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Grade:
        The length of the normal at any point on the catenary y=c{(^x/c + e^-x/c)/2} varies as 
8 years ago

Answers : (2)

SAGAR SINGH - IIT DELHI
879 Points
							

Dear student,

First of all find the normal by differentiating, normal.dy/dx=-1

8 years ago
vikas askiitian expert
509 Points
							

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

 

8 years ago
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