find the radius of curvature at point a,0 on the curve xy^2=a^2(a-x)
Answers
Answer:
here is your answer
Step-by-step explanation:
Given : curve x² + xy + y² = 4
To find : radius of curvature of curve: at radius of curvature of curve:
Solution:
x² + xy + y² = 4
2x + x(dy/dx) + y + 2y(dy/dx) = 0
=> (dy/dx)(x + 2y) = -(2x + y)
=> dy/dx = -(2x + y)/(x + 2y)
x = - 2 , y = 0
=> dy/dx = -(-4 + 0)/(-2 + 0) = - 2
2x + x(dy/dx) + y + 2y(dy/dx) = 0
=> 2 + x(d²y/dx²) + dy/dx + 2y((d²y/dx²) + 2(dy/dx)(dy/dx) = 0
putting x = - 2 , y = 0 & dy/dx = - 2
=> 2 -2(d²y/dx²) - 2 + 0 + 2(-2)(-2) = 0
=> d²y/dx² = 4
R = ( √ (1 + (dy/dx)²)³ )/ (d²y/dx²)
=> R = √(( 1 + (-2)²)³) / (4)
=> R = √125 / 4
Radius of curvature of curve: = √125 / 4
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Answer:
here is your answer
Step-by-step explanation:
Given : curve x² + xy + y² = 4
To find : radius of curvature of curve: at radius of curvature of curve:
Solution:
x² + xy + y² = 4
2x + x(dy/dx) + y + 2y(dy/dx) = 0
=> (dy/dx)(x + 2y) = -(2x + y)
=> dy/dx = -(2x + y)/(x + 2y)
x = - 2 , y = 0
=> dy/dx = -(-4 + 0)/(-2 + 0) = - 2
2x + x(dy/dx) + y + 2y(dy/dx) = 0
=> 2 + x(d²y/dx²) + dy/dx + 2y((d²y/dx²) + 2(dy/dx)(dy/dx) = 0
putting x = - 2 , y = 0 & dy/dx = - 2
=> 2 -2(d²y/dx²) - 2 + 0 + 2(-2)(-2) = 0
=> d²y/dx² = 4
R = ( √ (1 + (dy/dx)²)³ )/ (d²y/dx²)
=> R = √(( 1 + (-2)²)³) / (4)
=> R = √125 / 4
Radius of curvature of curve: = √125 / 4