CHAPTER THREE APPLICATIONS OF THE DERIVATIVE by dsp14791

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									       Hare Krsna Hare Krsna Krsna Krsna Hare Hare
       Hare Rama Hare Rama Rama Rama Hare Hare
  Jaya Sri Sri Radha Vijnanasevara (Lord Krsna, the King of
                     Math and Science)
             KRSNA CALCULUS™ PRESENTS:


  CHAPTER THREE:
APPLICATIONS OF THE
    DERIVATIVE
             Edited on October 7, 2003
                HARE KRISHNA!

   Welcome back, dear students! I pray that you are all
    well.
   In order to understand this chapter, you will have to
    know the following:
   Definition of the derivative (at least the concept of it
    difference quotient.), derivatives of polynomials, trig
    functions, implicit differentiation, chain rule, and the
    rectilinear motion relationships (position, velcoity, and
    acceleration.)
   If you do not understand this, please review the last
    chapter (Chapter 2) and keep practicing that chapter
    until you understand those topics above.
WHERE WE HAVE GONE?

   As one mathematician said, “Calculus is a
    journey. A good tourist will think about the
    past here and use it to him or her in the
    future.”
   Before we move onto the next chapter, let’s
    take a step back and see what we have
    covered so far, just to help us remember
    important things.
SUMMARY OF PREVIOUSLY
STUDIED MATERIAL
   In chapter 1, we had studied limits. An informal and
    most commonly used definition of the limit is simply
    the question, “When I make x closer and closer and
    closer to some number c, what f(x) am I
    approaching to?”
   Asymptotes were explored using limits. Limits help
    determine end-behavior.
   Continuity was briefly mentioned. Continuity is
    present at x=c is f(c) exists, if the limit as x
    approaches c from both sides exist and are the
    same value, and if the limit as x approaches c and
    f(c) are equal.
SUMMARY

   With the help of limits, we were able to
    calculate instantaneous rates, and the slope
    of the line tangent to the curve at a certain
    point. By letting the change in x be
    infinitesimally small in the slope formula
    (difference quotient), we are able to get
    derivatives.
   A differentiable function is continuous
    function, but a continuous function is not
    always differentiable.
DERIVATIVES.

   Practice of derivative rules and formulas
    must be emphasized. Derivatives and
    Integrals (next chapter) will be the backbone
    to calculus.
   Definitely study the differentiation rules from
    the last chapter!!!!!!
APPLICATIONS OF THE
DERIVATIVE
   This chapter will underscore the uses of
    derivatives in the following cases.
   How to graph functions by knowledge of
    critical points and derivatives.
   How to differentiate inverse, exponential and
    logarithmic functions.
   Differentials
   Related rates and optimization problems
CURVE SKETCHING

   You can easily graph any function by
    knowing three things.
   1) ZEROS AND UNDEFINED SPOTS
   2) MAXIMUM AND MINIMUM POINTS
   3) CONCAVITY AND INFLECTION POINTS.
CRITICAL POINTS
   Let’s find the points where the function is zero or undefined.
   We want to graph f(x)=x3+2x2+x.
   Obviously, there are no undefined spots, due to the fact that this
    is a polynomial function. All polynomial functions are continuous,
    thus differentiable.
   To find the zeros, set the entire function to 0.
   0= x3+2x2+x  Original function
   0=x(x2+2x+1)  Factored
   0=x 0=(x+1)(x+1)  Factored the trinomial and set both factors
    equal to zero.
   x=0, -1  x is solved for.
   Thus x=0 and x=-1
MAXIMUM OR MINIMUM POINTS

   Here is the graph of the
    f(x).
  MAXIMUM AND MINIMUM
Notice  that if you draw
tangent lines at the
maximum and the
minimum, you will see that
you will get horizontal
lines. That means the
slope is zero. Thus, the
derivative is zero. In other
words, at a maximum or
minimum, the derivative is
zero.
MAXIMUM AND MINIMUM POINTS

