) = x3 – 5x2 +8x +9, use the differential to estimate value for f(3.74) Let’s take x=4 andx = - 0.26. Since f’(x) = 3x2 – 10x+8 , f’(4) = -16. So f(3.74) = f(4) + f ’(4)x = 25 +16(-0.26) = 20.84 Since f(3.74) = 21.23 we have % error  (0.39 / 21.23) = 1.84% REMARK: As you see when working with the differential df we are replacing f(x) by the equation of the tangent line at x and f (x+x)= f(x) + f ’(x) xis the (linear) approximation for f. Besides lim error = lim f- df = df – df. x0 x x0x dx dx = 0 which means, that for small values of x , df is a good estimation of f .

24. Ex 3: Use linear approximation to estimate the change of the volume of a cylinder when the radio increase from 5 in to 5.2 in and the height decreases from10 in to 9.7 in. ( Volume of a cylinder is given by V(r,h)= r2h) Find the relative error. V  dV =(2rdrh+r2dh) Now we replace r=5, h=10, dr=0.2 & dh = -0.3 dV= [ 2(5)(0.2)(10) + (5)2(-.3) = (20-3)]=17 =53.4 Relative error = 17. 250 = 6.8 % Ex 4: A hemispherical bowl of radius 10in. is filled with water to a depth of x in. Assume that you measure the depth of water in the bowl to be 4 inch. with a maximum error of 1/20 in. Estimate the maximum error in the calculated volume of water in the bowl. (Volume of a hemispherical is V=(3rx2 – x3)/3 ) V (x+x)-V(x)  V’(x)x = (60x – 3x2)/3·x = (20x – x2)·(1/20) = (80 – 16)/20 =   64/20 = 10.05in3 So, V(4)=(301.59 10.05)in3

25. If c is in a closed interval [a,b], then f( c) is called the minimum value of f(x) on [a,b] if f( c) ≤ f(x) for all x in [a,b]. Similarly, if d in [a,b], then f(d) is called the maximum value of f(x) on [a,b] if f(d) ≥ f(x) for all x in [a,b]. MAXIMA AND MINIMA OF FUNCTIONS ON CLOSED INTERVAL It can be proved that a continuous function on a closed interval [a,b] attains a minimum value f( c) and a maximum value f(d). Y Y f(d) f(b) f(c) f(a) X X d c b a b a We say that the value of f( c) is a local minimum value of f(x) if f(c) ≤ f(x) for all x sufficiently near c ; more precisely if there exists an open interval containing c where the inequality holds. A local maximum value of f(x) is defined similarly.

26. APLICATIONS OF FIRST AND SECOND DERIVATIVES Theorem: Let f be a differentiable function on (a,b) the 1) if f ’(x) is positive then f is increasing on (a,b) 2) if f ’(x) is negative then f is decreasing on (a,b) Proof: 1) Let x1 & x2 be any two numbers in (a,b) and x2 > x1 , using the Mean Value Theorem there exist a real number c in (x1 , x2 ) such that: f ’(c ) = [f(x2 ) – f(x1)] / [x2 - x1] + + But by hypothesis f ’(c ) & x2 - x1 are positive so f(x2)–f(x1) is positive. This means that f(x2) -f(x1)> 0 for any x1 & x2 in (a,b). In other words f is increasing on (a,b) Similarly 2) is true. Ex 1: Prove that f(x) = 2x3+6x -13 is and increasing for every real x Using the Theorem we only have to prove that f ’(x) is always positive Since f ’(x)=6x2+6 =6(x2+1) and x2+1 is always positive we get that f ’(x) > 0 on (- ∞ , + ∞).

