Law of Parallelogram of vectors
* Suppose an equation is R2 =a2 +b2+2absin2φ . How to convert it to an equation of law of parallelogram of addition of vectors ? we write R2 =a2 +b2+2absin2φ =R2 =a2 +b2+2abcos (π/2 - 2φ). Taking π/2 - 2φ =θ, R2 =a2 +b2+2abcosθ....(1) Assuming one end of Resultant vector being at origin, the other end is at a point x=a+bcosθ & y=bsinθ...(2) . What equation does the locus of the resultant vector satisfy ? From (2), eliminating θ, we get (x-a)2 +(y-0)2 =b2 which is the equation of a circle with origin at (0,0) and having radius b and center at(a,0).This equation resulted when a coincides with x-axis, a>b and the smaller vector b rotates around a (a being the axis). We can also coincide b with the x-axis and let a rotate around which creates a symmetrical equation by replacing a with b and vice versa. The equation becomes (x-b)2 +(y-0)2 =a2.The center lies at (b,0) with origin at (0,0)
* What curve does the resultant vector draw when a, b are constants and angle θ varies from 0° to 360° i.e completes one cycle. Since the equation carves out symmetrical shape from (0° to 180°) & (180° to 360°), we explore the behavior of the equation from 0° to 180° .
When the angle is 0 degree, R= a+b
When the angle is 90 degree, R= a2 +b2 .
When the angle is 180 degree, R=a-b
Total spectrum covered in x-axis (a+b)+(a-b)=2a
Corollary of above is x=a+bcosθ . Hence xmax =a+b at cosθ=1 or θ=0° and xmin =a-b at cosθ=-1 or θ=180°. Similarly y =bsinθ . Hence ymax =b at sinθ=1 or θ=90° and ymin =-b at sinθ=-1 or θ=270°
When θ=90° , y=b, x=a. Hence, the trajectory swept by the resultant traps an enclosed space which is a circle whose radius is b , and origin is (0,0) which is outside the circle and with near vertex at (a-b,0) and farthest vertex at (a+b,0) and center is at (a,0). Area trapped is πb2 and circumference is 2πb.
The above conclusion was because of equation (x-a)2 +(y-0)2 =b2. For the equation (x-b)2 +(y-0)2 =a2,i.e. when b is in x-axis and a is rotated around b, one can observe from the accompanying figure that the origin lies within the circle, every resultant intersects the circle at 2 points which are 180 degree apart. But , unlike the above case, there are no tangents from origin to the circle. Area trapped is πa2 and circumference is 2πa.
When (a/b) >π /2 .......area swept by by the resultant > area trapped by the position vector of resultant on rotation of b
When (a/b) =π /2 .......area swept by by the resultant = area trapped by the position vector of resultant on rotation of b
when π /2> (a/b)>=1 ....area swept by by the resultant is less than area trapped by the position vector of resultant on rotation of b
In all the above cases, area swept by by the resultant is less than area trapped by the position vector of resultant on rotation of a
2ab=πa2 => a/b= π / 2 which is not true
2ab=πb2 => (a/b)min= π / 2 for the equality
Difference of area trapped on rotation of a and b=Δab=πa2-πb2 =
(a)Minimum Value of Δab =0 when b=a.
(b) Δab =a2 (π - 4/π)=1.87a2 when b=2a/π=0.6365a
(c ) Δab is greater than πb2 but less than πa2 when b < 2a/π
The angle R makes with a is φ1 .
(d) Δab is directly proportional to ( a2-b2) which is discrete and not continuous since a, b are discrete.
tanφ1= bsinθ / (a+bcosθ)
The angle R makes with b is φ2 .
tanφ2= asinθ / (b+acosθ)
Now we can put a as a=(π/2)b*ncosθ1 where n is any integer n=1,2,3,..... and θ1 can take any value within specified range. we can refine this formula to a=(π/2)b[1+ncosθ1] where n=0,1,2,3,......We have thus quantized the value of a .
