Question

Calculate the centre of curvature of the curve y=x^2 at (1,0)​

Answer 1

Answer:

the answer is

Step-by-step explanation:

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Answer 2

Answer: The answer is 2.6

The radius of curvature of curve y = x^2 at (1,0) is 2.6.

Radius of curvature: In differential geometry, the reciprocal of curvature is denoted by the letter R, which stands for the radius of curvature. An approximation of a curve at a particular location is equal to the radius of the circular arc that travels through that point on the curve. The lowest diameter of a circle that may satisfactorily fit a surface's normal section or combinations of normal sections is referred to as the surface's radius of curvature.

Step-by-step explanation:

Step 1: The given data:

Equation of curvature, $y = x^2$

Points x = 1 and y = 0.

Step 2: Find the curvature

We know curvature, k(x) is given by,

$k(x) = \frac{y''(x)}{(1-y')^{\frac{3}{2}}}\\y'(x) = 2x\\y''(x) = 2\\k(x) = \frac{2}{(1+2x)^{\frac{3}{2}}}$

At x = 1

$k = \frac{2}{(1+2)^{\frac{3}{2}}}\\k = \frac{2}{(3)^{\frac{3}{2}}}\\k = 0.384$

Step 3: Find the Radius of curvature

Radius of curvature R is given by,

$R = \frac{1}{k}\\R = \frac{1}{0.384}\\R = 2.6$

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