27. Ex 2: Find the increasing and the decreasing open intervals for the function . f(x) = 6x5-240x3-750x+1 f ’(x)=30x4-720x2-750 = 30(x4-24x2-25) and we need study the sign of f ’ Factoring … f ’(x) = 30(x2-25)(x2+1) = 30(x+5)(x-5)(x2+1) Note that x2 +1 is always positive so 5 and -5 are the only critical numbers + + -6 0 6 -5 5 Take -6 as a testing value for the sign of f ’ on (- ∞ ,-5): f ’(-6) is + Take 0 as a testing value for the sign of f ’ on (-5 ,-5): f ’(0) is - Take 6 as a testing value for the sign of f ’ on (6, + ∞): f ’(6) is + So f is increasing on (- ∞ ,-5) & (6, + ∞) and decreasing on (-5,5)

28. Theorem: Suppose that f is differentiable at c in an open interval containing c. If f(c) is either a local maximum value or a local minimum value of f, then f ’(c)=0 Proof: Assume f is differentiable and has a local maximum at x=c. If f ’(c) > 0 then f is increasing and cannot have a local max. at x=c. Similarly if f ’(c) < 0 , then f is decreasing and cannot have a local min. So, f ’(c) has to be equal to zero REMARK: If a function f continuous on (a,b) has a local maximum or a local minimum at x=c for then: 1) f ’(c)=0 or 2) f ’(c) is undefined at x=c (f has a corner). The real numbers satisfying the above two conditions are called critical numbers. Theorem: Suppose that a continuous function f is defined on a closed interval [a,b], then the absolute maximum and the absolute minimum of f occur at a critical value or at the extreme of the interval.

29. Ex 1: Find the maximum and minimum values for the function f(x)= x3-3x2-9x+20 on the interval [-2,5]. First find f ’(x) … f ’(x) =3x2-6x-9 Since f ’(x) exists for every x in [-2,5], the critical values are given by f ’(x) =3x2-6x-9 =3(x+1)(x-3)=0  x= -1 or x=3 The maximum and minimum values are located at the critical numbers -1 or 3 or at the ends -2 or 5 Select the maximum and the minimum values using the table below. x f(x) Max {18, 25, -7, 25} = 25 -2 -1 3 5 f(-2)=18 f(-1)=25 f(3) = -7 f(5) = 25 Min {18, 25, -7, 25} = -7 So the maximum value is 25 and is obtained at x= -1 and x = 5. And the minimum value is -7 and is obtained at x=3.

30. Ex 2: We are given a fence of 200 ft long and we want to build a rectangular enclosure along a straight wall. If the side along the wall needs no fence, find the largest area of such rectangle. See figure on the right W A L L Let’s call x & y the sides of the rectangle as shown in the figure y y x So, length of the fence = x+2y =200  x=200-2y and area of the rectangle = xy = A  A = (200-2y)y=200y-2y2 We need to maximize A, so we need dA/dy … dA/dy = 200-4y From 200-4y =0 we get y=50 (the only critical number) From 200 = x+2y we get that 0 ≤ y ≤ 100 , so [0,100] is the interval So, the maximum area is the maximum value of A for y= 0,100 & 50 y A=200y-2y2 The maximum area is A=5000 ft2 . 0 100. 50 . 0 . 0 5000

31. Ex 2: A can (cylinder) with volume of 240 cm3 has to be constructed. Find the dimension of the cheapest (using the least material) can. (Assume that the radius r has to satisfy the inequality 1cm ≤ r ≤ 5 cm). See figure below. We have to minimize the total surface of the cylinder. Isolating h from V=240=r2h we get h= 240 / r2 r Replacing h in S formula we get S= 2(240 / r )+2r2 Find the critical numbers … dS/dr = - 480/r2 + 4r = 0 h So S=480/r + 2r2 and dS/dr = - 480/r2+4r Solving … - 480 +4r3= 0  r =(120/ )1/3  3.4cm Since r has to be in [1,5], making the table: V= r2h r 1.0 3.4 5.0 S(r) 486.3 213.8 253.1 S=2rh + 2r2 The radius of the cheapest can is r  3.4cm

32. LOCAL MAXIMUM AND LOCAL MINIMUM Let f(x) be a continuous function. We say that x=c is a critical value of f(x) if f ’(x) is 0 or if f ’(c) is not defined. In other word, if the slope of the tangent line to y=f(x), at x=c, is 0 or is not defined. We say that f(x) has a local maximum at x=c if there exists an open interval (c-h , c+h) where f(c) is the maximum value of f(x). In a similar way we define a local minimum. Ex 1: The graph of y=f(x) is shown below. Find : 1) critical values, 1) f ’(x)=0 at x=h with b  h  c , d, e & g 2) local max and min. Y f ’(x) DNE at x= a, b, c & f So, the critical values are x=a, b, c, d, e, g & h X a b h c d e f g 2) Local maximum at x =d & f Local minimum at x= a & e

