Showing posts with label differentiation. Show all posts
Showing posts with label differentiation. Show all posts

Find two positive numbers whose sum is 16 and sum of whose cube is Minimum


Show that of all the rectangles inscribed in a circle of given radius . The Square has maximum Area.


Solutions


Let ABCD be rectangle which is  inscribed in a given circle of radius ‘r’
Show that of all the rectangles inscribed in a circle of given radius . The Square has maximum Area.
And Let θ be the angle between side of rectangle and Diameter of given circle.


Therefore from right angled  Δ ABC ,

We have 
  AB  = AC cosθ          ∵ AC = 2r
Let A(x) be the area of Rectangle ABCD
∴ A(x) = AB × BC
    A = (2r cos θ)(2r sin θ )
    A =  4r2 sin θ cos θ
    A = 2r2  (2sin θ cos θ)
    A = 2r2  (sin 2θ )

⇒ 2r2 2 (cos 2θ ) = 0 ,As r2 is constant
⇒cos 2θ = 0
⇒cos 2θ =cos (π/2)
⇒ θ = π/4
 =4r2  (-2sin 2θ 



∴ A has Maximum value at θ = π/4



Find two positive numbers whose sum is 16 and sum of whose cube is Minimum

Solution

Let us consider two numbers x and 16- x .
Then transforming our problem to mathematical form which says “sum of whose cube”  as follows
A (x) =   x3 + (16 - x)3…….. (1)
Differentiating both sides w .r. t  “x” , we get



     X = 8
So  x  =  8 will be the 1st required numbers if Double derivatives of A  w. r. t  ‘x’ comes to be positive at x = 8.
Differentiate (2)  w. r. t. ‘x’  .




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How should the wire of 28 m be cut so that the combined area of the circle and square is as small as possible ?

Application of Derivative 

A piece of wire 28 cm long is to be cut into two pieces. One piece is to be made into a circle and another into a square. How should the wire be cut so that the combined area of the two figures is as small as possible?

Let the wire be cut at a distance of  x meter  from one end. Therefore then two pieces of wire be x m and (28-x) m.


Calculate Dimension of Circle and Square


Now 1st part be turned into a square and  the 2nd part be be made into a circle.

Since 1st part of the wire is turned into square. then its perimeter will be x m. 
So using formula of perimeter of square , we can calculate side of the square = x/4 m


Calculate Areas of Circle and Square


Therefore Area of square = (x/4)(x/4) sq m

                                     A1 = x2/16


And  when 2nd part of the wire is turned to circle, then its perimeter ( circumference ) will be 28 - x m. So using formula of perimeter of square , And if  "r" be  radius of the circle , Then
Circumference of circle =  2 π r =  (28-x)
 ∴  r = (28-x)/2π

We know that Area of Circle A2   = π r2  

                                     A2  π[(28-x)/2π]2  


Express Areas in terms of Function





To find value/s of x


Now to find the value of x for which this function A(x) is maximum or minimum ,put A(x) = 0



To Test the Minimum Value of  Function


Now we have the value of "x" on which either A(x) have maximum or minimum value . To check the maximum or minimum value we have to find A''(x) as follows






So A''(x) has positive value Therefore A(x) shall have maximum value at x = 112/(π + 4)

Hence two pieces of wire should be of length x m and (28-x) m

These pieces should be of length 112/(π+4) and 28π/(π + 4)


Verification



we can calculate the sum of these pieces , it must be 28 m


1st part     

   
112/(π+4) = 112/{(22/7)+4}=112×7/50 = 784/50


2nd part 


28π/(π + 4) = {28×22/7}/{(22/7)+4} = 88×7/50 = 616/50

Sum of Two Parts 


 112×7/50 + 28×7/50 = (784+616)/50
                                                                 
  = 1400/50= 28 m



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HOW TO PROVE TRIGONOMETRIC IDENTITIES || TRIGONOMETRY


Proof of trigonometric identities , trigonometric identities problems, proving trigonometric identities formulas,these trigonometric identities of class 10, fundamental trigonometric identities,trigonometric identities class 11 and its formation with the help of some examples.