   Since the f’(x)=0, we can solve for the maximum and minimum
    points.
   First, find the derivative of f(x)=x3+2x2+x.
   f’(x)= 3x2+4x+1  Differentiation using power rule and
    sum/difference rule.
   0 = 3x2+4x+1  Max/Min always have f’(x)=0.
   0=(3x+1)(x+1)Factoring to solve for x.
   0=3x+1 0=x+1  Setting factors equal to 0.
   x=-1/3 and x=-1  Solve for two answers of x.
   We know that x= -1/3 and x=-1 are the possible points, however,
    without a graph, we cannot determine which is the max and
    which one is the min.
SIGN ANALYSIS TEST FOR
MAX/MIN.
   You can do a sign chart and determine the sign between possible
    zeros. Look how it is set up.
   What the sign of the two factors to the right of x=-1/3?
   What is the sign of the two factors between -1 and -1/3?
   What is the sign of the two factors to the left of -1?
   Since they are factors, you multiply them. Account for overall sign.
DERIVATIVE OF SIGNS

   If the derivative goes
    from negative slope to
    zero to positive slope,
    then x is where the       zero
    minimum is.
   If the derivative goes
    from positive slope to    zero
    zero to minimum slope,
    then x is where the
    maximum is.
WHAT WE KNOW

   We know that this is a cubic function, with x=-
    1 and x=0 as the two zeros. No undefined
    points
   We know that the maximum is at x = -1 and
    the minimum is at x= -1/3.
    HORIZONTAL TANGENTS
   Be careful when solving for x
    when f’(x) is 0. Perhaps you will
    have such function where the
    slope is zero at a certain point,
    but that point is neither
    maximum nor minimum. y(x)=x3
    for example. Once you
    differentiate that, you get
    y’(x)=3x2. y’(x)=0. Therefore,
    you equate 0 and 3x2. You will
    see that x = 0. However, by
    doing a sign analysis chart, you
    will see that before and after 0,
    there is no sign change. The
    slope continues to be positive.
CONCAVITY

   Concavity is very difficult to define.
   A good way to define concavity is using a
    spoon. When you hold a spoon the right way
    (where the milk is in the spoon), the spoon is
    said to be concave up.
   When the spoon is down (when the milk is
    not in the spoon… the spoon is not in the
    right way), then the spoon is said to be
    concave down.
CONCAVITY

   An upward parabola is concave up. A downward
    parabola is concave down.
   The f(x) is above the tangent line when it is concave
    up. The f(x) graph is under the tangent line when it
    is concave down.
   The concavity is the rate of slope change. Hint hint!!
    RATE OF CHANGE!! The rate of change of the
    slope, means the derivative of the slope… The
    slope is the derivative of the function. So the
    concavity is the derivative of the derivative. The
    first derivative of slope and the second
    derivative of the function.
CONCAVITY GRAPHS
                                      Both function
Both function                         have negative
have positive                         slopes. f’(x)<0.
slopes. f’(x)>0.                      The blue
However, the                          function is
red function is                       concave up,
concave up,                           since it is
since it is                           above the
above the                             tangent line.
tangent line.                         The red
The blue                              function is
function is                           concave down,
concave down,                         since it is
since it is      RED       BLUE                          RED        BLUE
                                      below its
below its        f’(x)>0   f’(x)>0                       f’(x)>0    f’(x)>0
                                      respective
tangent lines. f’’(x)>0    f’’(x)<0
                                      tangent line.
                                                         f’’(x)<0   f’’(x)>0
ZERO CONCAVITY

