Question

find the radius of curvature for this equation x^3+y^3+2x^3-4y+3x=0
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Answer 1

GIVEN :–

• Equation –

$\\ \implies\tt {x}^{3} + {y}^{3} + 2 {x}^{3} - 4y + 3x = 0$

TO FIND :–

• Radius of curvature = ?

SOLUTION :–

$\\ \implies\tt {x}^{3} + {y}^{3} + 2 {x}^{3} - 4y + 3x = 0$

$\\ \implies\tt 3{x}^{3} + {y}^{3} + 3x - 4y= 0$

• We know that –

$\\ \implies{ \boxed{\tt Radius \: \: of \: \: curvature = \dfrac{ \bigg| 1 + \bigg(\dfrac{dy}{dx} \bigg)^{2} \bigg|^{ \frac{3}{2} }}{ \bigg| \dfrac{d^{2} y}{dx^{2}} \bigg| }}}$

• Now given equation –

$\\ \implies\tt 3{x}^{3} + {y}^{3} + 3x - 4y= 0$

$\\ \implies\tt {y}^{3} - 4y= - 3{x}^{3} - 3x$

• Differentiate with respect to 'x' –

$\\ \implies\tt 3{y}^{2} \dfrac{dy}{dx}- 4\dfrac{dy}{dx}= - 3(3 {x}^{2}) - 3$

$\\ \implies\tt (3{y}^{2} - 4)\dfrac{dy}{dx}= - 9 {x}^{2}- 3$

$\\ \implies\tt \dfrac{dy}{dx}= \dfrac{- 9 {x}^{2}- 3}{3{y}^{2} - 4}$

$\\ \large \implies{ \boxed{\tt \dfrac{dy}{dx}= \dfrac{9 {x}^{2} + 3}{4 - 3{y}^{2}}}}$

• Again Differentiate with respect to 'x' –

$\\\implies\tt \dfrac{d^{2} y}{dx^{2} }= \dfrac{(4 - 3{y}^{2})(18x) -(9 {x}^{2} + 3) \bigg( - 6y \dfrac{dy}{dx} \bigg)}{(4 - 3{y}^{2})^{2} }$

$\\\implies\tt \dfrac{d^{2} y}{dx^{2} }= \dfrac{(4 - 3{y}^{2})(18x) -(9 {x}^{2} + 3) ( - 6y) \bigg(\dfrac{9 {x}^{2} + 3}{4 - 3{y}^{2}} \bigg)}{(4 - 3{y}^{2})^{2} }$

$\\\implies\tt \dfrac{d^{2} y}{dx^{2} }= \dfrac{(4 - 3{y}^{2})^{2} (18x) + (9 {x}^{2} + 3)^{2} (6y)}{(4 - 3{y}^{2})^{3} }$

• Now put the values –

$\\ \implies\tt Radius \: \: of \: \: curvature = \dfrac{ \bigg| 1 + \bigg(\dfrac{9 {x}^{2} + 3}{4 - 3{y}^{2}}\bigg)^{2} \bigg|^{ \frac{3}{2} }}{ \bigg|\dfrac{(4 - 3{y}^{2})^{2} (18x) + (9 {x}^{2} + 3)^{2} (6y)}{(4 - 3{y}^{2})^{3} }\bigg|}$

$\\ \implies\tt Radius \: \: of \: \: curvature = \dfrac{ \bigg| 1 + \dfrac{(9 {x}^{2} + 3) ^{2} }{(4 - 3{y}^{2}) ^{2} }\bigg|^{ \frac{3}{2} }}{ \bigg|\dfrac{(4 - 3{y}^{2})^{2} (18x) + (9 {x}^{2} + 3)^{2} (6y)}{(4 - 3{y}^{2})^{3} }\bigg|}$

$\\ \implies\tt Radius \: \: of \: \: curvature = \dfrac{ \bigg| \dfrac{(4 - 3{y}^{2}) ^{2} + (9 {x}^{2} + 3) ^{2} }{(4 - 3{y}^{2}) ^{2} }\bigg|^{ \frac{3}{2} }}{ \bigg|\dfrac{(4 - 3{y}^{2})^{2} (18x) + (9 {x}^{2} + 3)^{2} (6y)}{(4 - 3{y}^{2})^{3} }\bigg|}$

$\\ \implies\tt Radius \: \: of \: \: curvature = \dfrac{ \bigg| \dfrac{[(4 - 3{y}^{2}) ^{2} + (9 {x}^{2} + 3) ^{2} ]^{ \frac{3}{2} } }{(4 - 3{y}^{2})^{3} }\bigg|}{ \bigg|\dfrac{(4 - 3{y}^{2})^{2} (18x) + (9 {x}^{2} + 3)^{2} (6y)}{(4 - 3{y}^{2})^{3} }\bigg|}$

$\\ \large \implies{ \boxed{ \tt Radius \: \: of \: \: curvature = \bigg|\dfrac{\{(4 - 3{y}^{2}) ^{2} + (9 {x}^{2} + 3) ^{2} \}^{ \frac{3}{2} }}{(4 - 3{y}^{2})^{2} (18x) + (9 {x}^{2} + 3)^{2} (6y)} \bigg|}}$