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The length of the normal at any point on the catenary y=c{(^x/c + e^-x/c)/2} varies as
Dear student, First of all find the normal by differentiating, normal.dy/dx=-1
Dear student,
First of all find the normal by differentiating, normal.dy/dx=-1
dear praveen , length of normal = y{1+(dy/dx)2}1/2 y = c{(ex/c+e-x/c)/2} ...................1 dy/dx = { (ex/c-e-x/c)/2} ...................2 length = y{1+(dy/dx)2}1/2 = y { 1+(ex/c-e-x/c)2}1/2 =y { (e2x/c+e-2x/c+2)/4}1/2 = y { ((ex+e-x)/2)2}1/2 = y { ex+e-x)/2 ..................3 from eq 1 putting (ex+e-x)/2 = y/c in eq 3 we get length = yy/c = y2/c so , length of normal varies as square of abcissa .... approve if u like my ans
dear praveen ,
length of normal = y{1+(dy/dx)2}1/2
y = c{(ex/c+e-x/c)/2} ...................1
dy/dx = { (ex/c-e-x/c)/2} ...................2
length = y{1+(dy/dx)2}1/2
= y { 1+(ex/c-e-x/c)2}1/2
=y { (e2x/c+e-2x/c+2)/4}1/2
= y { ((ex+e-x)/2)2}1/2
= y { ex+e-x)/2 ..................3
from eq 1 putting (ex+e-x)/2 = y/c in eq 3 we get
length = yy/c = y2/c
so , length of normal varies as square of abcissa ....
approve if u like my ans
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