Question

The radius of curvature of the curve y=x^3 at the point (1, 1) is

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Answer 1

Answer:

y=

the curve y=x^3 at the

Answer 2

Answer:

The radius of curvature of curve y = x³ at (1,1) is 4.667 units

Step-by-step explanation:

Given problem  

The radius of the curvature of curve y = x³ at point (1, 1)  

⇒ the formula radius of the curvature is given by

                                     R=  $\frac{[1 + (y')^{2} ]^{\frac{3}{2} } }{ | y"|}$    

here we need to calculate  y' and y"          

⇒ differentiating  y = x³ with respect to x

                     ⇒    y' = 3  (x³⁻²)  = 3x²                  

                     ⇒    y' = 3x²

⇒ differentiating  y' = 3 x² with respect to x

                    ⇒     y" =  3 ( 2 x²⁻¹ ) = 6 x¹

                    ⇒     y" = 6x  

y' = 3x² at point (1, 1) ⇒  y' = 3(1)² = 3

y" = 6x at point (1, 1) ⇒  y" = 6(1) = 6

⇒ radius of the curvature  R = $\frac{[1+ (y')^{2}]^{\frac{3}{2} } }{|y"|}$  

                                               =  $\frac{[1+ (3)^{2}]^{3/2} }{|6|}$  

                                               =   $\frac{[1+3^{3}] }{6}$

                                               =   $\frac{1+27}{6}$

                                               =   $\frac{28}{6}$ = 14/3 = 4.667 units