   Remember when the function, f(x), was at a
    maximum or minimum, the derivative, the
    slope, f’(x) was 0.
   When the slope is at its maximum or
    minimum, the derivative of the slope is 0.
   In order words, when f’(x) is at its maximum
    or minimum, then the derivative, f’’(x), the
    concavity will be 0.
   Concavity is the slope of the slope. !!
Side note: Vertical tangents.
   Let’s examine for a moment, the vertical
    line graph.
   There is no such function that when
    graphed has a vertical line.
   If you look at the slope of a line (Dy/Dx).
    You will see that the difference in y is some
    number. However, since there is no change
    in x, that will be equal to zero. n/0 is
    infinitely big that is undefined.
   However, there are functions with vertical
    tangents. The cube root of x at x=0 has a
    vertical tangent. Therefore, the derivative of
    the cube root function will be defined
    everywhere except x=0.
   Just be aware of that!
CONCAVITY AND INFLECTION
POINTS
   The point where the change in derivative sign
    (positive vs. negative slope) is called extrema (i.e.
    max/min points) f(max) and f(min), their derivatives
    at those points will both be 0.
   The point where the change in concavity occurs is
    called the inflection point.
   Since this is where the slope, f’(x), is the maximum,
    the steepest or the flattest, the derivative of slope,
    f’(x), will be 0.
   After solving for x and doing a sign analysis, you will
    find the inflection points.
FINDING INFLECTION POINTS
   The original function was x3+2x2+x.
   Since we want to find the inflection point, we must obtain the second
    derivative.
   f(x)= x3+2x2+x.  GIVEN
   f’(x)= 3x2+4x+1.  FIRST DERIVATIVE
   f’’(x)=6x+4  SECOND DERIVATIVE
   Since the inflection points are found by solving for x when f’’(x) = 0.
   0=6x+4 f’’(x) =0
   x= -2/3  SOLVED FOR X.
   Now we must do sign analysis.
   Everything right of -2/3 for the factor 6x+4 is positive. Everything left
    of it is negative. Thus, x=-2/3 is an inflection point for x3+2x2+x.
   Everything right of x=-2/3 is concave up. Everything the to left of x=-
    2/3 is concave down.
When sketching this curve…

   Make sure your zeros are at the x values.
   Make sure your max, mis, and horizontal
    tangents are drawn at the proper spots.
   Make sure your inflection point is accurately
    drawn.
   A good way of thinking about inflection points
    is when the tangent lines switch sides with
    respect to the graph.
YOUR GRAPH

                                                                                f’(x)>0 f’’(x)>0
                      f(x)=0
                      f’(x)=0
                      f’’(x)<0
                                 MAXIMUM                     ZERO
                   ZERO
                                                 f’(x)<0               f(x)=0
                                                 f’’(x)>0
                                                            f’(x)=0
                                 f’(x)<0                    f’’(x)>0
                                 f’’(x)<0
f’(x)>0 f’’(x)<0                                              MINIMUM
                                  INFLECTION
                                     POINT
                                            f’(x)<0
                                            f’’(x)=0
ROLLE’S THEOREM

   Some people hate calculus a lot. For example, they
    might like to find instantaneous velocity by saying
    distance/time. But we know that is not true.
   Rolle was a mathematician who tried to disprove
    calculus. He disliked it very much. He wanted to a
    lot of geometry and algebra. Well, you know, if you
    hate something a lot, it is just as much as loving it a
    lot. Similarly, Rolle hates calculus so much, that he
    had a theorem that is very helpful IN calculus.
   This theorem will not be tested very frequently in
    exams, but its best to know it anyway.
ROLLE’S THEOREM (MEAN VALUE
THEOREM)
   There will always be an
    average rate line that will be
    parallel to the tangent line     f (b)  f (a)
    of a certain point. This                        f ' (c )
    means they have the same
    slope.
                                         ba
   You can see that the
    average rate line between
    x=-2 and x=2 is parallel to
    the tangent line of x=0.
    Thus, you can conclude,
    that they have the same
    slope.
DIFFERENTIATION OF INVERSE
FUNCTIONS
   An inverse of a function is found by reflecting the
    graph over the line y=x.
   The function is found simply by switching the y with
    the x and vice verse.. For example: If y = 3x+2 with
    coordinates (0,2) as one possible point, then its
    inverse is x=3y+2 with coordinates (2,0). I just
    switched the x and y around. That’s all.
   Sometimes, a function may have an inverse that is
    not a function. y=x2 is function and x=y2 is not.
    Remember, a function is a function if for any x, there
    is one and only one y value. The vertical line test
    determines that. (Course III Info for N.Y.S students).
DIFFERENTIATION OF INVERSE
FUNCTIONS
   From pre-calculus, you know that if f(x) has its inverse
    g(x), then the composition f(g(x)) is x.
   If f(x) has g(x) as its inverse function, then the formula
    for finding the derivative of a function is….



                 dy       1
                    
                 dx f ' ( g ( x))
DERIVATIVE OF THE INVERSE OF
y=x2.
   f(x)                       f ( x)  x   2


   g(x)                       g ( x)  x
                                                 
   Composition of f and g                        2
   dy/dx of f(x)              f ( g ( x))  x
   Composition of f’ and g.   f ' ( x)  2 x
                               f ' ( g ( x))  2 x
   Inv. Deriv. Formula        dy            1
                                     
                               dx f ' ( g ( x))
                               dy         1
   Substitution.                    
                               dx 2 x
CHECK
   You can check it also using the power rule.

