Question

Find the curvature of the curve\(y = {x^4}\)at the point\(\;\left( {{\bf{1}},{\bf{1}}} \right)\).

Short answer

The curvature at the point\((1,1)\)is\(\frac{{12}}{{{{17}^{3/2}}}}\) .

Step-by-step solution

Definition of the curve

A smooth drawn line or figure on a plane with turns or a bend is called a curve.

Evaluating the derivative.

For the curve which is given by the equation \(y = {x^4}\)we can write it in the vector form by defining the parameter \(x = t\):

\(r(t) = \left( {t,{t^4},0} \right)\)

We have \(\begin{array}{l}{r^\prime }(t) \times {r^{\prime \prime }}(t) = \left( {1,4{t^3},0} \right) \times \left( {0,12{t^2},0} \right)\\ = \left( {0,0,12{t^2}} \right)\end{array}\)

Thus the curvature is

\(\begin{array}{l}\kappa = \frac{{\left| {{r^\prime }(t) \times {r^{\prime \prime }}(t)} \right|}}{{{{\left| {{r^\prime }(t)} \right|}^3}}}\\ = \frac{{\sqrt {0 + 0 + 144{t^4}} }}{{{{\left( {\sqrt {1 + 16{t^6}} } \right)}^3}}}\\ = \frac{{12{t^2}}}{{{{\left( {1 + 16{t^6}} \right)}^{3/2}}}}\end{array}\)

The point \((x,y) = (1,1)\)is corresponding to\(t = 1\), thus the curvature at\((1,1)\)is

\(\frac{{12 \cdot {1^2}}}{{{{\left( {1 + 16 \cdot {1^6}} \right)}^{3/2}}}} = \frac{{12}}{{{{17}^{3/2}}}}\) .