33. Remark : The above example shows that local maximum and minimum occur at the values of x that are critical values ( the reverse is not true). Test for Local Extrema: Let f(x) be continuous on an interval I and differentiable there except possibly at the interior point c of I: 1) f ’(x) 0, for x< c & f ’(x) 0for x>c, then f(c ) is local minimum of f(x). 2) f ’(x)0,for x<c & f ’(x)0 for x>c, then f(c) is local maximum of f(x). 3) f ’(x) has the same sign both to the left and to the right of c, then f(c) is neither a maximum nor a minimum value of f(x). Ex 1: Find local Extrema for y= x2(x-3)2/3 2x(x-3)+x2(2/3) . (x-3)1/3 y’ = 2x(x-3)2/3 + x2(2/3)(x-3)-1/3 = 1) y’ = 0  6x2-18x+2x2 =0  x= 0 or x=9/4 =2.25 Critical numbers 0, 2.25 &3 2) y’ DNE  (x-3)1/3=0  x=3 Study of sign for y’ … + + Using -1,1,4 & 10 as checking points…. -1 1 4 10 0 3 8.6

34. So using the Test for Extrema we get: Local minimum at x=0 & x=9/4 Local Maximum at x=3 Y Note that at x=3 the graph has a cusp, see its shape on the right. X 3 Ex 2: Find local Extrema for y= (x-3)3(x2-4x-4) f ’(x) = 3(x-3)2(x2-4x-4)+(x-3)3(2x-4) = (x-3)2(3x2-12x-12+2x2-4x-6x+12 ) = (x-3)2(5x2-22x) =x (x-3)2(5x-22) From f ’(x) = 0 we get the critical values 0,3 & 4.4 Find the sign of y’ … + + Using the Test … 0 3 4.4 Local maximum at x=0 Neither a Local maximum nor a minimum at x=3 Local minimum at x=4.4 See the graph in the next slide

35. Y Neither a local max nor a local min Note that f ’(3) exists e.g. the graph has an horizontal tangent line at x=3 but it has neither a local maximum nor minimum at there. See the shape of the graph on the right X 4.4 3 Second Derivative Test for Extrema: Suppose that f is twice differentiable on an open interval I containing a critical value c at which f ’(c)=0. Then 1) If f ”  0 on I, then f(c) is the local minimum number of f(x) on I. 2) If f ”  0 on I, then f(c) is the local maximum number of f(x) on I. Proof: We will prove only part 1). Since f ’(c)=0 and f ”(x)0 on I  f ’(x) is increasing on I, we may conclude that f ’(x)  0 for x  0 and f ’(x) 0 for x  0. Using the First Derivative Test, we conclude that f(c) is a local minimum value

36. Ex 1: Find the minimum of f(x)=x2+128x-1 for 0 x+∞ 1) Find critical values. f ’(x) = 2x-128x-2 = 0  2x3 -128 =0  x=4 2) Find second derivative f ”(x) = 2 +256x-3 3) Applying Second Test … Since f ”(4) = 6  0 on (0, +∞) , we know that f(4) = 16+32=48 is the maximum value of f(x) on (0, +∞) CONCAVITY POINTS AND INFLECTION POINTS Theorem: Suppose that a function f is twice differentiable on an open interval I and T is tangent line to the graph y=f(x) at (a,f(a)) then 1) If f ”(a)  0 then tangent line T at a lies above T . We say that the function f is concave upward ) 2) If f ”(a)  0 then tangent line T at a lies below T. ( We say that the function f is concave downward) Let f be a continuous function. We say that (c,f(c)) is an inflection point of f when for x c the concavity of f is the opposite of the one for x c. In other words, f has a change of concavity at x = c.