                                 1
         y xx                  2

                             1
         dy 1                           1
            x               2
                                 
         dx 2                        2 x
DERIVATIVE OF EXPONENTIAL
FUNCTIONS
   The natural exponential function, f(x)=ex, has the base e.
    e is a transcendental number (just like 16,108, p, f...). It
    is named after its founder, Leonard Euler. e is derived
    and found many ways. e is a very special number in
    calculus. You will see why in a moment. e is
    approximately 2.718… If you define it using limits, e = …



                                              x
                        1
              e  lim 1  
                  x 
                        x
DERIVATIVE OF THE
EXPONENTIAL FUNCTION
                                    f x  h   f x 
   Derivative definition     lim
                              h 0            h
   f(x)=ex                         e xh  e x
   Law of exponents,         lim
                              h 0       h
    ex+h=(ex)(eh)
                                    e xeh  e x
   Factor out ex.            lim
                              h 0        h
                                    e x (e h  1)
                              lim
                              h 0        h
   Previously proven limit         eh 1
                              lim            1
                              h 0     h
   Product Rule of limits    e x (1)
   Derivative                ex
THE DERIVATIVE OF THE
EXPONENTIAL FUNCTION
   This is one of the two functions whose derivative is
    its own function! d(ex)=ex! Amazing how
    transcendental functions work! Jaya! Jaya!




               d x
               dx
                  e e  
                       x
THE NATURAL LOGARITHM
FUNCTION y=ln x
   The natural logarithm is NOT taking a piece of log
    naturally from a tree, play rhythms on it! 
   The natural logarithm is the inverse of the natural
    exponential function. It has e as its base. There are
    many ways y=ln x is derived by.
   y=ln x is defined for all x>0.
   We can find the derivative using the inverse
    differentiation rule since ln x is the inverse of ex.
DIFFERENTIATING y=ln x.
   f(x)                                         f ( x)  e   x


   g(x)  inverse of f(x)                       g ( x)  ln x
    Composition of f(x) and g(x)
                                                 f ( g ( x))  e ln x


   dy/dx of f(x)
   Composition of f’(x) and g(x)                f ' ( x)  e x
   If f’(g(x)) = f(g(x)), and if f(g(x)) = x,   f ' ( g ( x))  e   ln x
                                                                            x
    then f’(g(x))=x.
                                                 dy       1
   Inverse Differentiation Rule                    
                                                 dx f ' ( g ( x))
   Replacing f’(g(x)) with x.                   dy 1
                                                    
                                                 dx x
THEREFORE…

   I find it pretty cool that an algebraic function like y
    = 1/x is the derivative of transcendental function
    y = ln x. Transcendental functions are functions
    that cannot be derived simply by algebra.



                 d
                    ln x   1
                 dx           x
INVERSE TRIGONOMETRIC
FUNCTIONS
   The inverse of the function y = sin x is y= arcsin x.
   NOTE: arcsin x is NOT a function. For example, arcsin (½) = p/3, 5p/6,
    -7p/6, -11p/6, so on… since one x value produced an infinite number of
    y values, you know that this is not a function
   However, Arcsin x IS a function (notice capital A, compared to previous
    lower-case a). Arcsin x is restricted from [-p/2, p/2], just as Arccos x is
    restricted from [0,p] and Arctan x is restricted from (-p/2, p/2). Note that
    the () parenthesis represent the open interval excluding endpoints.
    Remember than tan (p/2) = ∞.
   If you look at the inverse cosine function, you will notice that it is just the
    inverse sine function reflected over the x axis. Therefore, the slope, the
    derivative, of the inverse cosine function will be negative the slope,
    derivative, of the inverse sine function. Same thing will occur with
    tangent and cotangent, secant and cosecant functions.
   d(arcsin x)=-d(arccos x)
   d(arctan x)=-d(arccot x)
   d(arcsec x)=-d(arccsc x)
DIFFERENTIATION OF THE
ARCSINE FUNCTION
   f(x)                       f ( x)  sin x
   g(x)  inverse function    g ( x)  arcsin x
   Composition of f and g.    f ( g ( x))  sin(arcsin x)
   dy/dx of f
                               f ' ( x)  cos x
   Composition of f’ and g.
   Derivative formula for     f ' ( g ( x))  cos(arcsinx)
    finding inverses of        dy       1
    functions                     
                               dx f ' ( g ( x))
   Applying f’(g(x)).
                               dy          1
   What is cos(arcsin x)?        
                               dx cos(arcsinx)
The denominator

