Implicit Differentiation

Problem : 20) Find the second derivative of (d²y/dx²) ,
if (x³ - 3xy + y³ ) = 1


Solution : 


Differentiating, we get  3x² - 3x * dy/dx + 3y² * dy/dx ) = 0
=> dy/dx = ( y - x²) / (y² - x) - - - - - (1)

Now, note that:
(i) (y² - x) * (dy/dx) = ( y - x²) ;
(ii) ( y - x²) * (dy/dx) = (y - x²)² / (y² - x)
(iii) 3x²y - x⁴ - xy³ = -x (x³ - 3xy + y³ ) ==> -x

Differentiating (1) , we get,

[(y² - x) * (dy/dx - 2x) - (y - x²) * (2y * (dy/dx) - 1) ] / (y² - x)²
=> [(y² - x) * (dy/dx) - 2x * (y² - x) - 2y * (y - x²) (dy/dx) + (y - x²)] / (y² - x)²

Using (i) & (ii) ,
. . [(y - x²) - 2x *(y² - x) - 2y * {(y - x²)² / (y² - x)} + y - x² ] / (y² - x)²
. . [ 2y - 2x² - 2xy² + 2x² - 2y * {(y - x²)² / (y² - x)} ] / (y² - x)²

. . 2y [ (1 - xy) * (y² - x) - {y² + x⁴ - 2x²y} ] /(y² - x)³
. . 2y [y² - x - xy³ + x²y - y² - x⁴ + 2x²y ] /(y² - x)³
. .2y [ - x - xy³ + x²y - x⁴ + 2x²y ] /(y² - x)³
. . 2y [- x + (-x) * (x³ - 3xy + y³ ) ] /(y² - x)³

Using (iii) we have,
==> 2y * (-2x) /(y² - x)³
==> -4xy / (y² - x)³ ==>(Answer) 

Algebra

Problem : 19 )
if x + y = 1, and (x² + y²) (x³ + y³ ) = 26, then 
what is the value of (x² + y²) ?


Solution : 


Problem : Given x + y = 1 and (x² + y²) (x³ + y³ ) = 26

. .(x² + y²) (x³ + y³ ) = 26
=> [(x + y)² - 2xy] [(x + y)³ - 3xy (x + y)] = 26
=> (1 - 2xy) (1 - 3xy) = 26

=> 1 - 5xy + 6x²y² = 26
=> 6x²y² - 5xy - 25 = 0
=> 6x²y² - 15xy + 10xy - 25 = 0

=> (3xy + 5)(2xy - 5) = 0
=> xy = -5/3 and xy = 5/2

Now, x² + y² = (x + y)² - 2xy
=> x² + y² = 1 - 2xy
=> x² + y² = 1 - 2 * (-5/3)

=> 13 / 3 ==> (Answer)

Note : If we take xy = 5/2, x² + y² = -ve , which is not possible as sum of two real squares will never be negative

Inequality

Problem : 18 ) 

Prove the inequality (1 + a) (1 + b) (1 + c) ≥ 8 (1 - a) (1 - b) (1 - c)

a, b, c are positive numbers and their sum is 1. Also indicate when equality occurs.

Solution : 

Given : a, b, c > 0 and a + b + c = 1
=> (1 − a) > 0 ;(1 − b) > 0 ; (1 − c) > 0

Therefore, a = (1 − b − c ) ==> 1 + a = (1 − b) + (1 − c)
and. . . . . b = (1 − a − c ) ==> 1 + b = (1 − a) + (1 − c)
similarly, . c = (1 − a − b ) ==>1 + c = (1 − a) + (1 − b)

Using A.M.- G.M. Inequality ;
(1 + a) = (1 − b) + (1 − c) ≥ 2 √[(1 − b) × (1 − c)] - - -(1)
(1 + b) = (1 − a) + (1 − c) ≥ 2 √[(1 − a) × (1 − c)] - - - -(2)
(1 + c) = (1 − a) + (1 − b) ≥ 2 √[(1 − a) × (1 − b)] - - - - (3)

Multiplying (1), (2) and (3) we get,

(1 + a)(1 + b)(1 + c) ≥ 8 (1 − a)(1 − b)(1 − c)

Equality holds true if a = b = c = 1/3

Link to Y/A

Geometry : Circles touching externally

Problem : 17 ) Circles A, B, and C are externally tangent to each other and internally tangent to circle D. Circles B and C are congruent. Circle A has a radius of 1 unit and passes through the center of circle D. Find the radius of circle B.

