Question

Find the curvature of the curve y=x^3 at the point (1,1)

Answer 1

$\large\underline\purple{\bold{Solution :- }}$

Now,

We know that,

$\tt \:  \longrightarrow \: \boxed{ \bf \red{Curvature \: ( \kappa) \: = \dfrac{y_2}{ { \bigg(1 + {(y_1)}^{2} \bigg)}^{ \frac{3}{2} } } }}$

$\tt \:  \longrightarrow \: Now, \: y \: = \: {x}^{3}$

☆ Differentiate both sides w. r. t. x, we get

$\tt \:  \longrightarrow \: \dfrac{d}{dx}y = \dfrac{d}{dx} {x}^{3}$

$\tt \:  \longrightarrow \: \dfrac{dy}{dx} =y_1 = 3 {x}^{2}$

☆ On differentiate both sides w. r. t. x, we get

$\tt \:  \longrightarrow \: \dfrac{d}{dx} \dfrac{dy}{dx} = \dfrac{d}{dx} {3x}^{2}$

$\tt \:  \longrightarrow \: \dfrac{ {d}^{2}y }{ {dx}^{2} } = y_2 = 6x$

$\tt \:  \longrightarrow \: Curvature \: ( \kappa) \: = \dfrac{y_2}{ { \bigg(1 + {(y_1)}^{2} \bigg)}^{ \frac{3}{2} } }$

$\tt \:  \longrightarrow \: Curvature \: ( \kappa) \: = \dfrac{6x}{ { \bigg(1 + {(3 {x}^{2} )}^{2} \bigg)}^{ \frac{3}{2} } }$

$\tt \:  \longrightarrow \: Curvature \: ( \kappa) \: = \dfrac{6x}{ { \bigg(1 + {9x}^{4} \bigg)}^{ \frac{3}{2} } }$

☆ Now curvature at (1, 1) is

$\tt \:  \longrightarrow \: Curvature \: ( \kappa) \: = \dfrac{6 \times 1}{ { \bigg(1 + {9 \times (1)}^{4} \bigg)}^{ \frac{3}{2} } }$

$\tt \:  \longrightarrow \: Curvature \: ( \kappa) \: = \dfrac{6}{ { \bigg(1 + 9 \bigg)}^{ \frac{3}{2} } }$

$\tt \:  \longrightarrow \: \boxed{ \purple{ \bf \: Curvature \: ( \kappa) \: = \dfrac{6}{ { \bigg(1 0\bigg)}^{ \frac{3}{2} } } }}$