   To simplify the denominator to more practical terms,
    we know that y=arcsin x is the same as sin(arcsin
    x)= sin y = x. Then to solve for cos y.

       Pythagorean Trig Identity   sin y  cos y  1
                                        2       2


       sin y = x                   x  cos y  1
                                    2       2


       Solved for cos y.           cos y  1  x    2
AFTER APPLYING cos y IN THE
DENOMINATOR…
   We see that…

        d arcsin x     1
                      
             dx         1 x 2


        d arccosx        1
                      
             dx          1 x  2
  USING THE SAME PROCESS…

     If you use the derivative of an inverse rule for the
      arctangent and the arcsecant functions, you will see
      that



d
   arctan x     1          d
                                 arc sec x         1
dx               1 x 2       dx                x     x 1
                                                      2


d
   arc cot x    2 1       d
                                 arc csc x            1
dx                 1 x       dx                  x       x2 1
      MISCELLANEOUS THINGS TO
               KNOW
   Some textbooks will make great emphasis on this.
    Other textbooks won’t talk about it too much. Since
    the AP and many college calculus I courses do not
    test this topic greatly, I won’t discuss it in great
    depth.
   You can treat dy/dx as a fraction of the
    infinitesimally small change in y over the
    infinitesimally small change in x.
   You can consider y = x2,  dy = 2x dx since dy/dx
    = 2x. dy is said, “a differential amount of y.”
    Differential means an infinitesimal change in a
    specific direction.
EXAMPLE DIFFERENTIAL
PROBLEM
   If you take a cube with each side having
    length 8, what is the differential volume if you
    have a cube that has each side having length
    7.99?
PROBLEM WORKED OUT

   The volume of a cube is
    given

                                   V x
   dV/dx is the differential             3
    amount of volume over the
    differential amount of
    change in length. The
    change in length is simply     dV  3 x dx2
    8-7.99=.01=dx. Multiply
    both sides by it so you
    isolate dV. Then plug in       dV  3(7.99) (.01)
                                                  2

    7.99 for x, and plug in 0.01
    for dx and calculate the
    differential amount of
                                   dV  1.92
    volume.
                 APPLICATION

   At the Sri Sri Gaura
    Nitai temple, they want
    to make the rectangular
    temple room such that
    the perimeter is 1000
    feet. Find the maximum
    length and width so that
    everyone can have the
    fortune to the Sri Sri
    Gaura Nitai. In addition,
    find the area.
                OPTIMIZATION

   Optima- what???
   Optimization problems is a fancy way of saying
    maximum minimum problems.
   The best way to start out is to draw a picture to help
    yourself visualize the problem.
                    THEN..

   Since we have two variables, ℓ and w, we
    have to solve everything in terms of one
    variable.
   We know that A= ℓ w and Prect = 2 ℓ +2w.
   So we can solve for ℓ in either equation. Most
    preferably use the P equation.
                 Prect  2w
              
                      2
And finally…

   Substite the expression for l into the area
    formula.        Prect  2w
               
                         2
                A  w
                    Prect  2w 
                A             w
                         2     
                   1
                A  Prectw  w2
                   2
According to the problem…

   P = 1000 ft so…

                               A  500w  w   2




   But to find maximum w,      A  500w  w2
    the derivative will have    dA
    to equal 0.                      500  2w  0
                                dw
   Then solve for w.
                                500  2w
                                250 ft  w
Then…

   Plug w in for the              1000  2w
    expression for length.     
                                       2
   Coincidentially, the
    width and length are         500  w
    the same. Therefore,
    the temple room will be      500  250
    square so that
    everyone could see the       250 ft
    Lord Sri Sri Gaura Nitai
     Jaya!
FIND THE AREA…

   Simple.. The length is
    250 ft. The width is 250
    ft. Multiply them
    together..
   A= 625,000 sq ft.

