| Message |
|
|
|
Hello dear
you know very well that in dot product we always get a scalar quantity so there is no mean of directiin,bcz scalar quantities only have magnitude they dont have the direction.but in the case of cross product we get the vector quantity and we know that vector quantity always have magnitude as well as direction so that there is a role of direction in the case of cross product.
Cheers..................
|
|
|
|
|
hello dear...
go for Arihant ,they has preevious year paper of all the exams such as AIEEE,DCE,VIT,Manipal............
|
|
|
|
|
hello dear..
it is obviously a foggy day in which sound velocity is more than clear day.....
enjoy and chilllllllllllllllllllllllll
|
|
|
|
|
hello dear...
Dont worry...Just send them a mail to them,and try to call the conducting IIT.........and check regularly the status of your application form,Dont worry they wont reject the form in any case ,they will coordinate you thru mail or ne correspondence....
enjoy and chillllllllllllllllllllllllll
|
|
|
|
|
hello dear.........
{x/{x/(x sin x+cos x)}2
multiplying and dividing by (x cos x)
we get :
x sec x. x cos x / ((x sin x+cos x)2 dx
x sec x . x cos x / (x sin x+cos x)2 dx - (d/dx (x secx ) ) ( x cos x / (x sin x+cos x)}2 dx) dx
after solving above integrals......
- x sec x / (x sin x+cos x) + [ (x sin x+cos x) / cos2 x (x sin x+cos x) ] dx
- x sec x / (x sin x+cos x) + sec2 x dx
- x sec x / (x sin x+cos x) + tan x + c
|
|
|
|
|
hello dear...
= [ sin ( x - a ) / sin ( x + a ) ] 1/2
= [ sin ( x - a ) / sin ( x + a ) * sin ( x - a ) / sin ( x - a ) ]
= sin (x - a) / (sin2 x - sin2 a )1/2 dx
= cos a [sin x / (sin2 x - sin2 a )1/2 dx ] - sin a [cos x / (sin2 x - sin2 a )1/2 dx]
= cos a [sin x / (cos2 a - cos2 x)1/2 dx ] - sin a [cos x / ( sin2 x - sin2 a )1/2 dx]
in the first part put cos x = t and in the second part sin x = u
now solve it....
you will get the answer:
iam writing the finla answer::
- cos a . sin -1 (cos x / cos a) - sin a . log [sin x - (sin2 x - sin2 a)1/2 ]
|
|
|
|
|
hello dear..
this is quite simple question put..
cos x = (1 - tan 2 x / 2) / (1 + tan 2 x / 2)
sin x = 2 tan (x/2) / (1 + tan 2 x / 2)
1 + tan 2 x / 2 = sec2 x/2
put tan x/2 = t
now solve .try it .........if u are not able to get the solution, than nudge i will provide u complete solution but plz try it once
|
|
|
|
|
[ sin n-2(x) - cos n-2(x) ] / [sin n-2(x) . tan x + cos n-2(x) . cot x ] dx
[ sin x . cos x (sin n-2(x) - cos n-2(x) )] / [sin n(x) + cos n(x) ] dx
put :: sin n(x) + cos n(x) = t
n (sin n-1 x . cos x - cosn-1 x . sin x ) dx = dt
now:
1/n integral ( dt / t)
1/ n log [sin n(x) + cos n(x) ] + c
|
|
|
|
|
hello dear...
[sinn 2x / cos x ]1/2
(2 sin x)1/2
above is non integrable function ...acccording to methods prescribed for IITJEE course
|
|
|
|
|
helo dear...
Ranking varies from year to year na from surveying agency ...............so, no one quote a best IIT institute in country...........well all IITs are good enough and their departmental rank is different for each IIT
|
|
|
|
|
hello dear
the formula for energy per unit area per second is 1/2 (rho) c (omega)2 A2
pressure is directly proportional to (rho) c (omega)2 A2
|
|
|
|
|
Hello dear
there is very little difference betweeen IT and CSE,even in NIT's little difference,but if you can upgrade then do it.it is better for you.
ALL THE BEST................
|
|
|
|
|
hello dear..
well,amit saini has the proof of the theorem,weel dude,keep it up.........
hence iam concluding it..
according to bernaulli theorem:
p + 1/2 rho v2 + rho g h = constant
pressure energy + kinetic energy + potential energy = constant
|
|
|
|
|
hello dear....
Dont worry,physics need a good grasping over the subject and a good teacher is required to understand the subject,i must suggest you o for theory before attempting any question in them,if you have cammand over the theory ,you can easily make the question.
have deep knowledge of the theory and concepts and than proceed further.
|
|
|
|
|
hello dear........
36x2 + 100y2 = 3600
x2 / 100 + y2 / 36 = 1
a = 10 and b = 6 and e = 8/ 10
(8,18/5)
cordinates of foci are (ae,o) (-ae,0)
foci is (8,0) and (-8,0)
equation of the line with cordinats ::: (8,0) and (8,18/5)
equation of the line with cordinats ::: (-8,0) and (8,18/5)
the focal radii =
(a + ex) and (a - ex)
8 + 64/ 10 = r1
8 - 64/ 10 = r2
|
|
|
|
|
centre is(1,5), focus is (1,8) and d sum of d focal distances of a point on the ellipse is 12
focal distance 2 a = 12
a = 6....
(x-1)2 / a2 - (y-5)2 / b2 = 1
distance betweeen C and F = ae..
ae = 3..
e = 1/2
b2 = 27.
hence equation is :
(x-1)2 / 36 - (y-5)2 / 27 = 1
|
|
|
|
|
hello dear...
the cordinates of the foci are:
s = (ae,0)
si = (-ae,0)
vertex (0,b)
m1 = b / ae
m2 = -b / ae
tan 2@ = ( m1 - m2) / (1 + m1 m2)
value of (b is known)
tan 2@ = 2abe / (a2 - 2b2)
solve value of (a) thru above equation......
the equation of ellipse is :
x2 / a2 + y2 / b2 = 1
|
|
|
|
|
hello dear..
the cordinate of a focus = (ae,0)
the cordinate of a end of minor axis = (0,b)
(ae)2 + b2 = k2
distance between two foci is :
2 ae = 2h
length of semi axis are :
b = (k2 - h2)1/2
now solve for the value of (a) by the obtained value of (b) in the above equations
|
|
|
|
|
hello dear..
2 b2 / a = 4
ae = 3/2
now solve it :
divide the two equations
put e2 = 1 - (b/a)2
e = 3/ 7.....
nudge me if i did some mistake in solving
|
|
|
|
|
E-Store
