Calculus Problems

                   Introduction: The topic Calculus is a vast topic in Math which will help us to find the volume of a solid generated by a curve, area between the curves and various topics in physics which involve in rate of change of certain quantities. To learn all these we should be familiar with certain topics like, Relations, Functions, Types of functions, Limits and Derivatives of functions, Applications of Derivatives in different fields, Integration, Properties of Integration, Application of integration in various fields etc. Tutor Next is the organisation which conducts one on one tutoring which help you you to master these topics. In this section let us discuss with some of the problems in calculus and the various methods of solving them.


Calculus Problem Solver

Let us discuss certain problems on calculus from the basics.
1. Which of the following relations are functions.
    R1 = { ( 1, 2), ( 2, 3), ( 3, 4), ( 4, 5) }

R1
Every element of the domain has its image in the co-domain, therefore the above relation is a function
                               

    R2 = { ( 1, -1 ), ( 4, -2 ) , (0 , 0 ), ( 1, 1 ), ( 4, 2 ) }

  The element 1 has two images -1 and 1 and the element 4 has its 2 images -2 and 2 in the co-domain.
  Therefore the above relation is not a function.

2. A function is defined as follows. f : z ---> N

     f(x) = 3x + 5, find the range of { -4, 1, 4 }

   Solution:  We have f ( x) = 3x + 5
                                 f( 1) = 3 ( 1 ) + 5 = 3 + 5 = 8, which belongs to Natural numbers N
                               f( -4 ) = 3 ( -4 ) + 5 = - 12 + 5 = - 7 which is not a natural number.
                                f( 4 ) = 3 ( 4 ) + 5 = 12 + 5 = 17 , which is a natural number

              Therefore Range = { 8, 17 }
   

3. Let f : R ---> R and  f ( x ) = $\frac{ 2 } { x + 3 }$, find the domain and range of the function. Express your answer in interval notation.

  Solution: We have f ( x ) = $\frac{ 2 } { x + 3 }$

            The above function is a rational function.
            The function is not defined when the denominator x + 3 = 0
                                                                                (i. e ) x = -3
          Therefore the Domain consists of all real numbers except x = -3
                                                                               Domain = { x : x $\epsilon$  R - {3}  }
                                                           As x tends to infinity, y = -2/3
                                                          and for no value of x, y = 0, therefore y is not equal to 0.
                                  Therefore, the Range of the function = { y : y $\epsilon$ R - { 0 }   }

4. If f and g are functions such that f : A ---> B and g : B------> C.
    If A = { 1,2,3 }, B = { 1, 2, ,3 4, 5,6 } and C = { 10, 20, 30 }
    such that f ( 1 ) =2, f( 2) =3, f ( 3 ) = 4, g(2)=20, g(3)= 30 and g(4) = 40.
    Find g o f(1), (g o f)(2), (g o f )(3)


    Solution: We are given that, f : A ---> B and g : B------> C.
                                           If A = { 1,2,3 },
                                              B = { 1, 2, ,3 4, 5,6 }
                                        and C = { 10, 20, 30, 40, 50 }
                         such that f ( 1 ) =2, f( 2) =3,     f ( 3 ) = 4,
                                        g(2)=20, g(3)= 30 and g(4) = 40.
                                 g o f ( 1 ) = g [ f ( 1 ) ] = g ( 2 ) = 20
                                 g o f ( 2 ) = g [ f ( 2 ) ] = g ( 3 ) = 30
                                 g o f ( 3 ) = g [ f ( 3 ) ] = g ( 4 ) = 40

                    Therefore, g o f(1) = 20 , (g o f)(2) = 30, (g o f )(3) = 40

5. If f : R --> R and g : R----> R such that f ( x) = 3x and g ( x ) = x2 find f o g ( x) and g o f ( x ).


Solution: We are given that, f ( x) = 3x and g ( x ) = x2

                                   f o g ( x ) = f [ g ( x ) ] =f ( x2 ) = 3 ( x2 ) = 3 x2
                                   g o f ( x ) = g [ f ( x ) ] = g ( 3 x ) =  ( 3 x  ) 2 = 9 x2


6. Find the limit of the following functions.

   a. $\lim_{x -> 2}$ ( x2 - 3x + 4 )
      
