Question

Find the radius of curvature of √x+√y=1 at the point (1/4,1/4).

Answer 1

Given : curve √x+√y=1

To find : Radius of curvature  at the point (1/4,1/4)

Solution:

√x+√y=1

=> 1/2√x + (1/2√y )(dy/dx)= 0

=> dy/dx   =  -√y /√x

d²y/dx²  = (-√y(-1/2) /x√x   - (1/2√x.√y)dy/dx

=>  d²y/dx²  =  √y /2x√x   -( 1/2√x.√y) (-√y /√x)

=>  d²y/dx²  = √y /2x√x  + 1/2x

R =  (  1  + (dy/dx)²)^(3/2) / (d²y/dx²)

point (1/4,1/4)  =>  x = 1/4  , y = 1/4

√x = 1/2  ,  √y = 1/2  

dy/dx  at   (1/4,1/4)  =  -√y /√x   = -1

d²y/dx² at   (1/4,1/4)  =  2   +  2  = 4

 => R =  ( 1 +(-1)²)^(3/2) /( 4)

=> R = 2^(3/2) /( 4)

=> R = 2√2/4

=> R = 1/√2

radius of curvature of the curve =  1/√2

Learn More:

find the radius of curvature of the curve y2=x3+8at the point (-2,0 ...

https://brainly.in/question/4300854

Find radius of curvature of curve: x² + xy + y² = 4 at point (–2, 0 ...

https://brainly.in/question/18998082

brainly.in/question/19090143

Answer 2

ANSWER ⤵️

√x+√y=1

=> 1/2√x + (1/2√y )(dy/dx)= 0

=> dy/dx   =  -√y /√x

d²y/dx²  = (-√y(-1/2) /x√x   - (1/2√x.√y)dy/dx

=>  d²y/dx²  =  √y /2x√x   -( 1/2√x.√y) (-√y /√x)

=>  d²y/dx²  = √y /2x√x  + 1/2x

R =  (  1  + (dy/dx)²)^(3/2) / (d²y/dx²)

point (1/4,1/4)  =>  x = 1/4  , y = 1/4

√x = 1/2  ,  √y = 1/2  

dy/dx  at   (1/4,1/4)  =  -√y /√x   = -1

d²y/dx² at   (1/4,1/4)  =  2   +  2  = 4

 => R =  ( 1 +(-1)²)^(3/2) /( 4)

=> R = 2^(3/2) /( 4)

=> R = 2√2/4

=> R = 1/√2

$☺️☺️$