37. A useful Theorem for absolute max or min over an open interval Theorem: Let c in (a,b) and f(x) a continuous function on (a,b). If c is the only critical number for f(x) then the local maximum (minimum) is the absolute maximum (minimum). Proof: 1) Assume that at x=c f has a local minimum y On (a,c)& (cb) f’(x) is or increscent everywhere or increscent except at their inflexion points. In similar way can be proved the rest. x Ex: Find the absolute maximum of a c b . x2 . .1+x4 . x2 . .1+x4 f(x)= on (0,+∞) Look for critical numbers … 2x(1+x4)-x2(4x3) . (1+x4)2 2x(1-x4) .(1+x4)2  f ’(x)= = f ’(x) = 0 if x=0 or x=1 So on (0,+∞) the only critical number is x=1 It’s easy to see that at x=1 f(x) has a local min. So at x=1 f(x) has the absolute maximum on (0,+∞)

38. Proof: See figure on the right Y y=f(x) Since y=T(x)= f(a)+ f ’(a)(x-a) , we define g(x) = f(x)-T(x) then g ’(x) = f ’(x) – f ’(a) g(x) For x=a we get g’(a) = 0 i.e. x=a is a critical number for g(x). y=T(x) But g”(a) = f ”(a)  0, so at x=a, g(x) has a minimum value 0 (since g(a)=0) f(a) a x X So the graph of y=f(x) lies above of T Ex 2: For f(x) = x6-3x5 -20x4+90x3-135x2+2x+1, find the intervals that f is concave up , concave down, and identify the inflection points. f ’(x)= 6x5-15x4-80x3+270x2-270x+2 f ”(x) = 30x4-60x3-240x2+540x-270= 30(x4-2x3-8x2+18x-9) Possible rational zeroes for f ”(x) are 1, 3, 9 , working we get: f ”(x)=30(x-1)2(x+3)(x-3) +__+ Sign diagram for f ” ….. -3 1 3 Concave up on (-∞ , -3)(3 ,+ ∞ ) and Concave down on (-3,3). So, f(x) has inflection points at x=-3 , x=3

39. Ex 3: Find the dimensions of the rectangle with maximum area that can be inscribed in the ellipse 16x2+9y2=144 . y See figure on the right. 4 16x2+9y2=144 Let P=(x,y) be the vertex of the rectangle in the first quadrant. So x is in the interval (0,3). P(x,y) Area of rectangle is A=2x(2y)=4xy x Isolating y from the ellipse … y= (4/3) (9 -x2 )1/2 3 So A= (2x) 2(4/3) (9 - x2 )1/2 = (16/3)x (9 - x2 )1/2 A’= (16/3)(9 - x2 )1/2 + (16/3)x (½ )(9 -x2 ) -1/2 (-2x) A’= (16/3)(9 - x2 )-1/2 [9 - 2x2 ] A’=0  9-2x2 =0  x= 1.5√2 2.12 (is the only critical value in (0,3) ) Using the 1st derivative test … + - For x= 1.5√2 there is a local maximum. 2 2.12 2.2 But being this the only critical point in (0,3) it has to be a absolute maximum. The dimension of the rectangle are b=3√2 & h=4√2.

40. Ex 1: If the consumer demand for a commodity is D(p)= 10,000e-0.02p units per week, where p is the selling price , find the price that will maximize the consumer expenditure. ( Hint: E(p)= pD(p) ) E(p)=10000p e-0.02p E’(p)= 10000e-0.02p +10000pe-0.02p(-0.02) = 200e-0.02p(50-p) Since 200e-0.02p cannot be zero E’(p)=0  50-p=0  p= 50 E’’(p) = 200e-0.02p(-0.02)(50-p)+200e-0.02p(-1) E’’(50)= -200e-1< 0. So expenditure is maximized when p= $50 Ex 2: If a certain drug is injected intramuscularly, the concentration of the drug in the bloodstream after t hours will be C(t)= .75(e-0.2t –e-0.6t ). Find the time of maximum concentration. C’(t)= .75(-0.2 e-0.2t +0.6 e-0.6t ) =.15(-e-0.2t +3e-0.6t ) So C’(t) =0  -e-0.2t +3e-0.6t =0  3e-0.6t = e-0.2t  e-0.4t = 1/3 Solving for t … -0.4t = ln(1/3) = -ln3  t = 2.5ln3 = ln 35/2  2.75 Let find C’’ to apply the 2nd derivative Criteria for locals max and min C’’(t)=1.5 (0.2 e-0.2t – 1.8e-0.6t ) = 0.3e-0.2t (1- 9e-0.4t ) Since C’’(ln(35/2))  0, the maximum concentration occurs at t 2.75 hr.

