I was given the equation
$x=y^4$
and asked to find the points where the curvature was the biggest and the smallest. I know the curvature equation:
$κ(x)= \frac{|y″|}{\left(1+\left(y′\right)^2\right)^{\frac{3}{2}}}$
and I know $f(x)=y=\sqrt[4]{x}$
but I am not sure how to minimize or maximize the curvature. Any help would be great!
On
Hint:
Once you calculate $\kappa(x)$, you'll have a function that tells you the curvature of the graph of $f(x)$ for every value of $x$. You can then take the first derivative of $\kappa(x)$ and set it equal to $0$ like you would for finding the maxima/minima of any ordinary function.
Once you've got $x$-values of potential maxima/minima, you can plug them into $\kappa''(x)$ to check whether they are in fact maxima or minima.
Can you take it from here?
Parametrize $c(t) = (t^4,t)$, and compute $$\kappa(t) =\frac{\det(c'(t),c''(t))}{\|c'(t)\|^3} = \frac{-12t^2}{(16t^6+1)^{3/2}}. $$ Then $$\begin{align} \kappa'(t) &=\frac{-24t(16t^6+1)^{3/2}+12t^2 (3/2)(16t^6+1)^{1/2}(96t^5)}{(16t^6+1)^3}\\ &= \frac{-24t(16t^6+1)+(18\cdot 96)t^7}{(16t^6+1)^{5/2}} \\ &= \frac{1344t^7-24t}{(16t^6+1)^{5/2}} \\ &= \frac{24t}{(16t^6+1)^{5/2}} (56t^6-1)\end{align}$$From this you get the origin and two more points where $\kappa'=0$. Geometrically it is easy to see that the origin is a maximum and the other points are both local minima.