* Area swept by the resultant vector is the area of the triangle formed by vector a,b and R. This is absinθ/2. So total area swept is 4 times the area swept when θ moves from zero degree to 90 degree. Hence the equn.- A=dA=4/2 90°∫0°absinθ =-2ab[cosθ2 -cosθ1]90°0° =-2ab*(-1) =2ab (integral of sinθ is -cosθ). We can visualize the area as that of a rectangle with length 2a and breadth b or a circle with radius r= √[(2/π)(ab)] = √[(0.63694)(ab)] =√[(ab*sin2θ)]=sinθ*√(ab) where θ=0.924135 radian or 52.93° .
sinθ*√(ab) can be put as √[(asinθ)(bsinθ)]=√ [(projection of a on y-axis)*(projection of b on y-axis)] where we have reduced 2-D vector into 1-D vector (of course with loss of information).
* If we think of a rectangle with sides 2a,b, then we can inscribe an ellipse with major axis 2a and minor axis b with center coinciding with the center of the rectangle (considered as origin ). The equation of such an ellipse is x2/a2 +4y2/b2=1 and area is πab /2 =(π/2)ab=ab*cosec2θ where θ=0.924135 radian or 52.93° . Can we find a transformation matrix that transforms to the locus to the ellipse from the locus of resultant vector ?? and find the behaviour of resultant vector vis-a-vis the components.If we inscribe a circle , maximum area will be πb2 /4. The ratio of area of the recangle and that of the ellipse with major axis 2a & minor axis b is π irrespective of the value of a and b. The ratio of the area of ellipse to the circle inscribed in it shall be major axis/minor axis.
* Suppose we have different value of a,b such that ab=constant=c2 , then the average of a,b shall be the geometric mean c. Then the total area swept by resultant vector =2ab=2c2 =(√2c)*(√2c). If we construct a right angled isosceles triangle with side c, the hypotenuse shall be √2c and the square constructed on the hypotenuse will have the same area as total area swept by resultant vector.We can take
which are defined for all values of δ except δ=0;
At δ=0, we can also define the value with following trick:-
ab= ksinδ * k/sinδ =k (sinδ/δ) * k (δ/sinδ) . as δ->0, sinδ/δ =1 and hence ab=k2
a,b shall have the same sign,and extremum values are a=k , b=k & a=-k, b=-k.
as a+b= x+k2 /x = extremum value => d(x+k2 /x)/dx=0 or x=+k or -k.
Which means a can oscillate within the range δ=π/2 to -π/2 ( csinδ lies in[-k,+k]) and accordingly k/sinδ also varies .
a+b shall vary between 2k & -2k with condition that both shall have the same sign..
a b * x = ax+by
b a y bx+ay i.e. a vector say A with components as column matrix.
||A|| =√ [ a2x2 + b2y2 + 2abxy +b2x2 + a2y2 + 2abxy ] =√[(a2+b2)(x2+y2)+4abxy]=√[(x2+y2)+2ab*2xy ] as a2+b2=1
area =∫||A||da =∫ √[x2+y2+2xy*2a(1-a2)1/2] da.
If x=1 & y=1, area =∫ √[2+4a(1-a2)1/2] da=√2∫ √[1+2a(1-a2)1/2] da=√2∫ [(sinθ +cosθ)cosθ ] dθ by taking a=sinθ
So area =1/√2 [-cos2θ/2 +θ +sin2θ/2]π/20
area=√2∫ [(sinθ -cosθ)cosθ ] dθ=√2 [-cos2θ/2 -θ -sin2θ/2]π/20
* Average area swept by resultant vector is areaav =(1/2)ab0∫π/2sinθdθ / 0∫π/2dθ =ab/π. Had the area been swept in a linear manner, areaav =(0+2ab)/2 = ab.
Here the deviation =ab -ab/π =ab (1 - 1/π)
Again perimeter of area swept is taken as perimeter of parallelogram divided by 2 = a+b (this is not actually correct).
hence area swept / perimeter = absinθ / (a+b) = [ab/(a+b)] sinθ . Figure in red looks like 2 resistances connected in parallel.