41. Ex 4: A long gutter is to be made from a 12 inch-wide strip of metal by folding up the edges. How much of each edges should be folded up in order to maximize the capacity of the gutter. (Hint: maximizing the capacity means maximizing the cross-sectional area). See figure below Let x be the base and y the height of edge. So, we want to maximize the area A= xy y y area Isolating x from the restraint x+2y =12 … A=(12-2y)y = 12y -2y2 x A’=12 -4y  Critical value y=3 A’’ = -4  0  for y=3 the area has a local maximum. Since there is only one critical value, the local maximum is the absolute maximum. So the height of each edge y that maximizes the capacity is 3in.

42. Ex 5: When you cough you are using a high-speed stream of air to clear your trachea (windpipe). The radius of normal trachea is 3cm. When you cough the velocity of the air stream is given by V(r)=c(3-r)r2, where r is the radius of the contracted trachea and c is a positive constant (depending on the elasticity of the trachea). Find the radius r that maximizes V. V(r)=3cr2-cr3  V’(r)= 6cr-3cr2  critical values are 0 & 2 We discard r = 0 (has no meaning) and work on the interval (0,3) where r = 2 is the only one critical value. V’’(r)=6c-6cr  V’’(2) = -6c  0  V(r) has a local maximum at r =2 Since r=2 is the only critical value on (0,3) and is a local maximum, we conclude that V(r) has absolute maximum at r=2. So, the velocity of the airs tream is maximized when the radius of the contracted trachea is 2cm.

43. Ex 6: It costs a Car Company $ 18,000 to produce a car. The company’s weekly selling price function is p(x)=30,000 – 60x when exactly x items are sold. The fixed cost depends on the amount of production is $25,000. How many cars should be produced each week to maximize the company’s profit? Let x be the number of cars produced each week. The Cost is C(x)= 18000 x +25000 The Revenue is R(x)=(30000 – 60x )x So Profit P(x) = R(x)-C(x) = 30000x-60x2 – (18000x+25000) P(x) = -60x2 +12000x - 25000 P’(x)= -120x +12000  Critical value is x=100 Since P’’(x) = -120  0 , P(x) has a local maximum at x=100 But since this is the only critical value for P(x), it has to be a maximum The profit is maximized when 100 cars are produced weekly.

44. TRIGONOMETRIC FUNCTION DERIVATIVES lim sinx x0 x Using a geometric approach we will prove that = 1 The figure below shows a unitary circle with an acute central angle x (in rad) and its corresponding arc BC = x . D From figure: AC  arc BC  BD so sin x  x  tan x C Dividing by sin x .… tan x x 1 x 1 .. sin x cos x 1   sin x x O A B lim sinxx0x If x 0 … 1 1 lim sinx x0x So = 1 sin2x .sin x 1+cos x 1-cosx . sin x (1-cos x)(1+cosx) . sin x (1+cos x) = = REMARK: Since When x0 …. lim 1-cos x x0 x lim . sin x x0 1+cos x = = 0