       Solution:
$\lim_{x -> 2}$ ( x2 - 3x + 4 )  = 22 - 3 ( 2 ) + 4
                                                                       = 4 - 6 + 4
                                                                       = 2

   b. $\lim_{ x -> 2 }$   $\frac{( 3 x-6 )( 4 x + 5 )}{( x - 2 ) }$

        Solution:
$\lim_{ x -> 2 }$   $\frac{( 3 x-6 )( 4 x + 5 )}{( x - 2 ) }$
 
                                                            = $\frac{(6-6)(8+5)}{(2-2)}$

                                                            = $\frac{0}{0}$
        Therefore, $\lim_{ x -> 2 }$   $\frac{( 3 x-6 )( 4 x + 5 )}{( x - 2 ) }$

                                                            = $\lim_{ x -> 2 }$ $\frac{3(x-2)(4x+5)}{(x-2)}$

                                                            = $\lim_{ x -> 2 }$ $\frac{3(4x+5)}{1}$

                                                            = 3[ 4(2) + 5 ]

                                                            = 3 [ 8 + 5 ] = 3 [13 ] = 39


   c. $\lim_{x->-3}$   $\frac{(x^{2}-9)}{(x+3)}$

         Solution: $\lim_{x->-3}$   $\frac{(x^{2}-9)}{(x+3)}$

                                                           = $\frac{9-9}{-3+3}$ = $\frac{0}{0}$
         Therefore, $\lim_{x->-3}$   $\frac{(x^{2}-9)}{(x+3)}$

                                                           = $\lim_{x->-3}$   $\frac{(x^{2}-3^{2})}{(x+3)}$

                                                           = $\lim_{x->-3}$   $\frac{(x+3)(x-3)}{(x+3)}$

                                                           = $\lim_{x->-3}$  ( x-  3 )

                                                           = - 3 - 3 = - 6



   d.  $\lim_{x->\pi/2}$  $\frac{sin x + cos x}{sin x}$

        Solution:
$\lim_{x->\pi/2}$  $\frac{sin x + cos x}{sin x}$

                                                           = $\frac{sin(\pi /2)+cos(\pi /2)}{sin(\pi /2)}$ 

                                                           = $\frac{1+0}{1}$ = 1



Free Calculus Problem Solver

1. Find the Derivatives of the following function using derivative rules

   If f(x) =  2x + 5, find f ' ( x ) at x = 2.


      Solution: We have f (x ) = 2 x + 5
       f ' ( x ) = $\lim_{h->0}$ $\frac{f(x+h)-f(x)}{h}$

      f ' ( 2 )  = $\lim_{h->0}$ $\frac{f(2+h)-f(2)}{h}$

                  = $\lim_{h->0}$ $\frac{2(2+h)+5-(2(2)+5)}{h}$

                  = $\lim_{h->0}$ $\frac{4+2h+5-4-5}{h}$

                  = $\frac{2h}{h}$ = 2


2.   A man 6 feet high walks at a uniform rate of 4 miles per hour away from a lamp 20 feet high. Find the rate at which the length of the shadow increases.
Solution: Let us Study the following diagram.
Shadow Problem
  Let L be the lamp and AB the position of the man at t seconds. The length of his shadow is AC and let it be x feet.
  The rate at which he walks is given.
                    (i. e ) the rate at which AM = y is changing is given

                             i. e  $\frac{dy}{dx}$ = 4 miles per hour
                                            = $\frac{88}{15}$ feet per second

         $\Delta $ ABC and $\Delta $ LMC are similar.
                 Therefore, $\frac{CA}{CM}$ = $\frac{6}{20}$
                           i.e    $\frac{x}{x+y}$ = $\frac{3}{10}$
                           i.e         7x = 3y

               The rate at which the shadow is increasing is $\frac{dx}{dt}$
                                 Differentiating 7x = 3y with respect to t, we get,

                               7 $\frac{dx}{dt}$ = 3 $\frac{dy}{dt}$

                                         = 3 x $\frac{88}{15}$
                                 $\frac{dx}{dt}$
                                         = 3 x $\frac{88}{15}$ x $\frac{1}{7}$

                                         = $\frac{3\times 88}{15\times 7}$

                                         = $\frac{88}{35}$ feet per second

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