Solution : 

Radius (r) = 8/9 units.

here is the diagram in the link given below.
http://i854.photobucket.com/albums/ab104…

Note :
If two circles touch each other externally or internally then, the point of contact and the centers of the touching circles will be COLLINEAR.

Further, since the radius of circle A is 1 unit and it passes through the center of the bigger circle, therefore the radius of the bigger circle will be 2 units and hence DB = (2 - r).


Link to Y/A

Converse of Mid-Point Theorem

Problem 16 : ) Prove that :-
"The line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side and half of the other side"


Solution : 


Refer to the diagram given in the link below.
http://i854.photobucket.com/albums/ab104…

Let D is the mid-point of AB of Δ ABC and
DE drawn is parallel to BC.

To Prove That : E is the mid-point of AC.

Proof :
Extend DE and draw a line // BD ( or BA) which meets the extended DE at F .
=> BCFD is a //gm. {since opp. sides are parallel }
=> AD = DB = CF .

In Δs ADE and CFE,
1) ∠ADE = ∠CFE. . . . . . (Alt ∠s)
2) ∠AED = ∠ CEF . . . . . (Vert. opp. ∠s)
3) AD = CF

=> Δ ADE ≅ Δ CFE . . . . .(By A.A.S Test)
=> AE = CE and DE = EF . . . . . . .(C.S.C.T. )

Which shows that E is the mid-point of AC .
Further, BC = FD . . . {opp. sides are equal and parallel}
and DE = (1/2) FD
=> DE = (1/2) BC. ===> (Proved)


Link to Y/A

Trigonometry

Problem : 15) Prove that :
sinA + sin5A + sin3A + sin7A = 16 sin A cos²A cos² (2A) 


Solution :


Problem : sinA + sin5A + sin3A + sin7A = 16 sin A cos²A cos² (2A)

L.H.S. = sinA + sin5A + sin3A + sin7A
=> (sinA + sin7A) + (sin5A + sin3A) - { Rearranging the terms }
=> 2 sin 4A cos 3A + 2 sin 4A cos A
=> 2 sin 4A [ cos 3A + cos A ]

=> 2 sin 4A * 2 cos 2A cos A
=> 4 sin 2A cos 2A * 2 cos 2A cos A
=> 8 sin A * cos A *cos 2A * 2 cos 2A cos A
=> 16 sin A cos² A * cos² (2A) ==> R.H.S.

Note : 1) sin C + sin D = 2 * sin [(C + D)/2 ] * cos [(C - D)/2]
. . . . . 2)cos C + cos D = 2 * cos [(C + D)/2 ] * cos [(C - D)/2]
. . . . . 3) sin 2A = 2sin A * cos A


Link to Y/A

A farmer's Land compensation

Problem : 14 )  A villager has his land in the shape of quadrilateral ABCD. The government wants to procure some land from ONE corner of his plot. He obliges but with a condition that an EQUAL land should be given to him CONTIGUOUS to his plot and he must finally have his plot in the shape of a triangle. Analyze and show at to how he was compensated in the above agreed manner.

Solution : Please see the link given below

http://i854.photobucket.com/albums/ab104…


Here is the more elegant solution given by one of my most valuable contact Mr. Anil Bakshi .
area lost = area gained (as these lie between same parallels)

http://www.flickr.com/photos/53770577@N0…


Link to Y/A