45. DERIVATIVES OF TRIGONOMETRIC FUNCTIONS 1) (sin x)’ = cos x sin (x+x) – sin x . x sin x cos x +cos x sin x – sin x . x Since = sin x (cos x – 1) +cos x sin x . x = So when x0 … 0 1 lim cosx – 1 . x0x lim sin x. x0x So (sin x)’ = (sin x) + (cos x) = cos x 2) (cos x)’ = - sin x (cos x)’ = [sin(/2 – x)]’ = cos (/2 – x)] (-1) = - sin x 3) (tan x)’ = 1/cos2 x = sec2x 1-sin x 1+sinx Ex 1: Find y’ and y’’ for y= - 2 cosx (1+sinx)2 (-cosx)(1+sinx)-(1-sinx)( cos x). (1+sin x)2 y’ = = 2(2-sin x)(1+sin x)2 (2sinx)(1+sinx)2 - (-2cosx)[2(1+sinx)cos x]. (1+sin x)4 = y’’ = 3

46. Using Quotient Rule we get: 4) (sec x)’= sec x tan x 5) (cot x)’ = - csc2 x 6) (csc x)’ = - csc x cot x 1-sin x 1+sinx Ex 2: Let y= for 0≤ x ≤ 3/2 Find local extrema, concavity up and down intervals & inflection points. - 2 cosx (1+sinx)2 2(2-sin x) (1+sin x)2 From Ex 1 above we get y’ = y’’ = Since (1+sinx)2 ≥ 0 the sign of y’ depends only on -2cosx, so: + y is decreasing on (0, /2)) y is increasing on (/2,3/2) 0/2 3/2 So, y has a local min at x = /2 Since (1+sinx)2 ≥ 0 the sign y’’ deepens only on 2(2-sinx), so: + y is concave up on (0, 3/2) 0 Since y’’ > 0 on [0, 3/2] there is no inflection point.

47. Ex 3: The position function of a particle is given by s=(1+cos 3t)2 (1+sin3t)3, find the velocity of the particle when t =/6. s’ = [(1+cos 3t)2 ]’ (1+sin3t)3 + (1+cos 3t)2 [(1+sin3t)3 ]’ 2(1+cos 3t)(-sin 3t)(3) 3(1+sin3t)2 (cos 3t)(3) s’ = [2(1+cos 3t)(-sin 3t)(3) ](1+sin3t)3 +(1+cos 3t) 2 [3(1+sin3t)2 (cos 3t)(3)] Since for x= /6: sin(3t)= sin /2 = 1 & cos(3t)= cos /2 =0 v(/6 )= s’(/6) = - 6(1) (1+0) (1+1)3+ 9(1+0)2 (1+1)2 (0) = -48 +0 = - 48 Ex 3: The position function of a particle is given by s=(1+sin5t)2 (1-sin5t)2 Find the values of t in [0,/2] that the velocity is zero. First simplify the expression s= (1-sin2 5t)2 = cos4(5t) v=s’ = 4cos3(5t) (-sin 5t) (5) = -20 sin 5t cos3(5t) v=0  sin 5t =0 or cos 5t =0 sin 5t = 0  5t = 0  2k = 0 or 5t =   2k  t= 0, 0.2, 0.4 cos5t =0  5t = /2  2k = 0 or 5t = 3/2  2k  t = 0.1, 0.3 , 0.5 So v= 0 when t=0, 0.1, 0.2, 0.3 , 0.4, 0.5

48. RELATED RATES A related rates problem involves two or more quantities that vary with time and an equation that expresses some relationship between these quantities. In general, the chain rule is the key to find the related rates. Determine each of the following: (a) How fast is the top of the ladder sliding down the wall then? (b) At what rate is the area of the triangle formed by the ladder, wall, and ground changing then? (c) At what rate is the angle between the ladder and the ground changing then? Ex 1: A 15 ft ladder is leaning against a house when its base starts to slide away. By the time the base is 12 ft away from the house, the base is moving away at the rate of 6 ft/sec.

49. Sol: a) Using Pythagoras Theorem we get: Thus the top of the ladder is sliding down the wall at the rate of 6 ft/sec when the base of the ladder is 9 ft away from the house. b) Area of triangle = The area of the triangle is decreasing at the rate of 21 ft2/s when x = 12 ft.

50. (c) We need an equation that involve allow us to get d/dt Isolating d/dt … When x=12 the value of cos  is 12/15 and from b) dy/dt = - 8 ft/s Evaluating d/dt in that moment … The angle  is decreasing at the rate of 0.67 rad/s when x = 12 feet away from the house.