find the radius of curvature of a rectangular hyperbola xy= c²
Answers
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Answer:
Radius of curvature = 125/24
Step-by-step explanation:
The given curve is
xy = 12 ..... (1)
Differentiating both sides with respect to x, we get
x dy/dx + y = 0 ..... (2)
Again differentiating both sides with respect to x, we get
x d²y/dx² + dy/dx + dy/dx = 0
or, x d²y/dx² + 2 dy/dx = 0
or, d²y/dx² = - 2/x * dy/dx ..... (3)
From (2) and (3), using (1), we get
dy/dx = - y/x
d²y/dx² = 2y/x²
Therefore radius of curvature is
= {1 + (dy/dx)²}^(3/2) / (d²y/dx²)
= (1 + y²/x²)^(3/2) / (2y/x²)
= {(x² + y²)^(3/2) / x³} * (x² / 2y)
= (x² + y²)^(3/2) / (2xy)
= (3² + 4²)^(3/2) / (2 * 3 * 4) at (3, 4)
= 125 / 24 at (3, 4)
PLEASE MARK IT AS A BRAINLIEST AND FOLLOW ME
Answer:
Radius of curvature = 125/24
Step-by-step explanation:
The given curve is
xy = 12 ..... (1)
Differentiating both sides with respect to x, we get
x dy/dx + y = 0 ..... (2)
Again differentiating both sides with respect to x, we get
x d²y/dx² + dy/dx + dy/dx = 0
or, x d²y/dx² + 2 dy/dx = 0
or, d²y/dx² = - 2/x * dy/dx ..... (3)
From (2) and (3), using (1), we get
dy/dx = - y/x
d²y/dx² = 2y/x²
Therefore radius of curvature is
= {1 + (dy/dx)²}^(3/2) / (d²y/dx²)
= (1 + y²/x²)^(3/2) / (2y/x²)
= {(x² + y²)^(3/2) / x³} * (x² / 2y)
= (x² + y²)^(3/2) / (2xy)
= (3² + 4²)^(3/2) / (2 * 3 * 4) at (3, 4)
= 125 / 24 at (3, 4)