How to prove Identity


cos 6x = 32cos6 x - 48.cos4 x   + 18.cos2 x  - 6.cos2 x  - 1

Proof

1st of all  rewrite 3x as 3.2x

L.H.S. = cos 6x =  cos (3.2x) 

Now using the result cos 3θ = 4cos3 θ - 3 cos θ  -----(1)

Replacing θ as 2x in (1), we get 

L.H.S. = 4cos3 2x - 3 cos 2x  -----------(2)


Now using the result  1+ cos 2θ = 2 cos2 θ 

                                   ⇒ cos 2θ = 2 cos2 θ -1

Replacing cos 2x = 2 cos2 x -1 in (2), we get 

L.H.S.= 4 {2cos2 x -1}3 - 3 {cos2 x  -1}


Now using the result {a - b }3 = {a}3 - b }3  -  3{a }2 .b   + 3(a). b2

  cos 6x    = 4[ {2cos2 x  }3 - { 1 }3  -  3{2cos2 x  }2 .1   +3.(2cos2 x) .12 ] - 3 . {cos2 x  -1}


Taking the product of powers to simplify it

cos 6x  =   4[ 8cos6 x  - 1 - 12cos4 x  + 6cos2 x]  - 3{2cos2 x-1}

Multiply by 4 in 1st term and multiply by -3 in 2nd term

 cos 6x  = 32cos6 x  - 4 - 48cos4 x  + 24cos2 x  - 6cos2 x + 3

Adding the like powers terms and arranging in descending order

cos 6x   = 32cos6 x - 48cos4 x  + 18cos2 x  - 6cos2 x  - 1

Hence the Proof



Prove the Identity 

tan (2x) =  2tan x  1 - tan2 x 

Proof

We know that 


tan (A+B) =  tan A +  tan B1 - tan A tan B 

Put A = B  = x in above formula . then it becomes

tan (x+x) =  tan x +  tan x1 - tan x tan x 


tan (2x) =  2tan x  1 - tan2 x 
Hence the Proof


Prove that sin 2x = 2sin x cos x

Proof


As we know that sin (A + B) = sin A cos B + cos A sin B..  ...(1)

Put A = B  = x in ...   (1)

sin (x + x) = sin x cos x + cos x sin x

sin (2x) = sin x cos x +  sin x cos x

sin (2x) = 2 sin x cos x

Hence the Proof


Prove that cos 2x = cos2 x - cos2 x

Proof


As we know that cos (A + B) = cos A cos B - sin A sin B..  ...(1)
Put X = A = B in (1) , we get

cos (x + x) = cos x cos x - sin x sin x

cos 2x = cos2 x - sin2 x   

cos 2x = cos2 x - sin2 x   


Hence the Proof


                                                                                                                                                                   

 Using the result 
1+cos 2θ = 2cos2 θ
cos 2θ = 2cos2 θ -1 -------------(1)
Replacing θ with 2x in eq (1)
1+ cos 4x = 2cos2 2x
cos 4x = 2cos2 2x -1

Again using  cos 2θ = 2cos2 θ -1

cos 4x = 2Sq(2cos2 x -1) -1

It is the square of 2cos2 x -1

cos 4x = 2Sq(2cos2 x -1) -1

cos 4x = 2(4cos4 x +1 - 4cos2 x) -1

cos 4x = 8cos4 x +2 - 8cos2 x -1

cos 4x =  8cos4 x - 8cos2 x +1

Hence the Proof


What is the value of sin3x?



To find the value of sin 3x ,  use this formula which contain sin (A+B)
therefore sin (A+B) = sin A cos B cos A sin B——-(1)
put A = 2x and B = x in (1)
then Sin 3x = sin 2x cos x + cos 2x sin x

As we know that cos 2x = 1 - 2sin3 x and sin 2x = 2 sin x cos x


sin 3x = (2 sin x cos x) cos x + (1 - 2sin3 x ) sin x
sin 3 x = 2 sin x cos2 x + sin x -  2sin3 x

As we know that cos2 x = 1sin2 x

sin 3x= 2 sin x (1-sin3 x) + sin x - 2sin3 x
sin 3x = 2 sin x -2 sin3 x + sin x - 2sin3 x
sin 3x = 3 sin x - 4 sin3 x

Similarly we prove that cos 3x= 4 cos3 x - 3 cos x
For learning and memorising more trigonometric formulas

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