How to find the point where the curvature is maximum

Question

How to find the point on the curve $y=e^x$ at which the curvature is maximum.

from the graph of the above curve i think at (0,1) the curvature is maximum.Is it true? please help.

Answer 1

I will assume that you know the formula for curvature at $(x,f(x))$, for a well-behaved curve $y=f(x)$. It is $$\kappa(x)=\frac{|y''|}{(1 +(y')^2)^{3/2}}.$$

In our case, the derivatives are easy to compute, and we arrive at $$\kappa(x)=\frac{e^x}{(1+e^{2x})^{3/2}}.$$ We wish to maximize $\kappa(x)$. One can use the ordinary tools of calculus. It simplifies things a little to write $t$ for $e^x$. I imagine you can now complete the calculation. (The maximum curvature is not at $x=0$.)

Remark: One almost automatic reflex when we do the calculation is to maximize instead the square of the curvature. So we maximize $\frac{e^{2x}}{(1+e^{2x})^3}$, or, more simply, $\frac{t}{(1+t)^3}$. With this version, it is difficult even for me to make a mistake.

Answer 2

We have a formula to find the curvature function $k(x)$ for the graph of a given function $f(x)=e^x$:

$$k(x)=\frac{|f''(x)|}{(1+(f'(x))^2)^{\frac{3}{2}}}=\frac{e^x}{(1+e^{2x})^{\frac{3}{2}}}$$

Note that we have used the fact that $e^x>0$ for all x to remove the absolute value symbol.

Now we want to find out how large this function $k(x)$ can get. We can start our search for maxima of this function by studying the function's first derivative.

$$k'(x)=\frac{e^x(1+e^{2x})^{\frac{3}{2}}-e^x\frac{3}{2}(1+e^{2x})^{\frac{1}{2}}2x}{(1+e^{2x})^3}= \frac{e^x(1+e^{2x})^{\frac{1}{2}}((1+e^{2x})-3e^{2x})}{(1+e^{2x})^3}$$

$$=\frac{e^x(1+e^{2x})^{\frac{1}{2}}(1-2e^{2x})}{(1+e^{2x})^{3}}$$

We note that this is defined for all $x$, so the only critical points will occur where this is zero. If this is zero, then $$1-2e^{2x}=0$$ since $e^x>0$ for all $x$, and $(1+e^{2x})^{\frac{1}{2}}>0$ for all $x$.

Hence, the only critical point is at $$x=\frac{1}{2}\ln\frac{1}{2}$$.

Since $2e^{2x}$ is a strictly increasing function, we can see that $k'(x)$ is going to be negative for $x$ greater than the value we just found, and it will be positive for $x$ less than that value. In other words, we can conclude that this vlue of $x$ gives us the maximum value of $k(x)$.

The point on the curve where curvature is maximum is thus:

$$\left(\frac{1}{2}\ln\frac{1}{2},\frac{1}{\sqrt{2}}\right)$$