Question

Find radius of curvature of the
curve y = x3 at (2,8)​

Answer 1

Answer:

The radius of curvature of the curve at the point (2, 8) is 145.5. Hope this helps.

Answer 2

Answer:

145.503

Step-by-step explanation:

Radius of curvature of any curve $y= f(x)$ is given by

$R = \frac{\bigg[1+\left(\frac{dy}{dx} \right)^2\bigg]^{3/2}}{\bigg|\frac{d^2y}{dx^2} \bigg|}$

For $y=x^3$ :

$1).\,\,\frac{dy}{dx}\bigg|_{x=2} =3x^2\bigg|_{x=2}=12\\2).\,\,\frac{d^2y}{dx^2}\bigg|_{x=2} =6x\bigg|_{x=2}=12$

so we get,

$R= \frac{[1+(12)^2]^{3/2}}{12} \approx 145.503$