Curvature as a rate of change in slope While working with some  formulas on cantilever in mechanical engineering , in some places it assumes that
$\frac{1}{R} =\kappa = \frac{d^{2} y}{d x^{2}}$
But I know That
$k=\frac{y^{\prime \prime}}{(1+y^{\prime 2})^{\frac{3}{2}}}$
So It Is Very Counterintuitive to me , So Someone please help me
 A: Quoting from wiki:

When the slope of the graph... is small, the signed curvature is well approximated by the second derivative. More precisely, using big O notation, one has
$$k(x)=y''(1+O(y'^2)).$$
It is common in physics and engineering to approximate the curvature with the second derivative, for example, in beam theory or for deriving wave equation of a tense string, and other applications where small slopes are involved. This often allows systems that are otherwise nonlinear to be considered as linear.

A: So, In cantilever there is a catch: In ideal condition you can consider $\frac {dy}{dx} \neq 0$. However, in reality, the bend (the slope of tangent) to a cantilever beam is very very small and the square of that is almost zero so this works
You can neglect $\left(\frac {dy}{dx}\right)^2$
But you should NOT neglect $\color{red}{\frac {d^2y}{dx^2}}$ Why?
Because, $$\frac {d^2y}{dx^2} = \frac {\text{difference of slopes of tangents}}{\text{small value of x}}= \frac {\frac {dy}{dx}_{2} - \frac {dy}{dx}_1}{\delta x} \implies \frac {\text{small}_2 - {small}_1}{\text{small}} \ne \text{ small } or \text { may have significantly big value}$$
If your question is why the radius curvature is as the formula says then this might help you understand how we can use mathematics in real-life applications.

It was a time when I didn't know about the radius of curvature so I can say I coined this term as Instantaneous Radius of Curve:
I wanted to find the instantaneous radius of any curve so I defined it as

*

*Theory:
At the first, We find two normals $(N_1$ & $  N_2)$ to a curve and then the intersection of these normals;
If these straight lines(normals) tend to be the same then their point of intersection will be the point of instantaneous radius and the distance as the Instantaneous radius

$$\begin{align}
m_1 = -{\left(\frac {dy}{dx}\right)}^{-1}_{p1} \text{ and } m_2 = -{\left(\frac {dy}{dx}\right)}^{-1}_{p2}\\
\end{align}$$
$$a = \frac {dy}{dx} \text{ and } b = \frac {d^2y}{dx^2}$$
Now, $$m_1\times t_1 = -1 \text{ and } m_2\times t_2 = -1$$
$$y-m_1 \times x = y_1 - m_1 x_1 \text{ ..................equation of normal 1}$$
$$y-m_2 \times x = y_2 - m_2 x_2 \text{ ..................equation of normal 2}$$
$$\begin{align}
y =
& \frac {(m_2y_1 - m_1y_2)+(x_2-x1)\times m_1 \times m_2}{(m_2-m_1)}\\
& = \frac {\left(y_2t_2 - y_1t_1 + \Delta x\right)/ \Delta x}{\Delta t/\Delta x}\\
& =  \left[\frac {1+ G(x)}{f''(x)}\right] where, \int G(x)dx = y.f'(x)
\end{align}$$
Similarly, By changing the frame of axes, we find an equation in terms of $\frac {d^2x}{dy^2}$
$$\begin{align}
\text { and thus, get } \frac {x_1b- a - a^3}{b}  = x
\end{align}$$
and also using the below formula(used to find the 2nd derivative of x in terms of the derivates of y
)
$${\frac {d^2x}{dy^2} = -\left(\frac{dy}{dx}\right)^{-3}\times {\frac {d^2y}{dx^2}}}$$
$$\begin{align}
\left(l(R)\right)^2
& = (x-x_1)^2 + (y - y_1)^2\\
& = \frac {(a+a^3)^2 + (1+a^2)^2}{b^2}\\
& = \frac {(a^2 + 1)^3}{b^2}\\ 
\implies R = \frac {\left(1+\left(\frac {dy}{dx}\right)^2\right)^{\frac 32}}{\left(\frac {d^2y}{dx^2}\right)}
\end{align}$$

@GaneshKhadanga in reply to your comment:

*

*Why I'm solving for the value of $y$?:
I'm looking for the $y$-coordinate of the point of intersection of the normals.

$$\begin{align}
y =
& \frac {(m_2y_1 - m_1y_2)+(x_2-x1)\times m_1 \times m_2}{(m_2-m_1)}\\
& = \frac {\left(\frac{y_1}{m_1} - \frac {y_2}{m_2}\right) + (x_2 - x_1)}{\frac 1{m_1} - \frac 1{m_2}}\\
&\text { As, } m_1\times t_1 = -1 \text{ and } m_2\times t_2 = -1 \\
&\text {The product of slope of tangent and normal is negative one}\\
&\text{Here, by } \Delta t \text { I mean the difference of slopes of tangents }\\
& = \frac {\left(y_2t_2 - y_1t_1 + \Delta x\right)/ \Delta x}{\Delta t/\Delta x}\\
& =  \left[\frac {1+ G(x)}{f''(x)}\right] where, \int G(x)dx = y.f'(x)
\end{align}$$
Also, Why $\frac {\Delta t}{\Delta x} = f''(x)$
Here, I consider as if $\Delta x \to 0$ and it means what is the rate of change of the slopes of tangents e.i the rate of change of $\frac {dy}{dx}$
$\frac d{dx}\frac {dy}{dx} = \frac {d^2y}{dx^2}$