   Nice big temple ain’t
    it?? HARI HARI BOL!!!!
              RELATED RATES

   Remember how I constantly emphasized,
    RATE = DERIVATIVE
   Rate specifically deals with time.
   Here is an example:
   The diameter and height of a paper cup in the
    shape of a cone are both 4 in. and water is
    leaking out at the rate of 0.5 in3/ sec. Find the
    rate at which the water level is dropping when
    the diameter of the surface is 2 in.
VARIABLES SOUND NICE.. (laughs)

   V is volume in cubic
    inches                    1 2
   h is height.           V  πr h
   t is time                 3
                                      2
   Volume formula            1 h
    (known)                V  π  h
   2r = h, since             3  2
    diameter=h.
                              πh3
                           V
                              12
      Differentiate with respect to time
                                            πh 3
   Remember now… we are                 V
    differentiating with time!!! So,        12
    we must differentiate implicitly!!
    Remember chapter two??               dV  3πh 2  dh  πh 2  dh
   Since we know that the rate of            12  dt   4  dt
                                                            
    leak is ½ cu in/ sec, dV/dt is –     dt                  
    ½ since we are decreasing.
                                         dV      1
   Solve for the rate of change for
    the height.
                                            
                                         dt      2
   Since there is no variable for
    time, the rate of change for the     dh  1  4 
    height is constant for any value            2 
    of t.                                dt  2  πh 
   Replace the h with 2 in and the
    rate of the water level is           dh  1  4         1
                                                  
                                                    
                                         dt  2  π22 
    dropping with the height and
    time is 1/2p in/sec.                                    2π
SUMMARY

   We covered a good amount of material in this
    chapter. Let’s go over what we learned.
   You can find the maximum or minimum of f(x)
    by setting f’(x) = 0. Solve for x and do a sign
    analysis.
   You can find concavity of f(x) by setting
    f’’(x)=0. Solve for x and do a sign analysis.
DERIVATIVES                           
                                   d x
                                   dx
                                        e  ex

                                   d
                                       ln x   1
                                   dx            x
   We learned how to              d arcsin x      1
                                                  
    differentiate more functions         dx         1 x2
    using the inverse rule.        d arccosx         1
                                                  
   COMMIT THESE TO                      dx          1 x2
    MEMORY AS WELL AS
    THE FUNCTIONS IN
                                   d
                                      arctan x   1 2
                                   dx               1 x
    CHAPTER TWO!!! IF YOU
    DO NOT REMEMBER ANY
                                   d
                                      arc sec x   12
    FUNCTION’S DERIVATIVE,         dx                x x 1
    DO NOT GO ON TO
    CHAPTER FOUR!!
                                   d
                                      arc csc x    12
                                   dx                 x x 1
    COMMIT EVERYTHING TO
    MEMORY!!!!                     d
                                      arc cot x    21
                                   dx                 1 x
OPTIMIZATION/RELATED RATES

   For optimization problems, relate everything to one
    variable. For example, area and perimeter with sides
    a and b. Solve for a, for example, and plug the a
    expression in for the area formula so you have an
    area function A(b). Then differentiate that and set it
    equal to 0. DON’T FORGET SIGN ANALYSIS!!!
   For related rate problems, gather all of your
    variables together and use any geometry formulas
    you need. Differentiate with respect to TIME. You
    most likely will need to use implicit differentiation. It
   It helps in both scenarios to draw a picture of the
    problem to get a better understanding.
                      TEST

   Please visit the following website to take a
    practice exam and please look over the
    answers.
   If you have any difficulty, please e-mail me at
    kksongs_1@hotmail.com.
   Please read help statement prior to e-mailing
    me.
   Good luck! Hare Krsna!

								
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