Radius of curvature of curve y = x³ at (1,1) is
Answers
Step-by-step explanation:
Given:
\textsf{Curve is}\;\mathsf{y=x^3}Curve isy=x3
\textbf{To find:}To find:
\textsf{Radius of curvature of the given curve}Radius of curvature of the given curve
\textbf{Solution:}Solution:
\textsf{Consider,}Consider,
\mathsf{y=x^3}y=x3
\mathsf{\dfrac{dy}{dx}=3x^2}dxdy=3x2
\mathsf{\dfrac{d^2y}{dx^2}=6x}dx2d2y=6x
\mathsf{At\;(1,1),}At(1,1),
\mathsf{\dfrac{dy}{dx}=3(1)^2=3}dxdy=3(1)2=3
\mathsf{\dfrac{d^2y}{dx^2}=6(1)=6}dx2d2y=6(1)=6
\underline{\textsf{Radius of curvature}}Radius of curvature
\boxed{\mathsf{R=\dfrac{\left(1+\left(\dfrac{dy}{dx}\right)^2\right)^\frac{3}{2}}{\left|\dfrac{d^2y}{dx^2}\right|}}}R=∣∣∣∣∣dx2d2y∣∣∣∣∣(1+(dxdy)2)23
\implies\mathsf{R=\dfrac{(1+3^2)^\frac{3}{2}}{|6|}}⟹R=∣6∣(1+32)23
\implies\mathsf{R=\dfrac{(1+9)^\frac{3}{2}}{6}}⟹R=6(1+9)23
\implies\mathsf{R=\dfrac{10^\frac{3}{2}}{6}}⟹R=61023
\implies\mathsf{R=\dfrac{10\sqrt{10}}{6}}⟹R=61010
\implies\boxed{\mathsf{R=\dfrac{5\sqrt{10}}{3}}}⟹R=3510
Answer:
Radius of curvature of curve y = x³ at (1,1) is