Question

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

Answer 1

Answer:

curvature of the curve is 0

Answer 2

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

Now,

We know that,

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

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

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

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

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

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

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

$\tt \:  \longrightarrow \: y_2 = 6x$

☆ Now, Curvature at (1,1) is

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

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

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

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

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

$\tt \:  \longrightarrow \: Curvature \: ( \kappa) \: = \dfrac{6}{ { \bigg(10 \bigg)}^{ \frac{3}{2} } }$

$\tt \:  \longrightarrow \: Curvature \: ( \kappa) \: = \dfrac{6}{10 \sqrt{10} }$

$\tt \:  \longrightarrow \: Curvature \: ( \kappa) \: = \dfrac{3}{5 \sqrt{10} } \times \dfrac{ \sqrt{10} }{ \sqrt{10} }$

$\bf\implies \: \boxed{ \red{\tt \:  \: Curvature \: ( \kappa) \: = \dfrac{3 \sqrt{10} }{50} }}$